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[*] posted on 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]




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[*] posted on 24-7-2015 at 14:43


Quote: Originally posted by aga  

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... ;)




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[*] posted on 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.




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[*] posted on 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!




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[*] posted on 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 n<sub>2</sub> to n<sub>1</sub> (with n<sub>2</sub> > n<sub>1</sub>;) and energy will be emitted: ΔE = E<sub>n1</sub> - E<sub>n2</sub> < 0.

Quick side note: the mechanics of such transitions (from n<sub>2</sub> to n<sub>1</sub>;) 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 n<sub>1</sub> to 1, 2, 3 etc.

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

f<sub>1</sub> = h/(8mL<sup>2</sup>;) - h/(8mL<sup>2</sup>;)n<sub>2</sub><sup>2</sup>
= h/(8mL<sup>2</sup>;)[1 - n<sub>2</sub><sup>2</sup>]

For Series 2:

f<sub>2</sub> = h/(8mL<sup>2</sup>;)[4 - n<sub>2</sub><sup>2</sup>]

For Series n:

f<sub>2</sub> = h/(8mL<sup>2</sup>;)[n<sup>2</sup> - n<sub>2</sub><sup>2</sup>]

Of course h/(8mL<sup>2</sup>;) = E<sub>1</sub>.

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 E<sub>1</sub> = 6.8 eV is shown below.

Emission Spectrum 1D Box.png - 9kB


[Edited on 25-7-2015 by blogfast25]




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[*] posted on 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 ?




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[*] posted on 25-7-2015 at 10:01


Quote: Originally posted by aga  
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 ?



You would always need to know (or guess) the form of the energy function, say E<sub>n</sub> = F(E<sub>1</sub>,n) ("E<sub>n</sub> a function of E<sub>1</sub> and n"), as well as one of the n numbers in:

ΔE = E<sub>n1</sub> - E<sub>n2</sub> (= 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 = E<sub>1</sub>(n+1)<sup>2</sup> –E<sub>1</sub>n<sup>2</sup>

= E<sub>1</sub>(n<sup>2</sup> + 2n + 1 – n<sup>2</sup>;)

= E<sub>1</sub>(2n+1)

This kind of expression can be useful for a spectrocopist to determine E<sub>1</sub> with.


[Edited on 26-7-2015 by blogfast25]




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[*] posted on 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.




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[*] posted on 25-7-2015 at 11:23


The 'magik' of science! :D



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[*] posted on 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)<sup>2</sup> 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)<sup>2</sup> = 1, summation over all values of x.

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

∫ψ(x)<sup>2</sup>dx = 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:

Ψ<sub>n</sub>(x) = √(2/L) sin(nπx/L)

… and the resulting probability functions return the correct values.

Note that in the given example, A<sup>2</sup> = 2/L, has two roots:

A<sup>+</sup> = + √(2/L) and A<sup>-</sup> = - √(2/L)

And indeed both and –ψ both satisfy the SE. Textbooks sometimes note the wave functions as +/- ψ. The probability functions ψ<sup>2</sup> 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]




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[*] posted on 25-7-2015 at 21:37


Quote: Originally posted by aga  
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/n<sub>1</sub><sup>2</sup> - 1/n<sub>2</sub><sup>2</sup>;) for the frequencies of the spectral lines, where k is a constant. We now understand this by saying that E<sub>n</sub> = -k/n<sup>2</sup> are the energy levels of hydrogen.




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[*] posted on 26-7-2015 at 00:46


Incredible !

So if the spectrum emitted by a system (determined experimentally) coincides with the calculated En 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 ?




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[*] posted on 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.



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[*] posted on 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?



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[*] posted on 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.




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[*] posted on 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 ?



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[*] posted on 26-7-2015 at 14:51


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



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[*] posted on 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, m<sub>s</sub> = - 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]




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[*] posted on 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:

Ψ<sub>n</sub>(x) = √(2/L) sin(nπx/L)

Ψ<sub>m</sub>(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 E<sub>n</sub>;) or with m (having eigenvalue E<sub>m</sub>;) 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]




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[*] posted on 27-7-2015 at 09:11


Quote: Originally posted by blogfast25  
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 ?




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[*] posted on 27-7-2015 at 10:04


Quote: Originally posted by aga  
Quote: Originally posted by blogfast25  
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!




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[*] posted on 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.




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[*] posted on 27-7-2015 at 10:49


Quote: Originally posted by aga  
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 E<sub>n</sub>, and the wave function is the corresponding eigenfunction psi<sub>n</sub>. 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 E<sub>n</sub>.

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 c<sub>1</sub> psi<sub>1</sub> + c<sub>2</sub> psi<sub>2</sub>,
where c<sub>1</sub> and c<sub>2</sub> are two complex numbers whose squares add to 1. In that case the probability of measuring the energy and finding E<sub>1</sub> is |c<sub>1</sub>|<sup>2</sup>, and that of finding E<sub>2</sub> is |c<sub>2</sub>|<sup>2</sup>.
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.




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[*] posted on 27-7-2015 at 11:42


OK. So if a particle behaves according to two or more wavefunctions, let's say with products En being 10, 100 and 1000 for n=1,2,3
and Em 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.




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[*] posted on 27-7-2015 at 13:41


Quote: Originally posted by aga  
OK. So if a particle behaves according to two or more wavefunctions, let's say with products En being 10, 100 and 1000 for n=1,2,3
and Em 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.




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