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Author: Subject: Quantum Mechanics/Wave Mechanics in Chemistry Beginners Thread
aga
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[*] posted on 1-11-2016 at 13:45


If i've seen all THAT before, call me a cobblerwonky and slap my bodkin with a thrice folded cornichon !

Alternatively:-

Ah yes, of course.

I must take some time to consider such dazzlingly beautiful representations of awesome Mathematical Purity.

(buys enough time, hopefully, to run off and find the Mafs book, then frantically revise)




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[*] posted on 1-11-2016 at 15:29


Quote: Originally posted by aga  
to run off and find the Mafs book, then frantically revise)


Get over yourself! You know enough to revise this quite quickly!

Chop, chop... :cool:




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[*] posted on 1-11-2016 at 15:57


I had to look up what a cornichon is and I have no idea why it would be thrice-folded.


The mathematics here is rather beautiful and elegant. But the initial concepts of complex numbers involve the intersection of numerous different fields of mathematics as well as some counterintuitive ideas and some rather unfortunate terminology. Compounding this is the fact that computation with complex numbers manually quickly becomes its own special headache. (I distinctly remember asking my maths teacher the value of 2i and it taking him a long time to figure it out.)

The combined effect of this is that it easy to feel overwhelmed. But in reality it is not that daunting.




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[*] posted on 1-11-2016 at 16:11


Cornichons are just Gherkins.

Overwhelmed, a bit, over-imbibed, a lot.

Partake !

On the morrow i'll get some new car insurance.

After that, well, who knows ?




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[*] posted on 2-11-2016 at 09:50


Quote: Originally posted by j_sum1  
(I distinctly remember asking my maths teacher the value of 2i and it taking him a long time to figure it out.)



Even more "fun" are things like:

$$\sqrt [4]{-6}$$

But we're not going there. "Lite" here means addition, substraction, multiplication and division only! :)

[Edited on 2-11-2016 by blogfast25]




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[*] posted on 2-11-2016 at 11:52


The car insurance got complicated, a bit like where i'm heading on this first (presumably easy) question.

So far i got some pythagoras going on to give :-

$$ y = \sqrt { \frac {1}{2} } \sin{ \frac {\pi}{4} } $$

Is this going in remotely the right direction, or is it an overthunk, drunk and sunk direction ?

[Edited on 2-11-2016 by aga]




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[*] posted on 2-11-2016 at 12:22


Quote: Originally posted by aga  
The car insurance got complicated, a bit like where i'm heading on this first (presumably easy) question.

So far i got some pythagoras going on to give :-

$$ y = \sqrt { \frac {1}{2} } \sin{ \frac {\pi}{4} } $$

Is this going in remotely the right direction, or is it an overthunk, drunk and sunk direction ?

[Edited on 2-11-2016 by aga]


No, it is in fact correct but it's not the full solution.




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[*] posted on 2-11-2016 at 12:43


Phew ! I was once again worried that i'd gone all around the houses for no reason.

So,

$$ y = \sqrt { \frac {1}{2} } \sin{ \Big ( \frac {\pi}{4} \Big ) } $$

$$ x = \sqrt { \frac {1}{2} } \sin{ \Big ( 90 - \frac {\pi}{4} \Big ) } $$

The y part is the Imaginary bit, so, tentatively ...

$$ \sqrt { \frac {1}{2} } \sin{ \Big ( 90 - \frac {\pi}{4} \Big ) } + i \sqrt { \frac {1}{2} } \sin{ \Big ( \frac {\pi}{4} \Big ) } $$

The LaTeX is coming back, slowly ...




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[*] posted on 2-11-2016 at 15:24


Quote: Originally posted by aga  


$$ \sqrt { \frac {1}{2} } \sin{ \Big ( 90 - \frac {\pi}{4} \Big ) } + i \sqrt { \frac {1}{2} } \sin{ \Big ( \frac {\pi}{4} \Big ) } $$

The LaTeX is coming back, slowly ...


... is basically correct but slightly strangely expressed. But, don't mix (360) degrees wit radians. Instead of:

$$90-\frac{\pi}{4}\:\text{write}\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}$$
But that transformation wasn't even necessary:

$$c=r\cos\varphi+ri\sin\varphi=\sqrt{\frac12}\cos\frac{\pi}{4}+i\sqrt{\frac12}\sin\frac{\pi}{4}$$

And with:

$$\cos\frac{\pi}{4}=\sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}$$
$$\implies c=\frac12+\frac12 i$$

Quick backcheck:
$$r^2=x^2+y^2=\Big(\frac12\Big)^2+\Big(\frac12\Big)^2=\frac14+\frac14=\frac12$$

Thanks for playing!

Next installment tomorrow.



[Edited on 2-11-2016 by blogfast25]




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[*] posted on 2-11-2016 at 15:32


Mind: As boggled as a thing that can be boggled, but with extra boggle-ness.

Why was the expression 'strange' and where does 'c' some into it ?

Edit:

The notion of Radians did not enter my head, at all.

I recall that Raidans exist, but am not aware of what they are, apart from being an alternate way of expressing Degrees.

Gradians as well. Same story.

I would Google, but it's bed time, and Google does not always return the Best results.

[Edited on 2-11-2016 by aga]




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[*] posted on 2-11-2016 at 16:32


Quote: Originally posted by blogfast25  
Quote: Originally posted by j_sum1  
(I distinctly remember asking my maths teacher the value of 2i and it taking him a long time to figure it out.)



Even more "fun" are things like:

$$\sqrt [4]{-6}$$

But we're not going there. "Lite" here means addition, substraction, multiplication and division only! :)

[Edited on 2-11-2016 by blogfast25]

Division of complex numbers can be its own special kind of torture too. Fourth root of neg 6 is easy by comparison IMO. And yes, fun.




If you are interested, take a look at the latest offering from sum_lab:
A primer on metals and non-metals with at least one novel experiment.
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[*] posted on 2-11-2016 at 19:05


Quote: Originally posted by aga  

Why was the expression 'strange' and where does 'c' some into it ?

Edit:

The notion of Radians did not enter my head, at all.

I recall that Raidans exist, but am not aware of what they are, apart from being an alternate way of expressing Degrees.

Gradians as well. Same story.



c is just representation of a complex number.

Radians are NOT an "alternate way of expressing Degrees", rather it's the other way around. "Degrees" are a crappy non-math way of expressing angles: good for compasses and maps, rubbish for math.

In math a full circle is:

$$2\pi$$

So a quarter circle is a right angle or "90 degrees" or:

$$\frac{\pi}{2}$$
and half of that is "45 degrees" or:

$$\frac{\pi}{4}$$

Even "radians" is a misnomer: in the "2 pi system" angles are dimensionless numbers. They don't need a "name".

You invoked the famous trigonometric identity:

$$\sin\theta=\cos\big(\frac{\pi}{2}-\theta\big)$$

That was fine but not really needed.

"Gradiants"? I think you meant gradients, the basis of differential calculus.

Manana!




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[*] posted on 3-11-2016 at 14:33


Thanks for the explanations of Rads and Grads.

I really did not know what they meant.

Hasta mañana, Maestro.

(alt+164 for the n + tilde: changes it from 'n' to 'ny', hence 'manyana')





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[*] posted on 3-11-2016 at 16:59


Gradians are 400ths of a full circle.
They were an option on various calculators in the 1980s.

At the time I heard they were used by some engineers.

Your memory is not faulty here.
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[*] posted on 3-11-2016 at 17:07


Quote: Originally posted by Maroboduus  
Gradians are 400ths of a full circle.
They were an option on various calculators in the 1980s.

At the time I heard they were used by some engineers.

Your memory is not faulty here.


G*d yeah, I vaguely remember that now (not that I want to!) :D




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[*] posted on 4-11-2016 at 12:32
Complex numbers: complex conjugation


With c a complex number, c* is said to be its complex conjugate, if and only if:

$$c=a+bi \implies c^*=a-bi$$
$$c=a-bi \implies c^*=a+bi$$

The product of a complex number with its own complex conjugate is the modulus of the complex number (and thus a Real number):

$$c^*\times c=(a-bi)(a+bi)=a^2+abi-abi+(bi)^2=a^2-b^2i^2=a^2+b^2\in\mathrm{R}$$

If a number is Real, conjugation has no effect:

$$c\in\:\mathrm{R}\implies c=c^*\implies c^*c=c^2$$

Flash back/flash forward to quantum physics:

Higher up in this thread we saw that the probability distribution P(x) in QP is given by, with psi the wave function:
$$P(x)=\psi^*(x)\psi(x) \in\:\mathrm{R}$$

Probabilities are of course always Real numbers:

$$0\leq P \leq 1$$

If psi is a Real function:

$$\psi\in\mathrm{R}\implies \psi^*=\psi\implies P(x)=\psi(x)^2$$

The term:

$$\psi^*(x)\psi(x)$$

... is often referred to as the Norm.

Wave functions for which:

$$\int_{all\:\mathrm{ space}}\psi^*\psi dV=1$$

... are said to be Normalised.

Next up: complex numbers basic operations > vectors >...

[Edited on 5-11-2016 by blogfast25]




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[*] posted on 4-11-2016 at 14:21


To be sure, what does the sigma-like symbol mean ?
$$c\in\:\mathrm{R}$$
I guessed at 'in the realm of' but then it all went odd.




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[*] posted on 4-11-2016 at 15:25


Quote: Originally posted by aga  
To be sure, what does the sigma-like symbol mean ?
$$c\in\:\mathrm{R}$$
I guessed at 'in the realm of' but then it all went odd.


It technically means "in the set of" and R means "Real numbers" . Real numbers are just normal numbers that can be positive, negative and have as many decimal points as you like e.g. -1, 1.3, 3.14159...

So c∈R means c is a sensible, non-imaginary number. There are other letters you can use like N for "natural number", which is all the whole positive numbers and Z for integer, which can have negative numbers too.
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[*] posted on 4-11-2016 at 15:39


If even Numbers have limits, they cannot be a linear sequence, which, in turn, makes even Numbers meaningless

The status of Zero has always bothered me.

Ponderment required.




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[*] posted on 4-11-2016 at 16:44


Quote: Originally posted by aga  
If even Numbers have limits, they cannot be a linear sequence, which, in turn, makes even Numbers meaningless

The status of Zero has always bothered me.

Ponderment required.


Zero took a long time to get accepted: the other numbers didn't want to know it! :D But accepting it was a true revolution!

The first part of your ponderment is cryptic. Care to explain?




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[*] posted on 6-11-2016 at 14:51


Quote: Originally posted by blogfast25  
The first part of your ponderment is cryptic. Care to explain?

Sorry, i missed the question earlier.

It was some random drunken notion more than likely.

Something about Numbers spanning an infinity, yet allowing an infinity between each number.

0 to 1 (or -1) being an infinity, equal to, er, infinity, making 0 the only True number.

Every other number is infinitely distant from 0, and infinitely distant from the next number, making them all basically the same number.




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[*] posted on 6-11-2016 at 16:13


Quote: Originally posted by aga  
Quote: Originally posted by blogfast25  
The first part of your ponderment is cryptic. Care to explain?

Sorry, i missed the question earlier.

It was some random drunken notion more than likely.

Something about Numbers spanning an infinity, yet allowing an infinity between each number.

0 to 1 (or -1) being an infinity, equal to, er, infinity, making 0 the only True number.

Every other number is infinitely distant from 0, and infinitely distant from the next number, making them all basically the same number.

Don't contemplate infinity for too long. Especially when drunk. You know that Cantor went mad don't you? He was trying to resolve the very questions you have asked and eventually went nuts.

Not only are there an infinite number of numbers between any two given numbers (uncountably infinite actually), there are an infinite number of infinities; each one infinitely bigger than the one that preceded it.




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[*] posted on 7-11-2016 at 05:58
Complex numbers basic operations:


1. Addition and subtraction:

$$(a+bi)\pm(c+di)=(a\pm c)+(b\pm d)i$$

2. Multiplication:

$$(a+bi)\times(c+di)=ac+adi+bci+bdi^2=ac+(ad+bc)i-bd=(ac-bd)+(ad+bc)i$$

3. Division:

Division is carried out by multiplying denominator and numerator by:

$$\frac{c^*}{c^*}=1$$

$$\implies\frac{a+bi}{c+di}=\frac{a+bi}{c+di}\times\frac{c-di}{c-di}\tag{1}$$
$$=\frac{(a+bi)(c-di)}{c^2+d^2}=\frac{ac-adi+bci-bdi^2}{c^2+d^2}$$
$$=\frac{(ac+bd)+(-ad+bc)i}{c^2+d^2}=\frac{ac+bd}{c^2+d^2}+\frac{-ad+bc}{c^2+d^2}i$$

4. Exponential complex numbers:

Euler's formula:

$$e^{ai}=\cos a+i\sin a$$

Complex conjugation:

$$(e^{ai})^*=e^{-ai}$$

So:

$$e^{-ai}=\cos a-i\sin a$$

Exercise:

Using the 'trick' of (1), determine:

a)

$$\frac1c=\frac{1}{2-3i}$$

b)

$$\frac1i$$

Next up: vectors > functions as vectors > ...


[Edited on 7-11-2016 by blogfast25]




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[*] posted on 16-11-2016 at 11:43


$$\frac1c=\frac{1}{2-3i}$$
So
$$c = 2-3i$$
and
$$(2-3i) - (c - 0i) = 0$$
Let a = 2, b=3, c=c, d=0, then with rule (1)
$$(2-c) + (3+0)i = 0 = 2-c+3i$$
Which basically winds up back at the start !?!?

Confused.




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[*] posted on 17-11-2016 at 12:29


To be 100% honest, i do not understand the first queston, nor how (1) applies.



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