# Quantum oscillator algebra help

• B
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?

Is there a way I can preview how it looks before I post it? I clicked preview but it is still in code form.

d^2

\frac d^2{x}

testing Latex

Why doesn't Latex work?
Help.

PeterDonis
Mentor
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.

PeterDonis
Mentor
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.

stevendaryl
Staff Emeritus
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?

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?

stevendaryl
Staff Emeritus
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}##

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.

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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Anybody? Am I missing something altogether?

PeterDonis
Mentor
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?

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/

PeterDonis
Mentor
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.)

PeterDonis
Mentor
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.

PeterDonis
Mentor
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.)

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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stevendaryl
Staff Emeritus
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##.

PeterDonis
Mentor
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.

Oh, I see. He was just thinking backwards. Thanks.

Oh geesh! I thought it was by a mathematician. Well, I now know better.