Quantum oscillator algebra help

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  • #1
Vol
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Hi. I am working on the quantum harmonic oscillator Schrodinger's equation and need help with the algebra or whatever it is I am missing. Here are the 2 steps I can't understand:

d^2psi/dx^2 + (2mE/h^2 - m^2w^2/h^2 * x^2)psi = 0

You substitute this into it:

y = sqrt(mw/h)*x

I don't understand how you get

d^2psi(y)/dy^2 + (2E/hw - y^2)psi(y) = 0

I get a mw/h in front of the 2nd term I can't get rid of. If I multiply the whole thing with h/mw then I get a h/mw in front of the first term I can't get rid of. What am I doing wrong? Is it my bad algebra?
 

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  • #3
Vol
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Is there a way I can preview how it looks before I post it? I clicked preview but it is still in code form.
 
  • #4
Vol
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d^2

\frac d^2{x}

testing Latex
 
  • #5
Vol
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Why doesn't Latex work?
Help.
 
  • #6
PeterDonis
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Why doesn't Latex work?
Because you have to enclose it in either double dollar signs (if it's all in its own separate paragraph/equation, no text) or double hash symbols (if it's in line with text).

Here is how the latter works: ##d^2##. There are double hash marks on either side of the d^2 there.

Here is how the former works:

$$
\frac{d^2 x}{dt^2}
$$

There are double dollar signs around the fraction there.
 
  • #7
PeterDonis
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you have to enclose it in either double dollar signs (if it's all in its own separate paragraph/equation, no text) or double hash symbols (if it's in line with text).
The section in the help article on "Delimiting your LaTeX code" explains this in more detail.
 
  • #8
stevendaryl
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Hi. I am working on the quantum harmonic oscillator Schrodinger's equation and need help with the algebra or whatever it is I am missing. Here are the 2 steps I can't understand:

d^2psi/dx^2 + (2mE/h^2 - m^2w^2/h^2 * x^2)psi = 0

You substitute this into it:

y = sqrt(mw/h)*x
So you are replacing ##x## by ##y \sqrt{\frac{\hbar}{m\omega}}## (isn't that prettier?). But you also have to replace ##\frac{d}{dx}## by ##\sqrt{\frac{m\omega}{\hbar}} \frac{d}{dy}##. Did you do both replacements?
 
  • #9
Vol
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Why do we have to replace ##\frac d {dx}## by ##\sqrt \frac {mw} h## ##\frac d {dy}## ? I can see we are changing from dx to dy because we are substituting in ##y\sqrt\frac h {mw}## for x? But where does ##\sqrt \frac {mw} h## come from? Is it the chain rule?
 
  • #10
stevendaryl
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Why do we have to replace ##\frac d {dx}## by ##\sqrt \frac {mw} h## ##\frac d {dy}## ? I can see we are changing from dx to dy because we are substituting in ##y\sqrt\frac h {mw}## for x? But where does ##\sqrt \frac {mw} h## come from? Is it the chain rule?
Yes. By the chain rule, ##\frac{d\psi}{dx} = \frac{d\psi}{dy} \frac{dy}{dx}##

If ##x = \sqrt{\frac{\hbar}{m\omega}} y##, then ##y = \sqrt{\frac{m\omega}{\hbar}} x##. So ##\frac{dy}{dx} = \sqrt{\frac{m\omega}{\hbar}}##

So ##\frac{d\psi}{dx} = \sqrt{\frac{m\omega}{\hbar}} \frac{d\psi}{dy}##
 
  • #11
Vol
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Yes. By the chain rule, ##\frac{d\psi}{dx} = \frac{d\psi}{dy} \frac{
Yes. By the chain rule, ##\frac{d\psi}{dx} = \frac{d\psi}{dy} \frac{dy}{dx}##

If ##x = \sqrt{\frac{\hbar}{m\omega}} y##, then ##y = \sqrt{\frac{m\omega}{\hbar}} x##. So ##\frac{dy}{dx} = \sqrt{\frac{m\omega}{\hbar}}##

So ##\frac{d\psi}{dx} = \sqrt{\frac{m\omega}{\hbar}} \frac{d\psi}{dy}##
Thanks. I get it now. Then you cancel out all the mw/h with h/mw. I will be back with questions about the next step for a series solution.
 
  • #12
Vol
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OK, so I get:
##\frac {d^2\Psi(y)} {dy^2} + (\frac {2E} {hw} - y^2)\Psi(y) = 0##
Then,
##\Psi(y) = u(y)e^{\frac {-y^2}{2}}## is the general solution.
Totally lost. Where did ##\frac{-y^2}2## come from?
 
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  • #13
Vol
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Anybody? Am I missing something altogether?
 
  • #14
PeterDonis
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Then,
##\Psi(y) = u(y)e^{\frac {-y^2}{2}}## is the general solution.
How did you find this?

Totally lost. Where did ##\frac{-y^2}{2}## come from?
Have you tried plugging the general solution into the differential equation?
 
  • #15
Vol
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That is exactly my point. Looking at the given equation and seeing that ##\frac {2E} {hw}## is negligible compared to ##y^2## you are supposed to guess the general solution... that ##\Psi(y)## will be as ##e^{\frac {-y^2} 2}##. Totally lost. Here is where I found this problem:

physics.gmu.edu/~dmaria/590%20Web%20Page/public_html/qm_topics/harmonic/
 
  • #16
PeterDonis
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Looking at the given equation and seeing that ##\frac {2E} {hw}## is negligible compared to ##y^2## you are supposed to guess the general solution... that ##\Psi(y)## will be as ##e^{\frac {-y^2} 2}##.
Once again: have you tried plugging ##\Psi(y) = e^{\frac{-y^2}{2}}## into the differential equation? (Hint: to get the equation to work you will not be able to ignore the ##2E / h w## term.)
 
  • #17
PeterDonis
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Here is where I found this problem:

physics.gmu.edu/~dmaria/590%20Web%20Page/public_html/qm_topics/harmonic/
This reference might not be entirely reliable. It looks like someone's personal resource, not a textbook or peer-reviewed paper.
 
  • #18
PeterDonis
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have you tried plugging ##\Psi(y) = e^{\frac{-y^2}{2}}## into the differential equation? (Hint: to get the equation to work you will not be able to ignore the ##2E / h w## term.)
Actually, I mis-phrased this. I should have said that you will not be able to get ##\Psi(y) = e^{\frac{-y^2}{2}}## to be a solution of the differential equation without the ##2E / h w## term; i.e, it is not a solution of ##d^2 \Psi / dy^2 - y^2 \Psi = 0##. So I am dubious about the article's claim that you can "guess" ##\Psi(y) = e^{\frac{-y^2}{2}}## by ignoring the ##2E / h w## term for large ##y##. (Note that in your statement of what the article said, you left out the "for large ##y##" qualifier.)
 
  • #19
Vol
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Actually, I mis-phrased this. I should have said that you will not be able to get ##\Psi(y) = e^{\frac{-y^2}{2}}## to be a solution of the differential equation without the ##2E / h w## term; i.e, it is not a solution of ##d^2 \Psi / dy^2 - y^2 \Psi = 0##. So I am dubious about the article's claim that you can "guess" ##\Psi(y) = e^{\frac{-y^2}{2}}## by ignoring the ##2E / h w## term for large ##y##. (Note that in your statement of what the article said, you left out the "for large ##y##" qualifier.)
Yikes. But it is from a university site. Now I am really confused. But where did ##e^{\frac{-y^2}{2}}## come from? How did he come up with that guess?
 
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  • #20
stevendaryl
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Yikes. But it is from a university site. Now I am really confused. But where did ##e^{\frac{-y^2}{2}}## come from?
It's just a guess. If you let ##\Psi_0(y) = e^{-\frac{y^2}{2}}##, then

##\frac{d\Psi_0}{dy} = -y \Psi_0##
##\frac{d^2 \Psi_0}{dy^2} = (y^2 - 1) \Psi_0##

So ##\frac{d^2 \Psi_0}{dy^2} + (\frac{2E}{\hbar \omega} - y^2) \Psi_0 = (\frac{2E}{\hbar \omega} - 1) \Psi_0##

So ##\Psi_0## solves the equation when ##E = \frac{1}{2} \hbar \omega##.
 
  • #21
PeterDonis
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it is from a university site
It's someone's personal page provided by their university. Again, it's not a textbook or peer-reviewed paper.
 
  • #22
Vol
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Oh, I see. He was just thinking backwards. Thanks.
 
  • #23
Vol
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Oh geesh! I thought it was by a mathematician. Well, I now know better.
 
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