Sciencemadness Discussion Board - Quantum Mechanics/Wave Mechanics in Chemistry Beginners Thread

So, in this one dimentional box model (got to remember that) :

In the Ground state, the total possible energy in the system is defined by the square of Planck's constant divided by ( 8 time the mass of the particle times the square of the length of the box )

Can i go into an exicited state now ?

Excite away!

Q: If the ground state energy is 15.4 eV, what is E for n = 6?

Pocket calculators allowed. No mobile phones, please.

[Edited on 23-7-2015 by blogfast25]

aga - 23-7-2015 at 12:38

Confooseled for as second there and was going to bleat on about not knowing m or L values, and being evil and everything.

Then it became clear that 15.4 = [h2/(8mL2)] n2 where n=1

i.e. the whole h2/(8mL2) schmozzle = 15.4

so the answer should be 15.4 n2 with n = 6

= 15.4 * 62
= 15.4 * 36
= 554.4 eV

Edit :

added the interim step so i can refer back to it when sober

[Edited on 23-7-2015 by aga]

blogfast25 - 23-7-2015 at 12:46

'Evil' I am (Mwhahaha!)

But 554.4 electron volts (eV) it is indeed!

The quantization means that a 'quantum car' would have a bumpy ride. A Classic car can take on any value of kinetic energy (a so-called 'energy continuum'). Your Fiat Quantum would move at 1 mph or 5 or 10 or 15 mph but not at 1.2 or 11.5 or 14.9 mph. Only the SE quantized levels would be allowed.

aga - 23-7-2015 at 12:56

Woohoo !

Thanks very much for posing a maths question i could actually answer !

This explains everything.

I do indeed own a Fiat Quantum, and n=3 is actually it's top speed.

blogfast25 - 23-7-2015 at 13:23

I do indeed own a Fiat Quantum, and n=3 is actually it's top speed.

You need a Time Dependent SE to get it into higher n! I can do you mate's rates, if you want?

[Edited on 23-7-2015 by blogfast25]

aga - 23-7-2015 at 13:50

A very generous offer.

With my new found maths powers i just removed a lot of 'm' (who really needs back seats ?) and replaced the factory-fitted Planck with a much bigger one and now it's a lot faster.

The wierd thing is that it is harder to find in the car park, as i am Uncertain of where it is.

blogfast25 - 23-7-2015 at 14:01

A very generous offer.

With my new found maths powers i just removed a lot of 'm' (who really needs back seats ?) and replaced the factory-fitted Planck with a much bigger one and now it's a lot faster.

The wierd thing is that it is harder to find in the car park, as i am Uncertain of where it is.

Nice.

Paradoxically perhaps, removing lots of m makes it go faster, yet the Fiat Quantum drives more bumpily. Any ideas why? For 5 points...

aga - 23-7-2015 at 14:28

First instinct was to shout 'Wavenumber' and hope to get full points and a lollipop.

Had to do some calculationings to actually see.

As mass Increases, the change in E_n to E_n+1 Decreases.

Lower mass means a Higher jump in energy levels from one state to the other.

Time to go out on a limb ( 30 year + dormant brain cells activated - Mr Capps would be proud if still living, and if it's right):

The first integral of E_n is inversely proportional to m.

blogfast25 - 23-7-2015 at 15:33

Lower mass means a Higher jump in energy levels from one state to the other.

Time to go out on a limb ( 30 year + dormant brain cells activated - Mr Capps would be proud if still living, and if it's right):

Correct. As m increases the E levels become closer and closer together, eventually resulting in a continuum. A very slow but smooth ride!

Cinqo puntos, Senor. But you still have to buy your own beer... :cool:

aga - 23-7-2015 at 23:21

Amazing.

It's possible i'll remember some trigonometry as well.

blogfast25 - 24-7-2015 at 05:59

Here’s another problem.

SCRAPPED as per AAHD pointing out an error. See new version below.

[Edited on 24-7-2015 by blogfast25]

annaandherdad - 24-7-2015 at 07:44

Here’s another problem.

A scientist studies a QS and from theory finds its energy quantization rule to be:

En = E1 / ( n + 1 ), with n the quantum number 1, 2, 3, 4, ... and E1 the ground state energy.

Do you really mean this? If I substitute n=1, I get E1 = E1/2, hence E1=0, hence all En=0.

blogfast25 - 24-7-2015 at 07:52

Quote: Originally posted by annaandherdad

Do you really mean this? If I substitute n=1, I get E1 = E1/2, hence E1=0, hence all En=0.

Oooopsie. You are indeed correct. My bad.

[Edited on 24-7-2015 by blogfast25]

blogfast25 - 24-7-2015 at 08:06

Ok, problem revised:

A scientist studies a QS and from theory finds its energy quantization rule to be:

En = E1 / n2, with n the quantum number 1, 2, 3, 4, ... ("E1 divided by the square of n")

Spectroscopically he determines that ΔE = E3 - E2 = - 20 eV ("minus 20 eV").

What is the ground state E1 of the QS?

[Edited on 24-7-2015 by blogfast25]

aga - 24-7-2015 at 12:18

En = E1 / n2

ΔE = E3 - E2 = -20 eV

OMG. It's like a muscle that has not been flexed for so long, it's gone all crystallised.

It's a couple of simultaneous equations !

E3 = E1 / 9   so (*) 9 E3 = E1

E2 = E1 / 4   so 4 E2 = E1

E3 - E2 = -20

E3 = E2 - 20

substituting E3 for E2 equivalence in (*)

9 ( E2 - 20 ) = 4 E2

9 E2 - 180 = 4 E2

9 E2 - 4 E2 = 180

5 E2 = 180

E2 = 36 eV

ΔE = E3 - E2 = -20 eV

E1 = 36 - 20 = 16 eV WRONG cos the delta is MINUS

E1 Ground State = 56 eV

Edit:

Did the crystals fall off properly ?

[Edited on 24-7-2015 by aga]

blogfast25 - 24-7-2015 at 12:31

Good effort, aga, but you’re ‘overthinking it’:

ΔE = E3 - E2 = E1/9 - E1/4 = (4E1 – 9E1

/ 36 = - 5E1/36 = - 20 eV

E1 = 144 eV

E2 = 144/4 = 36 eV

E3 = 144/9 = 16 eV

ΔE = - 20 eV

aga - 24-7-2015 at 12:42

E3 = E1 / 9   so (*) 9 E3 = E1

...

substituting E3 for E2 equivalence in (*)

9 ( E2 - 20 ) = 4 E2

It would help a lot if i read what i typed.

DOH !

Edit:

Double DOH. My initial assumption is obviously wrong as well.

The delta E is not linear between n values AND I have already seen and understood that just yesterday.

Feck.

[Edited on 24-7-2015 by aga]

[Edited on 24-7-2015 by aga]

blogfast25 - 24-7-2015 at 12:52

Only those who make nothing make no mistakes.

Before I move on to the next instalment, just for our ‘math delights’, I give you:

Mini-interlude: an alternative way of determining the E levels for a particle in a 1D box:

For such a particle, U(x) = 0 and the SE is:

- (ћ2/2m) δ2ψ(x)/δx2 = Eψ(x)

Which I’ll re-write slightly as:

- aψ” = Eψ (with a = ћ2/2m)

From higher up we know that:

Ψn(x) = √(2/L) sin(nπx/L)

It’s easy to show that Ψ’n(x) = √(2/L) (nπ/L) cos(nπx/L) [first derivative]

And thus Ψ”n(x) = - √(2/L) (nπ/L)2 sin(nπx/L) = - (nπ/L)2 Ψn(x) [second derivative]

Insert into the SE and Ψn(x) drops out:

a (nπ/L)2 = En and with a = ћ2/2m:

En = n2h2/(8mL2

The part - aψ” is often called a 'Quantum Operator'.

[Edited on 24-7-2015 by blogfast25]

aga - 24-7-2015 at 13:03

Incredible as it may sound, i follow all of that apart from the 'derivative' bit.

What exactly is a First and Second (presumably Third and so on) derivative ?

[Edited on 24-7-2015 by aga]

blogfast25 - 24-7-2015 at 13:19

Incredible as it may sound, i follow all of that apart from the 'derivative' bit.

What exactly is a First and Second (presumably Third and so on) derivative ?

[Edited on 24-7-2015 by aga]

Have a look at the earlier part of the course:

http://www.sciencemadness.org/talk/viewthread.php?tid=62973&...

Does this answer your question, at least partly?

blogfast25 - 24-7-2015 at 13:26

Say y = α sin(βx)

Then y’ = α β cos(βx)

And y” = - α β2 sin(βx) = - β2 y

aga - 24-7-2015 at 13:38

Thanks for the link (blush)

To be honest, my current environment is less than optimally conducive towards sustained concentration.

Would it be fair to compare Derivatives to what i imagine are Integrals ?

E.g. speed, accelleration, rate of change of accelleration and so on ?

blogfast25 - 24-7-2015 at 14:13

Derivatives ate the opposite of integrals, thus VERY closely related.

y === derivation ===> y'

y' === integration ===> y

==========================

Example: (derivatives)

A car travels on the x-axis. x represents position. Speed is v:

v =dx/dt (first derivative of x in time t)

Acceleration is a:

a= dv/dt = d2x/dt2

Derivatives show up when there is change. E.g. in QP because ψ(x) changes with x.

Wife having another little aga?

aga - 24-7-2015 at 14:18

Hardly likely that there will be another agajunior.

Age, surgery and good sense ensure that.

Aga's Owner, Agaitas and Hounds/beasts in general plus Beer are the main distractants.

Edit:

Apologies for asking a question to which you already explaned the answer to earlier in the thread.

[Edited on 24-7-2015 by aga]

[Edited on 24-7-2015 by aga]

blogfast25 - 24-7-2015 at 14:43

Apologies for asking a question to which you already explaned the answer to earlier in the thread.

Try keeping that dementia in check!

Wot was we talking about again? Totally gone...

aga - 24-7-2015 at 14:54

I think we were looking for a Quantum of Solace, which i found under the sofa along with a fossilised bit of cake.

Cake. Mmm Cake. Cake is nice.

I'll put the kettle on.

blogfast25 - 24-7-2015 at 15:20

You boil your beer? You only need to do that before fermentation, you know?

Next instalment on the morrow. The cliffhanger continues!

blogfast25 - 24-7-2015 at 17:57

QMSMers who have ever dropped a few grains of kitchen salt into a hot flame (or watched their potatoes boil over!) will have seen how that colours the flame strongly yellow. This is due to emission of photons from excited electrons in the sodium atoms of the NaCl, returning to the ground state. Other elements emit other colours, see ‘flame tests’. I’ll briefly illustrate how emission spectra work using our particle in a 1D box.

Emission Spectrum of a particle in a one dimensional box (Pi1DB)

When a particle in a one dimensional box is in an excited state (n > 1) it can ‘fall’ back to a lower state, for example from n2 to n1 (with n2 > n1

and energy will be emitted: ΔE = En1 - En2 < 0.

Quick side note: the mechanics of such transitions (from n2 to n1

and their probabilities are described by the Time Dependent Schrodinger Equation and will not be treated here.

Assuming the particle is an actual electron, the energy emission is in the form of a photon (electromagnetic radiation like light, X-rays, IR or UV) with energy E = hf.

Thus hf = ΔE

This schematic should be useful (emission AND absorption):

http://hyperphysics.phy-astr.gsu.edu/hbase/mod5.html#c2

Spectroscopically, the transitions (and corresponding photon emissions), are often organised in ‘series’. Here, somewhat arguably, we can discern Series 1, 2, 3 etc by setting n1 to 1, 2, 3 etc.

The frequencies f for a photons emitted in Series 1 then become:

f1 = h/(8mL2

- h/(8mL2

n22
= h/(8mL2[1 - n22]

For Series 2:

f2 = h/(8mL2[4 - n22]

For Series n:

f2 = h/(8mL2[n2 - n22]

Of course h/(8mL2 = E1.

Note also that for emitted photons c = λf (c = speed of light, λ = wavelength of photon, f = frequency of photon), so that f can be converted to λ easily.

A calculated emission spectrum of a Pi1DB, for a ground state of E1 = 6.8 eV is shown below.

[Edited on 25-7-2015 by blogfast25]

aga - 25-7-2015 at 09:08

So, let me get that straight ... if you can accurately measure the wavelength of the radiation from a system collapsing from one energy state to another, you can calculate the E values of all levels ?

How can you determine what the state change is in the first place ?

i.e is it going n₅ to n₄ or n₅ to n₁ ?

blogfast25 - 25-7-2015 at 10:01

You would always need to know (or guess) the form of the energy function, say En = F(E1,n) ("En a function of E1 and n"), as well as one of the n numbers in:

ΔE = En1 - En2 (= hf)

The ground state of the hydrogen atom had been determined by spectroscopy, some years before Schrodinger solved the SE for the H atom.

In the case of a Pi1DB, the difference between two adjacent energy levels n and n+1 is always:

ΔE = E1(n+1)2 –E1n2

= E1(n2 + 2n + 1 – n2

= E1(2n+1)

This kind of expression can be useful for a spectrocopist to determine E1 with.

[Edited on 26-7-2015 by blogfast25]

aga - 25-7-2015 at 10:28

Deduce the working of the universe and the Energy of a particle by maths and the colour ?

That is truly remarkable.

blogfast25 - 25-7-2015 at 11:23

The 'magik' of science!

blogfast25 - 25-7-2015 at 18:27

Next instalment:

Normalisation of Wave Functions (Pi1DB)

Note that the wave functions discussed above have already been normalised but I want to briefly expand on the so-called Normalisation Requirement separately.

As we know, ψ(x)2 represents the probability P(x) of finding the particle in the location x.

We also know that the particle in a one dimensional box is always found inside the box (as it cannot escape from it due to the infinitely high potential energy) and the boundaries x = 0 and x = L.

In mathematical terms that means that to sum total of all values of P(x) over the interval [x = 0, x = L] must be 1, as this represents the probability of finding the particle inside the box. In mathematical terms this means:

Σ P(x) = Σ ψ(x)2 = 1, summation over all values of x.

Because ψ(x), and thus by extension also ψ(x)2, is a smooth and continuous function, the summation Σ is replaced by integration:

∫ψ(x)2dx = 1, integrated between x = 0 and x = L.

In the following example:

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/pbox.html...

… the normalisation requirement is used to determine the factor A (for ‘amplitude’ of the wave function) to be √(2/L). This guarantees that the resulting wave functions:

Ψn(x) = √(2/L) sin(nπx/L)

… and the resulting probability functions return the correct values.

Note that in the given example, A2 = 2/L, has two roots:

A+ = + √(2/L) and A- = - √(2/L)

And indeed both +ψ and –ψ both satisfy the SE. Textbooks sometimes note the wave functions as +/- ψ. The probability functions ψ2 are not affected by the sign either, of course.

The normalisation ‘procedure’ needs to be applied to all wave functions of bound particles, in all quantum systems. The mathematics of the procedure depends a lot on the QS in question.

[Edited on 26-7-2015 by blogfast25]

annaandherdad - 25-7-2015 at 21:37

So, let me get that straight ... if you can accurately measure the wavelength of the radiation from a system collapsing from one energy state to another, you can calculate the E values of all levels ?

How can you determine what the state change is in the first place ?

i.e is it going n₅ to n₄ or n₅ to n₁ ?

The wavelength of the spectral line gives the difference in energies between the two states. Different spectral lines correspond to different pairs of states. For example, there is one spectral line on going from E2 to E1, another from E3 to E1, another from E3 to E2, etc. It's possible to disentangle this information and to get the energy levels themselves, up to an arbitrary choice for one specific level. For example, if you just choose E1=0, then you can use the spectral information to find all the other levels. You don't need a formula for the energy levels to do this.

Actually the process can be more complicated, because some transitions between some pairs of levels are very weak, or almost not present. That is, one must work with partial information.

Before the advent of quantum mechanics spectroscopists had figured out that the frequencies of spectral lines could be expressed as the differences of a smaller set of frequencies, what we now know are the energy levels (converting energy to frequency by E=hf). The energy levels themselves didn't seem to follow any simple mathematical pattern, with the exception of hydrogen, where Balmer found the formula f = k(1/n12 - 1/n22

for the frequencies of the spectral lines, where k is a constant. We now understand this by saying that En = -k/n2 are the energy levels of hydrogen.

aga - 26-7-2015 at 00:46

Incredible !

So if the spectrum emitted by a system (determined experimentally) coincides with the calculated E_n values with no ouliers), that's kind of a proof that the assumptions of QM are correct, and that matter does in fact obey those laws ?

annaandherdad - 26-7-2015 at 06:58

Yes, getting the spectrum right was one of the main objects in the early history of quantum mechanics. The first major triumph in this regard was Bohr's theory of 1911 (thereabouts) which derived the Balmer formula for hydrogen from some assumptions about quantized classical orbits. Bohr's method however did not work for helium or any other atom, that required the full quantum mechanics of Schro"dinger and Heisenberg and didn't come until 1927 (thereabouts). By examining spectroscopic data, theorists discovered electron spin and the exclusion principle, they explained the fine structure of atomic spectra as relativistic corrections, etc etc.

Magpie - 26-7-2015 at 08:59

Can electron spin be described in classical terms? If not, what is it other than -1/2 or +1/2?

annaandherdad - 26-7-2015 at 09:11

Quote: Originally posted by Magpie

Can electron spin be described in classical terms? If not, what is it other than -1/2 or +1/2?

The existence of spin was never suspected before quantum mechanics came along, and it's usually regarded as a fundamentally quantum phenomenon, with no classical analog. Sometimes it is said that classical mechanics becomes valid when quantum numbers are large, and spin (electron spin, anyway) has a quantum number that never goes above 1/2. So in that respect you'd have to say that it is purely quantum mechanical.

Nevertheless there are classical models that capture some of the features of quantum spin. A very crude one is to imagine that the electron is a charged sphere, spinning on an axis. Because the rotation carries the charge around in circles, it creates a magnetic field. It is the electron's magnetic field that allows spin to be detected and measured. Whatever the model, it is a fact that an electron has both an electric field and a magnetic field.

aga - 26-7-2015 at 12:40

So if in QM, a particle cannot have an energy state other then 1,2,3 etc, how come it needs to be 'bent' to make a 0.5 state ?

annaandherdad - 26-7-2015 at 14:51

I don't know what you mean by "bent", or what a 0.5 state is. Please explain.

blogfast25 - 26-7-2015 at 15:28

aga, re, the 'spin' of an electron, it's something that you just kind of have to accept.

Quote:

In quantum mechanics, the spin–statistics theorem relates the spin of a particle to the particle statistics it obeys. The spin of a particle is its intrinsic angular momentum (that is, the contribution to the total angular momentum that is not due to the orbital motion of the particle). All particles[citation needed] have either integer spin or half-integer spin (in units of the reduced Planck constant ħ).

https://en.wikipedia.org/wiki/Spin%E2%80%93statistics_theore...

Electrons have an intrinsic (regardless of physical system they find themselves contained in) Spin Quantum Number, ms = - ½ or + ½. So simply consider an electron to be a Quantum System with only two allowed Quantum States.

Spin was discovered in the Stern Gerlach Experiment:

https://en.wikipedia.org/wiki/Stern%E2%80%93Gerlach_experime...

[Edited on 27-7-2015 by blogfast25]

blogfast25 - 26-7-2015 at 17:13

I hope people don’t mind too much but I want to push on, as there is still a lot of material to cover. With this instalment we’re past about 75 % of Part 1.

Thank you for reading so far.

Wave function orthogonality

Two mathematical functions f(x) and g(x) are said to be orthogonal on an interval a <= x <= b, if:

∫f(x)g(x)dx = 0 (integrated between x = a and x = b)

Take two wave functions (of the Pi1DB) with differing energy levels:

Ψn(x) = √(2/L) sin(nπx/L)

Ψm(x) = √(2/L) sin(mπx/L)

Provided n is not equal to m (remember that both n and m are positive, non-zero integers!) the energy levels (the eigenvalues associated with these wave functions) are not the same.

It can be proved that:

∫ √(2/L) sin(nπx/L) √(2/L) sin(mπx/L) dx = 0 (integrated between x = 0 and x = L), because:

∫sin(nπx/L) sin(mπx/L) dx = 0 (integrated between x = 0 and x = L)

(The proof can be found here: http://tutorial.math.lamar.edu/Classes/DE/PeriodicOrthogonal...)

This means that for the particle in a one dimensional box two non-equal wave functions are always orthogonal. It turns out that this is the case for all wave functions of bound particles. The physical interpretation of this is that the particle can be in the eigenstate associated with n (having eigenvalue En

or with m (having eigenvalue Em

but not in both states at once. Of all the possible states n = 1, 2, 3,… the system can only be in one eigenstate at a given moment t.

And this has also consequences for quantum chemistry, folks!

[Edited on 27-7-2015 by blogfast25]

aga - 27-7-2015 at 09:11

Thank you for reading so far

Thank You for posting so far, and not giving up the will to live !

Looking up what Orthogonal means, i'm still a bit confused.

Does this mean that say two particles with differing wave functions cannot occupy the same energy level at the same time ?

... or that a single particle can be described by two different wave functions ?

blogfast25 - 27-7-2015 at 10:04

Thank you for reading so far

Thank You for posting so far, and not giving up the will to live !

Looking up what Orthogonal means, i'm still a bit confused.

... or that a single particle can be described by two different wave functions ?

The meaning of words depends enormously on context. I don't know what you've been looking up.

A single particle can have a whole bunch of wave functions ψ and assorted eigenvalues E but it can only be in ONE of these eigenstates at once. This may sound self-evident but the orthogonality property proves this.

Forget about multi-particle systems for now, they'll come out of your ears in Part 2!

aga - 27-7-2015 at 10:27

so if a single particle can be described by two (or more) wavefunctions, then it cannot behave according to both functions at the same time ?

i.e. it's energy state will be defined by one wavefunction only, and never both at the same time.

annaandherdad - 27-7-2015 at 10:49

so if a single particle can be described by two (or more) wavefunctions, then it cannot behave according to both functions at the same time ?

i.e. it's energy state will be defined by one wavefunction only, and never both at the same time.

If a particle is in a state of definite energy, then the energy must be one of the eigenvalues En, and the wave function is the corresponding eigenfunction psin. In this case the probability of finding the particle in any region of space is constant in time, and if you measure the energy you always get the value En.

But a particle need not have a definite energy, just as it need not have a definite position or momentum. A particle can be in a linear combination of two energy eigenstates, for example, its wave function could be c1 psi1 + c2 psi2,
where c1 and c2 are two complex numbers whose squares add to 1. In that case the probability of measuring the energy and finding E1 is |c1|2, and that of finding E2 is |c2|2.
These probabilities are independent of time. But the probability of finding the particle in a given region of space is now a function of time.

The most general state of the system is a linear combination of all the energy eigenfunctions. In this way the particle can behave like all the energy eigenfunctions all at once.

Such states can be created by optical pumping, in which an system in its ground state is lifted into an excited state. While this is going on, the system is in a linear combination of the ground state and the excited state. The reverse happens when a state decays; during the decay process the wave function is a linear combination of the excited state and the target state (a state of lower energy).

Another way to do this is to take a system in a definite energy state, and then change the Hamiltonian, for example, by turning on an electric or magnetic field. The poor system, which was happily in an eigenstate of the old Hamiltonian, suddenly finds that the Hamiltonian and its eigenstates have changed, so it is no longer in an energy eigenstate of the new Hamiltonian.

aga - 27-7-2015 at 11:42

OK. So if a particle behaves according to two or more wavefunctions, let's say with products E_n being 10, 100 and 1000 for n=1,2,3
and E_m products being 5, 500 and 1500 eV, then all you're saying is that it will never be at 100 and 1500 eV at the same time.

Or something like that. Kinda makes sense.

annaandherdad - 27-7-2015 at 13:41

OK. So if a particle behaves according to two or more wavefunctions, let's say with products E_n being 10, 100 and 1000 for n=1,2,3
and E_m products being 5, 500 and 1500 eV, then all you're saying is that it will never be at 100 and 1500 eV at the same time.

Not sure what you mean by products. If 100 and 1500 are two energy levels (in some units) then the system can certainly be in a linear combination of the two corresponding wave functions (that is, energy eigenfunctions). If it were, and you made a measurement of energy, you'd get 100 with some probability, and 1500 with some other probability.

aga - 27-7-2015 at 14:12

ok. But never 750 for example, if the particle were pegged to the values that the two wavefunctions predict.

blogfast25 - 27-7-2015 at 14:24

AAHD:

With all due respect but you are sort of mudding the waters here a bit. We have been specifically keeping everything limited to particles in a definite, say 'steady state', eigenstate, and not how a particle may transit from one eigenstate to another (e.g. n to m). The math of these transitions is definitely outside the scope of this little excursion into QM/WM/QC...

Thanks nonetheless.

[Edited on 27-7-2015 by blogfast25]

aga - 27-7-2015 at 14:35

@AAND i appreciate your input : it is great to see commentary on stuff i barely grasp.

I suppose teach is right though, gotta see the Way first before walking, and Running is not an option just yet

blogfast25 - 27-7-2015 at 14:45

Quantum physics has become a vast, vast paradigm. On holiday recently I read 'Quantum Mechanics - the theoretical minimum' by Leonard Susskind (& Art Friedman) and had to study it rather than 'read' it.

To newcomers it must be a bewildering experience!

annaandherdad - 27-7-2015 at 15:17

ok. But never 750 for example, if the particle were pegged to the values that the two wavefunctions predict.

That's right. One of the postulates of quantum mechanics is that when you measure an observable, the possible outcomes are the eigenvalues of the corresponding operator. Here we're talking about energy, but the same applies to position, momentum, angular momentum, spin, charge, etc.

Also, quantum mechanics allows you to compute the probabilities of the various outcomes. In general the probabilities are functions of time, but not when the observable (such as energy in an isolated system) is conserved.

aga - 27-7-2015 at 15:32

Woohoo !

Thanks AAHD. It appears as though i almost understood something !

Happy happy day !

I'll have another beer to celebrate.

blogfast25 - 27-7-2015 at 17:45

The last in the series of properties of wave functions and some interim conclusions:

Parity of wave functions

In the case of a particle in a one dimensional box the coordinate system was chosen so that the box existed (U(x) = 0) between x = 0 and x = L. But had we chosen the coordinate system so that U(x) = 0 for x = - L/2 to x = + L/2, the shapes of the resulting wave functions (and eigenvalues E) would have been identical.

Note however that with the ‘new’ coordination system a centre of symmetry arises at x = 0 (at half the length of the box): looking to the left or the right from that point the physical system looks the same. The wave function formula needs to be slightly adjusted to accommodate the new coordination system, as:

Ψn(x) = √(2/L) sin[nπ(x + L/2)/L]

It can be shown that:

Ψ*(x)Ψ(x) = Ψ*(-x)Ψ(-x)

Or for Real wave functions:

Ψ2(x) = ψ2(-x)

Thus the probability functions show symmetry with respect to x = 0, the point of symmetry. This can be generalized to the statement that wave functions must have a definite parity with respect to symmetry operations in the physical problem, provided the potential energy function U(x) of the system is also symmetric with respect to the symmetry centre. In math terms a function U(x) symmetrical to x = 0 means U(x) = U(-x).

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/pfbox.htm...

An excellent and more mathematical explanation of Quantum parity can be found here:

http://physics.stackexchange.com/questions/13980/definite-pa...

Conclusions so far (particle in a one dimensional box)

• The Schrodinger equation is satisfied by suitable wave functions ψ and assorted eigenvalues E (the combination is also called an ‘Eigenstate’).
• The energies E are quantized and can be characterized by a quantum number n.
• The energy E cannot be exactly zero.
• The square of the wave functions are probability functions.
• The smaller the confinement, the larger the energy required.
• The wave functions are orthogonal.
• The wave functions show definite parity.

[Edited on 28-7-2015 by blogfast25]

Darkstar - 28-7-2015 at 01:58

As AAHD explained, a particle can indeed behave according to more than one wave function at the same time. What you're getting at is essentially what's called superposition.

For example, let's take the momentum basis. If the particle were in a definite momentum eigenstate, its momentum would be one of the possible eigenvalues and its wave function would be the corresponding eigenfunction. On the other hand, if the particle were NOT in some definite momentum eigenstate, then its momentum state is instead expressed as a superposition of all the possible momentum eigenstates. In other words, the particle is basically in a sort of state where it has all possible momentum simultaneously. This is why you may sometimes hear people claim that subatomic particles can "be in more than one location at the same time."

Now, if you were to actually attempt to measure/observe that particle's momentum, you would not find it in superposition. Instead, you would find it in a single, definite eigenstate of momentum, which was selected randomly with some probability from the superposition of possible eigenvalues. This is called wave function collapse.

Since Professor Blogfast is likely about to blow his top by now, you should probably get back to your main studies. Maybe a little later when he's not looking we can see how all of this relates to Heisenberg's uncertainty principle! (i.e. the more momentum eigenstates in the superposition, the more localized the particle's position becomes and the less defined its momentum gets, and vice-versa. thus the position state for a perfectly localized particle would actually be a superposition of ALL values for momentum, from -∞ to ∞ and everything in between)

annaandherdad - 28-7-2015 at 07:00

Darkstar---precisely.

aga - 28-7-2015 at 12:14

if the particle were NOT in some definite momentum eigenstate, then its momentum state is instead expressed as a superposition of all the possible momentum eigenstates.

In effect, the maths are continually 'bent' to fit with observed phenomena.

That tends to suggest that the maths do not work.

I suspect there's a well-established counter argument.

blogfast25 - 28-7-2015 at 13:23

In effect, the maths are continually 'bent' to fit with observed phenomena.

That tends to suggest that the maths do not work.

I suspect there's a well-established counter argument.

There would be but you need to be more precise about the 'bent math'.

What AAHD and Darkstar have been talking about hasn't actually been expressed in mathematical form here. What hasn't been rendered explicit ('precise', 'defined', 'unequivocal') yet cannot have been 'bent' yet.

[Edited on 28-7-2015 by blogfast25]

aga - 28-7-2015 at 13:57

True and i promised not to, until the end, so i won't.

blogfast25 - 28-7-2015 at 16:26

Erratum – Probability and Probability Density

The functions (for a Pi1DB):

Pn(x) = Ψ<ub>n2(x) = (2/L) sin2(nπx/L)

... have previously been described erroneously as ‘probability functions’ when really they should have been described as 'probability densities’ (also known as 'probability distribution', syn.). The difference is not mere semantics. Like an object can have a mass of 1 kg, that same object can have a density many times higher than 1 kg/m3.

The previous assertion that Pn(x) is the probability of finding the particle at position x is also wrong.

The actual probability of finding the particle can only be determined on an interval of x, for instance Δx = x2 - x1 (see the attached schematic, below) and is then calculated from:

P(x2 ,x1 = ʃ Ψ2(x)dx, integrated between x2 and x1 (for Real functionsΨ)

For a relatively narrow interval Δx, P(x2 ,x1

is approximately the area of the rectangle in the schematic below, or:

P(x2,x1

≈ Ψ2(x1,2

Δx (assuming that Ψ2(x1

≈ Ψ2(x2

)

For systems involving electrons the probability density is is often referred to as the electron density (we’ll encounter that term frequently in the future). In areas where the electron density (Ψ2

is high, the probability of finding electrons is high.

Apologies for the error!

[Edited on 29-7-2015 by blogfast25]

Darkstar - 30-7-2015 at 02:59

If anyone's interested, I made a table of contents to make navigating a little easier. The larger this thread gets, the more scattered the information will become and the harder it'll be to find specific things. This thread already has over 160 posts, and I imagine (and hope) it will continue to grow over the years. With a table of contents, readers will be able to quickly locate the topic(s) they are interested in learning more about without having to sort through page after page to find it. Hopefully this will make the thread slightly less intimidating for newcomers. (as if the material wasn't already intimidating enough!)

If it's alright with blogfast, maybe a moderator can add this to the first post? We can update it periodically as needed.

*NOTE* - Fixed it so that it works no matter what your "posts per page" is set to. I suggest all future additions be done this way as well. Don't copy the link to a post out of your actual address bar as it has a reference to the page it is on, which will vary from user to user depending on their settings. Instead, hover over the "posted on..." link above the target post and just copy that link directly.

Quick Navigation

Part I – Basic Wave Mechanics

Additional Contributions

[Edited on 7-30-2015 by Darkstar]

blogfast25 - 30-7-2015 at 05:54

That is very useful Darkstar, thank you very much!

blogfast25 - 31-7-2015 at 09:45

Before we start discussing the hydrogen atom, let’s have a look at another ‘simple’ QS for bound particles.

The Quantum Harmonic Oscillator

The Quantum Harmonic Oscillator (QHO) is an approximation of vibrational modes of a diatomic molecule like H2 or HCl. Unlike in the example of a Pi1PD, U(x), the potential energy function, is not zero, but it is symmetric (with respect to x = 0).

General outline:

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hosc.html...

Wave functions and probability densities:

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hosc5.htm...

Note that there is a finite probability that the oscillator will be found outside the "potential well" indicated by the smooth curve. This is forbidden in Classical physics. It’s a phenomenon called ‘Quantum tunneling’.

Eigenvalues E:

En = (n + ½) ω ћ

(With ћ = h/(2π) (“h bar”) and n the Quantum Number of the QHO)

Note that here n = 0 is allowed because E0 = ½ ω ћ, that is non-zero. Ψ0 is also non-zero and thus a Real solution to the SE.

Application: the Vibration/rotation spectrum of HCl

The absorption spectrum of HCl:

http://hyperphysics.phy-astr.gsu.edu/hbase/molecule/vibrot2....

The central gap is due to the transition n = 0 to n = 1 for the HQO that is HCl. The other equidistant peaks are due to HCl also being a Quantum Rotator.

[Edited on 1-8-2015 by blogfast25]

blogfast25 - 2-8-2015 at 10:15

Operators: Classic and Quantum

(All in one dimension (x))

For Classical systems the system's properties are defined as follows.

Classical:

Position = x

Velocity: vx = dx/dt

Momentum: px = mvx = m dx/dt

Kinetic energy: E = 1/2 mv2

Total energy: E = 1/2 mv2 + U

Quantum:

Very high up in this course I wrote:

Quote:

The wave function ψ contains all measurable information about the particle.

There exist a number of relationships (which won’t be derived here) that allow to ‘extract’ that measurable information from the wave function.

(Time independent, x dimension)

Velocity: v = p/m = - (iћ/m) δψ(x)/δx

Momentum: p = - iћ δψ(x)/δx

Kinetic energy: - (iћ2/2m) δ2ψ(x)/δx2

Total energy: - (iћ2/2m) δ2ψ(x)/δx2 + U(x)ψ(x)

If we define an Operator as a mathematical operation we apply to the wave function, then we can define the following Quantum Operators (operators are noted in bold and italic here):

Quantum Momentum operator p = - iћ δ/δx

Quantum Velocity operator v = p/m

Hamiltonian operator (total energy) H = - (iћ2/2m) δ2/δx2 + U(x)

The time-independent Schrodinger Equation can thus be re-written as:

Hψ = Eψ

... with ψ an eigenfunction and E an eigenvalue.

For a general Quantum operator Q:

Qψ = qψ

... with ψ an eigenfunction and q an eigenvalue.

So, observable information about the QS can be obtained by applying quantum operators to the wave function ψ.

Expectation Values:

Position expectation value:

The expectation value < x > can be interpreted as the average value of x that we would expect to obtain from a large number of measurements. Alternatively it could be viewed as the average value of position for a large number of particles which are described by the same wave function.

< x > = ʃ ψ*x ψdx, integrated over - ∞ to +∞

For Real wave functions:

< x > = ʃ xψ2dx, integrated over - ∞ to +∞

For the Pi1DB (infinite potential well):

< x > = L/2. The average of many position measurements yields L/2.

For momentum, the expectation value is:

 = ʃ ψ*pψdx, integrated over - ∞ to +∞

With p the Quantum momentum operator.

For Real wave functions:

 = ʃ ψpψdx, integrated over - ∞ to +∞

It can be shown that for eigenstates (of bound particles), = 0. And because = m< v >, the expectation value < v > = 0 also.

Does this mean that the particle is motionless? No, not all. Take the example of a Pi1DB. The particle is analogous to a prisoner in a cell: the prisoner can move about in the confined space of the cell but he’s not going anywhere. His average velocity is effectively 0. If for the Pi1DB < v > > 0 then the particle would effectively be escaping, so < v > = 0 is the ‘natural’ result.

[Edited on 3-8-2015 by blogfast25]

aga - 2-8-2015 at 11:55

Woah !

Hang on a sec Blogfast.

Give us at least time to read the blackboard, please !

blogfast25 - 2-8-2015 at 13:50

Did you perhaps miss the one but last session?

aga - 2-8-2015 at 14:16

Yep.

(blush)

blogfast25 - 2-8-2015 at 14:53

No worries. You'll have plenty of tellyless evenings to figure it all out!

You know: '999 channels and never anything on' and all that...

Darkstar - 3-8-2015 at 02:08

Perhaps he can make up for it by correctly answering the following, which will serve as extra credit. All students are welcome (and encouraged) to try, of course, so please don't hesitate if you think you know the answer. All I ask is that you don't cheat and look it up! Also, I would appreciate it if the more knowledgeable readers would remain silent. Let the beginners figure this one out.

Note that there is a finite probability that the oscillator will be found outside the "potential well" indicated by the smooth curve. This is forbidden in Classical physics. It’s a phenomenon called ‘Quantum tunneling’.

Consider for a moment how exactly this phenomenon, known as "quantum tunneling," could be applied to chemistry, particularly radioactivity and the radioactive decay of elements with unstable nuclei. For our purposes, let's consider only the decay process known as alpha decay, a form of cluster decay where a nucleus emits an alpha particle (two protons and two neutrons--basically a helium nucleus). This decay process is ultimately the result of a Coulomb repulsion between the alpha particle and the rest of the nucleus, both being positively-charged, that pushes the two away from one another. The problem, however, is that at the short distances we're dealing with, the attraction caused by the strong force completely overpowers any electromagnetic repulsion between the alpha particle and the rest of the nucleus. This creates a potential well that effectively traps the alpha particle.

So here's the question: If the alpha particle is initially trapped inside the nucleus and forbidden to escape, similar to how the particle in the box is trapped and forbidden to escape, how is it that the alpha particle is eventually able to overcome the strong force and leave the potential well, all without ever actually having enough energy to escape the nucleus in any classical sense? In other words, how might quantum tunneling be used to explain this?

Hint: Unlike the particle in the box, the potential well the alpha particle is trapped in has finite potential walls. Now recall wave-particle duality and Heisenberg's uncertainty principle.

blogfast25 - 3-8-2015 at 05:33

Hint: Unlike the particle in the box, the potential well the alpha particle is trapped in has finite potential walls. Now recall wave-particle duality and Heisenberg's uncertainty principle.

Darkstar:

Would you consider authoring a small segment on the Uncertainty Principle? Perhaps one that relates confinement energy to well size? As we know confinement energy is more or less inversely proportional to well size.

blogfast25 - 8-8-2015 at 07:42

Last stop before the hydrogen atom. Darkstar should feel free to add other examples of the uncertainty principle.

The Uncertainty Principle

The position and momentum of a particle cannot be simultaneously measured with arbitrarily high precision. There is a minimum for the product of the uncertainties of these two measurements. There is likewise a minimum for the product of the uncertainties of the energy and time.

σx σp > ћ/2
σt σE > ћ/2

This is not a statement about the inaccuracy of measurement instruments, nor a reflection on the quality of experimental methods; it arises from the wave properties inherent in the quantum mechanical description of nature. Even with perfect instruments and technique, the uncertainty is inherent in the nature of things.

σx, σp, σt and σE are standard deviations. The smaller σ is, the more accurately the property (x, p, t or E) is known.

Application to a Pi1DB:

Higher up we saw that the confinement energy of a bound particle increases as the size of the box decreases. We can show this to be the direct consequence of the Uncertainty Principle.

For a particle in a box x = 0 to x = L it can be shown that:

σx = L/√12

(Proof: http://hyperphysics.phy-astr.gsu.edu/hbase/math/stdev.html#c...)

With the uncertainty principle:

σp > ћ/(2 σx

= ћ√12/2L

It can also be shown that σp2 = < p2 >

The minimum kinetic energy (KE) for confinement is thus:

< KEmin > = < p2 >/2m

Reworked we get:

< KEmin > = 3 ћ2/(2mL2

Or: < KEmin > = 3 h2/(8π2mL2

E1 = h2/(8mL2

This means effectively that E1 > < KEmin > ("the ground state energy is larger than the minimum kinetic energy demanded by the Uncertainty Principle").

So this means that the calculated eigenvalue E1 (ground state) is in agreement with the Uncertainty Principle. By contrast, a ground state of E1 = 0 would be in flagrant violation of it.

Further reading: on the page below the author(s) calculate the ground state of the hydrogen atom using nothing more than the Uncertainty Principle!

http://quantummechanics.ucsd.edu/ph130a/130_notes/node98.htm...

[Edited on 8-8-2015 by blogfast25]

Dan Vizine - 10-8-2015 at 17:54

I'll be happy if I finally understand Hamiltonian operators....you've got my interest!

blogfast25 - 10-8-2015 at 18:04

THE Hamiltonian operator (total energy of a quantum system) isn't too bad. Quantum operators in their full glory and properties... now that IS serious stuff!

[Edited on 11-8-2015 by blogfast25]

blogfast25 - 14-8-2015 at 10:24

We’re nearing the last of Part 1, which mainly concerns itself with the QM of the Hydrogen atom. So I bring you another mathematical (mini-)interlude, as well as the first part of the hydrogen atom’s treatment. Enjoy!

Another math mini-excursion: Cartesian and Spherical Coordinates

Because the hydrogen atom is roughly spherical the Schrodinger equation is ‘easier’ to solve when anchored in a spherical coordinate system than a Cartesian one.

A Cartesian coordinate system is composed of three perpendicular axis X, Y and Z that intersect at the point of origin O. The position of any point in space with respect to the Cartesian coordinate system is unequivocally determined by (x,y,z).

Now look a Spherical Coordinate System, here ‘superimposed’ on the Cartesian system:

https://en.wikipedia.org/wiki/Spherical_coordinate_system

The Spherical system uses r, θ and φ to define the point’s position in space:

1) Radial distance r is the distance between the point and the origin O (0,0,0).
2) Polar angle θ is the angle between the projection of the line that connects the point and O onto the XY plane and the X axis.
3) Azimuthal angle φ is the angle between the line that connects the point and O and the Z axis.

So the point’s position in space is now defined unequivocally by (r,θ,φ) (angles are expressed in Radian: 360 degrees is 2π)

The relationships between x, y, z and r, θ, φ are:

x = r sinθ cosφ
y = r sinθ sinφ
z = r cosθ

Using these relationships, any function F(x,y,z) can be transformed into a new function R(r,θ,φ).

And here’s a little teaser on just about how useful this can all be.

The Cartesian coordinate based equation for a sphere with radius R and centre point in O is:

R2 = x2 + y2 + z2, quite a mouthful, huh?

And in spherical coordinates?

r = R, and that’s it!

You can prove this by plugging the relationships for x, y and z into the previous equation, if you’re that way inclined!

++++++++++++

The Hydrogen Atom’s Schrodinger Equation

In a mono-electronic atom the potential energy function of the electron is determined by the electrostatic attraction exerted by the positively charged nucleus on the negatively charged electron. It depends only on the distance r between the electron and the nucleus:

U(r) = - e2/(4πε0r)

U(r) is a radially symmetrical function and ideal for treatment in spherical coordinates.

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydsch.ht...

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydsch.ht...

The transformation of coordinate system (Cartesian to Spherical), with some VERY nifty rearranging, splits the SE into three parts: the Radial Equation, the Colatitude Equation and the Azimuthal Equation:

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydsch.ht...

Quantum numbers

Similarly the hydrogen atom’s wave function itself is transformed into the product of three functions:

Ψ(x,y,x) === > Ψ(r,θ,φ) = R(r) P(θ) F(φ)

With each part (radial, colatitudinal, azimuthal) are associated Quantum Numbers:

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydsch.ht...

R(r): the Principal Quantum Number n = 1, 2, 3, 4,…

P(θ): Orbital Quantum Number l = 0, 1, 2, … n - 1

F(φ): Magnetic Quantum Number ml = - l, - l + 1, to + l

In addition, electrons have an intrinsic (regardless of physical system they find themselves contained in) Spin Quantum Number, ms = - ½ or + ½.

http://hyperphysics.phy-astr.gsu.edu/hbase/spin.html#c1

[Edited on 14-8-2015 by blogfast25]

Darkstar - 18-8-2015 at 09:09

Sorry for the delayed response; it's been a busy last couple of weeks for me. I see you went ahead and introduced the uncertainty principle, so that's good. I may write a brief segment later to specifically address some of the most common misconceptions about the uncertainty principle. For example, the uncertainty having nothing to do with the measurement instrument itself, or how the uncertainty principle is not about single measurements and/or single particles, but is instead a statistical statement about subsequent measurements of identically-prepared systems.

Of all the concepts in QM, it seems like HUP is the one students have the most trouble understanding. It is easily one of the most widely misunderstood concepts in modern physics, and I think a lot of it is due to the way HUP is frequently taught at the undergraduate level, especially in chemistry and other non-QM courses. I remember briefly covering the uncertainty principle back in General Chemistry I, only it was incorrectly taught as something more akin to the observer effect. Instead of explaining the uncertainty when attempting to measure conjugate pairs simultaneously as being an inherent property of matter due to the very nature of the physical universe itself, it was instead explained as though the uncertainty is due to the actual act of measurement, which alters what is being observed by interacting with it.

And speaking of QM misconceptions, don't even get me started on the biggest one of them all: that wildly popular pre-QM myth known as wave-particle duality--a non-existent concept that just won't die despite being outdated for nearly 100 years. The truth is that while quantum objects do sometimes behave like waves and other times like particles, the vast majority of the time they behave like neither. And when a particle does "behave like a wave," it isn't literally some real, physical wave that actually propagates through three-dimensional space, but rather one that propagates through an abstract mathematical concept known as Hilbert space. There's a reason you rarely ever see "wave-particle duality" mentioned anywhere in advanced or highly-technical QM books!

And, yes, the above part, while technically true, was mostly tongue-in-cheek and is intended to be taken with a grain of salt.

blogfast25 - 18-8-2015 at 09:58

And speaking of QM misconceptions, don't even get me started on the biggest one of them all: that wildly popular pre-QM myth known as wave-particle duality--a non-existent concept that just won't die despite being outdated for nearly 100 years. The truth is that while quantum objects do sometimes behave like waves and other times like particles, the vast majority of the time they behave like neither. And when a particle does "behave like a wave," it isn't literally some real, physical wave that actually propagates through three-dimensional space, but rather one that propagates through an abstract mathematical concept known as Hilbert space. There's a reason you rarely ever see "wave-particle duality" mentioned anywhere in advanced or highly-technical QM books!

My impression, as a perennial (and very modest level) student of QM, is that the original de Broglie hypothesis is viewed in Diracian (Hilbert space) QM as nothing more than a quaint, antiquated and poor analogy.

It appears that in the Complex space, Classical wave mechanics and Quantum wave mechanics share quite a bit of mathematics but that this is shown to be almost no more than a ‘coincidence’, when one moves the treatment to matrix mathematics in the Hilbert space.

Re. QM in the Hilbert space I’m still very much of a dilettante, to be completely honest. Having been ‘brought up’ on a diet of matter waves (electrons) I will find it hard to ‘let go’. Someone will have to detain me in the ‘Hilbert Re-education Boot Camp’, I think! It must be infuriating for the Youngens to see the Olds still fiddle with their charming but somewhat obsolete ‘waves’.

It's kind of ironic that one of the heroes of QM, Erwin Schrödinger himself, was a great believer in these matter waves, or at least for some time.

I can’t really see Hilbert space mathematics taught in middle schools any time sun though, so very sharply abstract it is.

I guess we need to take a leaf out of Scott Aaronson's book!

What do you think?

[Edited on 18-8-2015 by blogfast25]

aga - 18-8-2015 at 11:07

The failure of all models to accurately represent reality is a thing that bothers me a lot, which is my main interest in understanding what is already known.

Not doing so great in that so far, but am back on the homework waggon.

aga - 18-8-2015 at 11:24

So here's the question: If the alpha particle is initially trapped inside the nucleus and forbidden to escape, similar to how the particle in the box is trapped and forbidden to escape, how is it that the alpha particle is eventually able to overcome the strong force and leave the potential well, all without ever actually having enough energy to escape the nucleus in any classical sense? In other words, how might quantum tunneling be used to explain this?

Hint: Unlike the particle in the box, the potential well the alpha particle is trapped in has finite potential walls. Now recall wave-particle duality and Heisenberg's uncertainty principle.

Best guess is that it leaps the containments walls, uncertainly.

To be honest, the Maths have already slipped beyond my grasp, so i'll go back to page 4 or 5 and start again.

Darkstar - 21-8-2015 at 16:16

What do you think?

Can't say I blame you for having trouble letting go. I tend to think that it's actually quite helpful--almost necessary, even--to first understand QM in terms of "matter waves" before seriously trying to make sense of some of the more modern interpretations. So while the original de Broglie hypothesis may very well be an outdated concept as far as modern physics is concerned, I honestly can't imagine ever having learned QM without it.

At the end of the day, I try not to get too hung up on all the technicalities. I simply stick to what makes the most sense to me and just acknowledge that it may not always be entirely correct.

Best guess is that it leaps the containments walls, uncertainly.

Correct, although "leap" is probably not the best way to describe it. In quantum tunneling, the particle, for lack of a better word, goes through the barrier not over it, something that is forbidden classically. It would be like a rolling ball somehow making it to the other side of a hill despite not actually having enough energy to roll up and over it.

In the case of alpha decay, because there's a non-zero probability that the alpha particle will be found outside of the nucleus, given enough time, it eventually will be. When that happens, if the alpha particle is sufficiently far enough away from the nucleus, the electromagnetic repulsion between the two will become more significant than the attraction caused by the strong force, allowing the alpha particle to escape.

aga - 21-8-2015 at 16:24

My basic problem is that QM is internally inconsistent, in that the Assumptions are not really on-track, Conditions were Lifted to make it (the maths) work and it's Popular amongst Mathematicians.

Granted : it's good for some things, however does not model Reality accurately.

Damn.

I said it and it's not really Over.

Classical and Quantum are not really at odds at all.

The fundamental problem is that they both offer a View of Reality in mathematical terms, and neither view is actually reality, as Reality is not bound by our understanding of it.

[Edited on 22-8-2015 by aga]

blogfast25 - 21-8-2015 at 17:18

aga:

I'm hesitating between a time-consuming, all-out, 'high brow' rebuttal of your last post and the futility of such a rebuttal. We seem to have gone full circle as this is no more than a slightly more polite rehash of your prior statements re. QM (remember? The ones that prompted to organise this series of basic 'lectures' to begin with?)

On balance I'll keep it short:

Quantum Mechanics as a theory does not rely on aga-like observers for its validity/invalidity.

blogfast25 - 21-8-2015 at 17:45

Going back to Darkstar's alpha emission problem, think particle in a box with finite potential wall (the potential V here is the Strong nuclear force that keeps nuclei together, very simply put):

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/pbox.html...

Such an alpha particle has limited (but NON-ZERO) probability density outside of the Classically allowed region (x=0, x=L), so that a detector can (occasionally) pick it up outside of the nucleus (the potential well). It has 'tunnelled'.

[Edited on 22-8-2015 by blogfast25]

aga - 22-8-2015 at 07:17

That can't have been me !
Must have been HD using my password in order to tarnish my flawless professionalism and reputation as a clear thinker.

So the tunnelling phenonenon is that with non-infinite bounds, the particle can actually be 'seen' at some point beyond those bounds, albeit rarely.

Is that right ?

Darkstar - 24-8-2015 at 13:19

My basic problem is that QM is internally inconsistent, in that the Assumptions are not really on-track, Conditions were Lifted to make it (the maths) work and it's Popular amongst Mathematicians.

Granted : it's good for some things, however does not model Reality accurately.

I think you're seriously underestimating just how well QM does model reality. Just so it's clear, quantum mechanics is, bar none, THE most accurate description of nature we have. Nothing else even comes close to modeling reality as accurately as QM/QFT does. Is it perfect? No. While QM/QFT may accurately explain interactions between the strong force, weak force and electromagnetic force at the scale of atoms and subatomic particles, it completely fails to explain gravity at distances that small.

So no, QM doesn't model reality 100% correctly, but it's pretty damn close. A theory that literally models reality perfectly would require unifying QM (strong force, weak force and electromagnetism) and GR (gravity) to produce a Theory of Everything.

Quote:

Classical and Quantum are not really at odds at all.The fundamental problem is that they both offer a View of Reality in mathematical terms, and neither view is actually reality, as Reality is not bound by our understanding of it.

As I mentioned here and here, the classical world we observe is just an emergent property of the underlying quantum world, caused by a rapid loss of coherence in open quantum systems via interactions with their environment. At the end of the day, classical mechanics is really nothing more than an approximation to QM. The universe is ultimately quantum at its core--everything is governed by and can be explained by QM, both at the microscopic and macroscopic scale. The reason we use classical mechanics for macroscopic systems is because it's equally accurate when it comes to large objects (see the correspondence principle) only much simpler and easier to use.

What are at odds are QM and GR.

Quote:

So the tunnelling phenonenon is that with non-infinite bounds, the particle can actually be 'seen' at some point beyond those bounds, albeit rarely.

Is that right ?

Correct. But whether or not it's "rarely" found beyond those bounds really just depends on how small the probability of finding it there is. This is why the half-life of some radioactive elements (in the context of alpha decay) are longer or shorter than others. If the probability is extremely tiny (i.e. a more stable nucleus), the half-life of the element will be longer and finding the alpha particle outside of a given nucleus would be somewhat rare. On the other hand, if the probability is relatively high (i.e. a less stable nucleus), the half-life of the element will be much shorter, meaning each individual atom is much more likely to decay at any given moment.

To kind of visualize what is going on, imagine the alpha particle rattling around inside the nucleus and every so often smashing into the "containment" wall and then bouncing off. Now imagine that each time the alpha particle hits the wall, there's always a non-zero chance that it will tunnel out of the nucleus and escape. Thus the higher the probability of that happening, the less time it's likely going to take.

[Edited on 8-24-2015 by Darkstar]

aga - 24-8-2015 at 13:33

Thankyou for the clarifications - it is actually very helpful to have your input.

Appologies to yourself and blogfast for posting nonsense in such an informative thread that it should be stickied.

blogfast25 - 24-8-2015 at 14:39

Hyperphysics gives a very nice explanation (much better than my finite potential analogy, actually) of alpha decay with QM tunnelling:

http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/alptun.ht...

Follow the links to see how alpha decay half-lives can be modelled with quantum mechanical tunnelling. Math included!

Quantum mechanics? What is it good for? ABSOLUTELY NOTHING!

[Edited on 24-8-2015 by blogfast25]

aga - 25-8-2015 at 08:43

So, to be clear, Classical physics says that the particle can never escape the bounds set, however observations show that it does escape, therefore the Classical model simply does not work 100%.

QM allows for the escape, via the small, yet Finite Probability that the particle is seen escaping.

OK. If that's a fair precis of Quantum Tunelling, i may well be back on track.

blogfast25 - 25-8-2015 at 09:20

To put it maybe slightly more subtly, the Classical particle can only escape if it's energy is higher than the barrier potential Vb. If its energy (e.g. V1) is lower than the barrier potential Vb, the Classical particle is strictly confined to [-R,+R].

But for the Quantum particle, even if its energy is only V1, it can tunnel through the potential barrier and has a small but finite and NON-ZERO probability to be found outside the Classical area [-R,+R].

The schematic below is for a one dimensional potential well with a potential energy function V(x) comparable to that of an alpha particle in a nucleus:

It might also be useful to remind us of the Correspondence Principle. As we make the system larger (larger and larger R) the system starts resembling Classical mechanics: tunnelling decreases (until it ceases altogether), for instance. But for the alpha decay problem we are at extremely small scale: femtometers!

[Edited on 25-8-2015 by blogfast25]

aga - 25-8-2015 at 11:59

I suppose the notion that a Quantum Particle is never motionless is another key difference between classical and quantum theories.

In your diagram (ignoring wavefunctions for the moment) a moving particle with V1 will occasionally bounce off the precise point on the left hand barrier to clear the right hand barrier, putting it outside the cell.

(right handed thinking)

Edit:

In essence the QM model takes dynamics into account to a degree.

[Edited on 25-8-2015 by aga]

blogfast25 - 25-8-2015 at 12:26

A particle like an electron can be motionless but the particles we're interested in here, for our purposes in this course, are always bound. Bound by a potential well, the nucleus of an atom or the nuclei of atoms (molecules, see later). And bound particles have a ground energy that is ALWAYS non-zero. Non-zero ground energy means the particle has kinetic energy and thus that it moves somehow.

It's tempting to imagine that movement in the 'Classical way' but that doesn't really work. For a particle in a 1D box with potential well (for instance) we can calculate an expectation value (see quite a bit higher up) < v > for speed and it's zero (note that the expectation value is only an average). But we know it moves because it has kinetic energy. Weird but true!

[Edited on 25-8-2015 by blogfast25]

aga - 25-8-2015 at 13:19

Not weird at all in the agaspace model.

Still no maths yet relating to that, so pointless.

Hmm. Funny thing is that a dimentionless Point can't exists in agaspace.

Ignore all that. Back to the proper stuff.

blogfast25 - 25-8-2015 at 15:56

Hydrogen Atom Wave Functions

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydwf.htm...

Tick the boxes to change n.

The Radial wave functions Rn,l(r) can also found here, scroll down for graphical representations of R and R2 (the probability functions).

http://quantummechanics.ucsd.edu/ph130a/130_notes/node233.ht...

Quantum numbers and Orbital naming

While the mathematical descriptions of hydrogen’s electron waves are referred to as the wave functions, their ‘physical’ manifestations are called orbitals. Four main types of orbitals can be discerned: s, p, d and f:

n = 1, l = 0 === > 1s orbital

n = 2, l = 0 === > 2s orbital
n = 2, l = 1 === > 2p orbital

n = 3, l = 0 === > 3s orbital
n = 3, l = 1 === > 3p orbital
n = 3, l = 2 === > 3d orbital

n = 4, l = 0 === > 4s orbital
n = 4, l = 1 === > 4p orbital
n = 4, l = 2 === > 4d orbital
n = 4, l = 3 === > 4f orbital

As indicated above, for l > 0, ml = - l to + l. p, d and f type orbitals therefore have suborbitals:

p: l = 1, ml = - 1, 0, + 1. There are 3 types of p suborbitals.
d: l = 2, ml = - 2, - 1, 0, + 1, + 2. There are 5 types of d suborbitals.
f: l = 3, ml = - 3, - 2, - 1, 0, + 1, + 2, + 3. There are 7 types of f suborbitals.

To get a good idea of the shapes of these orbitals, I warmly recommend ‘Orbitron’ (a screen shot is attached below):

http://winter.group.shef.ac.uk/orbitron/AOs/1s/index.html

Play around with this to explore and suborbitals, and in the next instalment I will discuss these shapes a little more.

[Edited on 26-8-2015 by blogfast25]

aga - 1-9-2015 at 11:58

So, Lunch has finished.

Where were we at ?

So far :-

QM model for particle in a 1 dimensional box.
Schrodinger's equation.
Mention of Eigenstates
Wavefunction encapsulates all of a particle's properties.
Quantum Tunelling.
QM used to plot electron orbital probabilities when expanded to 3D.

Is there a reason why the orbitals are not simply named a,b,c,d etc ?

[Edited on 1-9-2015 by aga]

blogfast25 - 1-9-2015 at 12:42