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j_sum1
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I have the cat skin. I still don't know if the cat is alive or dead though. 
On the topic of thread contributions, I think it is much better in this case to keep the signal to noise ratio high. And with that, I will shut up
for a while.
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blogfast25
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As long as you're not looking at her, she's both dead and alive! 
[Edited on 22-7-2015 by blogfast25]
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annaandherdad
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Quote: Originally posted by Cheddite Cheese  | I have taken courses in linear algebra and vector calculus (but not Q.M.) so I know basically what the equation means mathematically. My question is:
what does the constant factor (-Hbar^2/2m) physically mean?
[Edited on 15-7-2015 by Cheddite Cheese] |
This constant factor is multiplied times the differential operator d^2/dx^2, that is, the term in the Schro"odinger equation is -(hbar^2/2m) x (d^2
psi/dx^2). This is regarded as (1/2m) times the square of the momentum operator p, where p is the differential operator -i hbar d/dx, that is, the
term in the Schro"dinger equation is
-(hbar^2/2m) x (d^2 psi/dx^2) = (hbar/2m) x (-i hbar d/dx)^2 psi
= (p^2/2m) psi
(This is easier to understand if you write it out on the blackboard).
In classical mechanics, things you observe (energy, momentum, angular momentum, position of planet x) are just numbers, which may be functions of time
in a particular system. In quantum mechanics, when you observe something (like the position or energy of an electron) you get a number, just like in
classical mechanics, but unlike classical mechanics, the number you get isn't necessarily the same if you make the identical measurement on
identically prepared systems. That is, there is a statistical distribution of the answers.
To calculate the statistical distribution of answers you need the linear operator associated with the observable. Every observable corresponds to a
definite linear operator. The linear operator acts on wave functions psi and maps them into new wave functions. As I mentioned in the last
paragraph, the linear operator corresponding to momentum is -i hbar d/dx.
How do we know this is the momentum operator? It was a guess of deBroglie. Einstein had been saying that a photon of energy E and momentum p was
associated with a wave of frequency omega and wavenumber k, where E=hbar omega and p =hbar k. (omega and k are measured in radians; omega is
radians/sec and k is radians/cm or radians per meter if you prefer). So deBroglie said, maybe this applies to massive particles as well. If so, a
massive particle (like an electron) of energy E and momentum p must correspond to a wave of frequency omega=E/hbar and wave number k=p/hbar. That
is, the wave must be
psi(x,t) = exp[i(kx -omega t)] = exp[i(px - Et)/hbar]
Now, deBroglie said, if we let the operator -i hbar d/dx act on this wave, we get
(-i hbar d/dx) psi(x,t) = p psi(x,t),
that is, the operator acting on the wave is the momentum times the wave. Historically this was how the momentum got associated with this particular
differential operator. By the same logic, the energy is associated with the operatior (i hbar d/dt). (These are really partial derivatives, since
psi depends on both x and t.)
To go back, the operator p^2/2m, which appears in one term in the Schro"dinger equation, corresponds to the classical number p^2/2m (where now p is a
value, not an operator), which is otherwise mv^2/2 because p=mv, which is otherwise the kinetic energy. So, to answer your question, what this factor
means physically, it means that this term in the Schro"dinger equation is the kinetic energy term. In fact, the Schro"dinger equation is an
operator encoding of the classical energy relation,
p^2/2m + V(x) = E
Any other SF Bay chemists?
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blogfast25
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Thanks AAHD:
I was trying to avoid Quantum Operators because they're hardly simple, for those with minimum math knowledge.
I might dedicate a small segment to them at the end of Part 1.
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aga
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| Quote: | | psi(x,t) = exp[i(kx -omega t)] = exp[i(px - Et)/hbar] |
Phew. Head. Over. Whoosh !
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blogfast25
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It looks a lot worse without proper scientific notation.
Gimme five.
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aga
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High Five Bro (?)
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blogfast25
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psi(x,t) = exp[i(kx -omega t)] = exp[i(px - Et)/hbar]
Ψ(x,t) = e<sup>i(kx – ωt)</sup> = e<sup>i(px – Et)/ћ</sup>
It’s the Time Dependent Schrodinger equation for a free particle:
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/scheq.htm...
But in AAHD's formula the factor A (amplitude) is missing...
Neat, huh? [cough!]
[Edited on 22-7-2015 by blogfast25]
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aga
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Oh ! I see now !
[/lie mode off]
Hokay.
No. Just No.
I am very pleased that other people understand this very well, and can have fun and games with it.
Personally, i do not follow the maths at all, and i assumed the idea of this thread was to Teach rather than to be a discussion group for advanced
theoretical mathematicians.
Yes, i am a drunken dullard because i do not grasp the idea even though i have been shown pitchforks, and should be left in the gutter where i belong.
If some Good Smaritan could summon some extra goodness, perhaps they could explain in simple terms what the Psi function is attempting to model, and
how on earth you get from kx - ... to wx - ...
For starters, Schroedinger's equation for a time dependant particle in a matchbox is attempting to describe what exactly, how, and what does each term
actually mean ?
[Edited on 22-7-2015 by aga]
[Edited on 22-7-2015 by aga]
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blogfast25
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Or with Euler:
Ψ(x,t) = Ae<sup>i(kx – ωt)</sup> = A cos(kx – ωt) + Ai sin(kx – ωt)
Much better, huh? 
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blogfast25
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| Quote: Originally posted by aga |
Personally, i do not follow the maths at all, and i assumed the idea of this thread was to Teach rather than to be a discussion group for advanced
theoretical mathematicians.
|
That's a tad unfair. I'm keeping it as math lite as possible. Hard to do with QP.  
As I wrote above: I want to avoid the TDSE as much as possible. Forget about this interlude. 1,2,3... click and you're back in the room!
[Edited on 22-7-2015 by blogfast25]
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aga
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To re-iodiotate,
Please Explain.
Psi function of x and t =
unexplained A mutiplied by presumably Euler's number
to the power of the square root of -1 multiplied by some random k minus some small buttocks multipled by t.
Obviously the following error or failure to understand is laughable !
fcs = (uint32_t *) &buf[len];
*fcs = swaporder(crc32(len-RTAP_SIZE, &buf[RTAP_SIZE]));
Belly laughs all the way when the Fool did not realise he should do a simple cast of the buf pointer to a const * uchar_8 ! How could he NOT know
!?!?! Gfaw Gfaw.
Oh the joy that the partial mastery of a single paradigm must bring.
Now, where were we with a simpler explantion of the actual terms being used in that equation, and how they were arrived at ?
Edit:
said pitchfork(x,y) instead of (x,t)
[Edited on 22-7-2015 by aga]
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blogfast25
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OK, G-ddammit, that's it, detention for you:
Copy this text FIVE times, in LONGHAND:
https://en.wikipedia.org/wiki/Free_particle#Non-Relativistic...
Oral examination on the material tomorrow! 
[Edited on 22-7-2015 by blogfast25]
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annaandherdad
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| Quote: Originally posted by aga |
Personally, i do not follow the maths at all, and i assumed the idea of this thread was to Teach rather than to be a discussion group for advanced
theoretical mathematicians.
|
Sorry, aga, I was answering CC's question. I think different people here have different backgrounds. The formula, which bf25 made look much nicer,
is the formula for a wave in complex notation. It's pretty standard in the theory of waves. deBroglie used it to connect the properties of a wave
(its frequency and wave length) to the properties of the associated particle (its energy and momentum). I'll explain it if you want.
It was deBroglie's dissertation. Later Debye (a noted physicist at the time) commented that if there was a wave there had to be a wave equation.
This motivated Schro"dinger to find his equation. It wasn't straightforward, because he tried to find the relativistic version of the wave equation
first, and that's much more complicated. He gave up for a while, but later returned to the problem, this time looking for the non-relativistic
version, which is the version usually known today as "the" Schro"dinger equation. The wave I wrote down is a solution for the case of a free
particle.
Any other SF Bay chemists?
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aga
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Sorry for the outburst.
One was simply trying to make clear how little can be understood when there's a lot of Running before Walking.
Explaining would be very helpful, but please try to use simple terms, such as k is a constant you'll just have to accept, x is
horizontal position in space etc.
As a noob, i do not immediately recognise Qdangle as anything at all.
@Blogfast25
Morning Orals have been banned since 2002 following the Jenkins vs Blogfast24 affair if you may recall, Sir25.
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blogfast25
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aga:
The physical meaning of kx – ωt you should be able to grasp from the wiki link provided, including k and ω.
This should also help:
https://en.wikipedia.org/wiki/Matter_wave#de_Broglie_relatio...
And maybe this too:
https://en.wikipedia.org/wiki/Wave#Phase_velocity_and_group_...
I do seriously suggest to leave all this aside for now. In chemistry (and that IS Part 2 of this little resume, if you recall) the
TDSE is much less important.
In the example of a quantum system featured so far (particle in 1D box) time plays no part: the (time independent) SE explores the 'steady state'
possible states of the system: E<sub>1</sub>, E<sub>2</sub>, E<sub>3</sub>, etc (and the associated wave functions
Ψ<sub>n</sub>(x)) . We're not considering any CHANGES or how these occur. Change inevitably involves time (t) and the solving of the
TDSE. Considering the added complexity of making the system dynamic (changeable in time) it's counter-productive here (but vital for a FULL
QP description of a system).
Hope you understand.
[Edited on 23-7-2015 by blogfast25]
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aga
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I think i was missing what the Result means.
Please stamp on any errors here :-
Ψ(x,t) computes a positive real number indicating the probability of a particle existing at position x at time t, given that it's Amplitude and
frequency are known.
Ψ(x,t) = Ae<sup>i(kx – ωt)</sup> or much better
Ψ(x,t) = A cos(kx – ωt) + Ai sin(kx – ωt)
where :-
x = horizontal position
t = time
A = amplitude
ω = angular frequency (how fast the particle is spinning) given by 2πf where f = frequency in Hz
k = the 'wave number' = 2π/λ.
λ = c/f where c = the speed of light.
This was helpful :-
http://www.conspiracyoflight.com/Schrodinger/Schrodinger.htm...
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blogfast25
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Ooopsie, not sure where this repetition came from.
[Edited on 23-7-2015 by blogfast25]
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blogfast25
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aga:
One error: λ = c/f
That is true for electromagnetic waves, like light. But 'electron waves' are de Broglie waves. For matter waves (de Broglie waves) in general:
λ = h/p with p = mv
Now go back a bit to this:
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/pbox.html...
In the box, the matter waves are 'standing waves'. What are the wavelengths λ?
Well, you can literally see this from the schematic:
for n = 1, λ = L/2
for n = 2, λ = L
for n = 3, λ = 3L/2
for n = 4, λ = 2L
for n, λ = nL/2
That was a good link, aga. I can see from that link I will be able to skip the treatment of the hydrogen atom altogether!
I know you're finding all of this hard but when I had Erwin Schrodinger and Werner Heisenberg in my class all those years ago, they found it hard
too. Not to mention the endless food fights between them!
[Edited on 23-7-2015 by blogfast25]
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aga
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Phew !
For absolute clarity :-
h being Plancks' constant right ?
and Ψ(x,t) does calculate a real number indicating the probability of finding the widget at x, t ?
Sorry if this is like wading through treacle for you, just that if i don't actually understand it, i just have to ask questions, or i never will
understand it and be utterly lost.
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blogfast25
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| Quote: Originally posted by aga |
Phew !
For absolute clarity :-
h being Plancks' constant right ?
and Ψ(x,t) does calculate a real number indicating the probability of finding the widget at x, t ?
Sorry if this is like wading through treacle for you, just that if i don't actually understand it, i just have to ask questions, or i never will
understand it and be utterly lost.
|
h is Planck's constant, indeedy.
The probability of finding a particle at x,t is given by: P(x,t) = Ψ(x,t)Ψ<sup>*</sup>(x,t)
With Ψ<sup>*</sup>(x,t) the Complex Conjugate of Ψ(x,t).
For a free moving particle Ψ(x,t) is a complex function but as we've seen above for any complex number a:
a<sup>*</sup>a is always a Real Number. P(x,t) = Ψ(x,t)Ψ<sup>*</sup>(x,t) always returns a Real Number. Probabilities are
of course real numbers (0 <= P <= 1).
Applied to a free moving particle:
Ψ(x,t) = Ae<sup>i(kx – ωt)</sup> = A cos(kx – ωt) + Ai sin(kx – ωt)
Ψ(x,t) Ψ<sup>*</sup>(x,t) = A<sup>2</sup>
Another thing that might help you is this: simple sinusoidal waves:
https://en.wikipedia.org/wiki/Wave#Sinusoidal_waves
In essence a wave is a harmonic oscillation, travelling through space (x) at constant speed.
Take a tuning fork, for instance. Hit the fork and it starts oscillating at the pitch (frequency) of its note. The oscillating fork 'pushes' against
the elastic air and a sound wave travels with that pitch, at the speed of sound.
[Edited on 23-7-2015 by blogfast25]
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aga
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Ah Right. I missed the i hiding next to the A before the sin, so Ψ(x,t) isn't a real number.
OK. Thanks. By jove i got it !
[Edited on 23-7-2015 by aga]
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blogfast25
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You'll be relieved that the next instalment is lighter. [cough! Must take some Benyllin]
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Darkstar
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| Quote: Originally posted by aga | | Yes, i am a drunken dullard because i do not grasp the idea even though i have been shown pitchforks, and should be left in the gutter where i belong.
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The fact that you're even trying to in the first place already puts you light years ahead of the general population. No one said these concepts were
simple or easy to understand. Don't be so hard on yourself.
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blogfast25
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Certificates of attendance will be handed out at the end of the course. Unless I lose the will to live before it!
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