B Quantum oscillator algebra help

[Vol]

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?

 

[PeterDonis]

Hi @Vol and welcome to PF!

Your posts will be more readable if you use the PF LaTeX feature. You can find help on it here:

https://www.physicsforums.com/help/latexhelp/

 

[Vol]

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

 

[Vol]

d^2

\frac d^2{x}

testing Latex

 

[Vol]

Why doesn't Latex work?
Help.

 

[PeterDonis]

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]

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]

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?

 

[Vol]

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]

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 $\frac{d\psi}{dx} = \sqrt{\frac{m\omega}{\hbar}} \frac{d\psi}{dy}$

 

[Vol]

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.

 

[Vol]

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?

 

[Vol]

Anybody? Am I missing something altogether?

 

[PeterDonis]

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?

 

[Vol]

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]

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]

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]

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.)

 

[Vol]

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?

 

[stevendaryl]

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]

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.

 

[Vol]

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

 

[Vol]

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