# Quantum 4

• Mar 6th 2008, 12:11 AM
DiscreteW
Quantum 4
Given that a particle is illustrated by $\displaystyle \psi(x,t) = Ce^{\frac{-x^2}{2}}e^{\frac{-iEt}{\bar{h}}},$

1.) Normalize the wave func.

2.) Find $\displaystyle <x>$ and $\displaystyle <\hat{p}>$

3.) Sketch $\displaystyle |\psi(x,t)|^2$ and explain what the plot illustrates.

------------------------

For 1.), we need this condition to be met:

$\displaystyle \int_{-\infty}^{\infty}|\psi(x,t)|^2 dx = 1$

So we have:

$\displaystyle \int_{-\infty}^{\infty} |\psi(x,t)|^2 dx = \int_{-\infty}^{\infty} |Ce^{\frac{-x^2}{2}}e^{\frac{-iEt}{\bar{h}}}|^2 dx = 1,$

$\displaystyle \displaystyle \implies C^2 \int_{-\infty}^{\infty} \left( {e^{-1/2\,{x}^{2}}} \right) ^{2} \left( {e^{{\frac {{\it iEt}} {h}}}} \right) ^{-2} dx = 1$

$\displaystyle \displaystyle \implies C^2\left({e^{-2\,{\frac {{\it iEt}}{h}}}}\sqrt {\pi }\right) = 1 \implies C = {\frac {1}{\sqrt {{e^{-2\,{\frac {{\it iEt}}{h}}}}\sqrt {\pi }}}}$

Not sure if I did this correctly. Used Maple to help me through it. I know we have a value for h that I can plug in, but what about the E and t?

For 2.),

$\displaystyle \displaystyle <x> = \int_{-\infty}^{\infty} |\psi(x,t)|^2 x dx \implies \int_{-\infty}^{\infty} |Ce^{\frac{-x^2}{2}}e^{\frac{-iEt}{\bar{h}}}|^2 x dx$

$\displaystyle \displaystyle \implies C^2\int_{-\infty}^{\infty} \left( {e^{-1/2\,{x}^{2}}} \right) ^{2} \left( {e^{{\frac {{\it iEt}}{h}}}} \right) ^{-2} x dx = C^2\int_{-\infty}^{\infty} \left( {e^{-1/2\,{x}^{2}}} \right) ^{2}x \left( {e^{{\frac {{\it iEt} }{h}}}} \right) ^{-2} dx = 0$

$\displaystyle C = 0$

Again, not sure if this is right. I had difficulty with finding $\displaystyle <\hat{p}>$.

I know:

$\displaystyle <\hat{p}> = \int_{-\infty}^{\infty} dp a(p)pa^{*}(p)$

$\displaystyle \psi(x) = \frac{1}{\sqrt{2\pi h}}\int_{-\infty}^{\infty} dp a(p)e^{\frac{ipx}{h}}$

$\displaystyle a(p) = \frac{1}{\sqrt{2\pi h}}\int_{-\infty}^{\infty} dx\psi(x)e^{\frac{-ipx}{h}}$

although what do I use?

Then, for 3.), that is graphing it, can't plot it with so many variables, as

$\displaystyle |\psi(x,t)|^2 = \left( {e^{-1/2\,{x}^{2}}} \right) ^{2} \left( {e^{-{\frac {{\it iEt}}{h}}}} \right) ^{2}$.

Thanks topsquark (or whoever else knows this stuff).
• Mar 6th 2008, 01:52 AM
sandstorm
Hello,

For number 1, when you square a wave function, you don't just multiply it with itself, rather you multiply it by its complex conjugate.

That'll allow you to solve numbers two and three aswell.
• Mar 7th 2008, 02:34 PM
DiscreteW
Quote:

Originally Posted by sandstorm
Hello,

For number 1, when you square a wave function, you don't just multiply it with itself, rather you multiply it by its complex conjugate.

That'll allow you to solve numbers two and three aswell.

Ugh, thanks. I'll correct it.
• Mar 7th 2008, 02:59 PM
DiscreteW
Okay, how is this:

We have: $\displaystyle \int_{-\infty}^{\infty} |\psi(x,t)|^2 dx = \int_{-\infty}^{\infty} |Ce^{\frac{-x^2}{2}}e^{\frac{-iEt}{\bar{h}}}|^2 dx = 1,$

$\displaystyle \int_{-\infty}^{\infty}\left[Ce^{\frac{-x^2}{2}}e^{\frac{-iEt}{\bar{h}}}\right]\left[Ce^{\frac{-x^2}{2}}e^{\frac{iEt}{\bar{h}}}\right]\, dx = 1$

$\displaystyle \int_{-\infty}^{\infty} {C}^{2} \left( {e^{-1/2\,{x}^{2}}} \right) ^{2}{e^{-{\frac {{\it iEt}}{h}}}}{e^{{\frac {{\it iEt}}{h}}}}\, dx = 1$

$\displaystyle \int_{-\infty}^{\infty} {C}^{2} \left( {e^{-1/2\,{x}^{2}}} \right) ^{2}\, dx = 1$

$\displaystyle \int_{-\infty}^{\infty} {C}^{2}e^{-x^2} \, dx = 1$

$\displaystyle c^2\sqrt{\pi} = 1$

$\displaystyle \implies c = \pm \frac{1}{\pi^{\frac{1}{4}}}$

Yay? Nay? I was fortunate that the t dropped out, or I wouldn't know how to solve it. I'm going to hope the same happens when I solve for <x> and <p>.

EDIT: Okay, solved for <x>, although not sure if it's right.

$\displaystyle \displaystyle <x> = \int_{-\infty}^{\infty} |\psi(x,t)|^2 x dx \implies \int_{-\infty}^{\infty} |Ce^{\frac{-x^2}{2}}e^{\frac{-iEt}{\bar{h}}}|^2 x dx$

$\displaystyle \int_{-\infty}^{\infty} {C}^{2}e^{-x^2}x \, dx$

$\displaystyle = 0$

Can anyone confirm if this is right?

I need help with the set-up for <p> please!

Are we meant to use the C I found in the first part for <x> and <p>? How should I attack the sketching part? I have the problem with there being too many variables.
• Mar 7th 2008, 05:44 PM
topsquark
Quote:

Originally Posted by DiscreteW
Okay, how is this:

We have: $\displaystyle \int_{-\infty}^{\infty} |\psi(x,t)|^2 dx = \int_{-\infty}^{\infty} |Ce^{\frac{-x^2}{2}}e^{\frac{-iEt}{\bar{h}}}|^2 dx = 1,$

$\displaystyle \int_{-\infty}^{\infty}\left[Ce^{\frac{-x^2}{2}}e^{\frac{-iEt}{\bar{h}}}\right]\left[Ce^{\frac{-x^2}{2}}e^{\frac{iEt}{\bar{h}}}\right]\, dx = 1$

$\displaystyle \int_{-\infty}^{\infty} {C}^{2} \left( {e^{-1/2\,{x}^{2}}} \right) ^{2}{e^{-{\frac {{\it iEt}}{h}}}}{e^{{\frac {{\it iEt}}{h}}}}\, dx = 1$

$\displaystyle \int_{-\infty}^{\infty} {C}^{2} \left( {e^{-1/2\,{x}^{2}}} \right) ^{2}\, dx = 1$

$\displaystyle \int_{-\infty}^{\infty} {C}^{2}e^{-x^2} \, dx = 1$

$\displaystyle c^2\sqrt{\pi} = 1$

$\displaystyle \implies c = \pm \frac{1}{\pi^{\frac{1}{4}}}$

Yay? Nay? I was fortunate that the t dropped out, or I wouldn't know how to solve it. I'm going to hope the same happens when I solve for <x> and <p>.

EDIT: Okay, solved for <x>, although not sure if it's right.

$\displaystyle \displaystyle <x> = \int_{-\infty}^{\infty} |\psi(x,t)|^2 x dx \implies \int_{-\infty}^{\infty} |Ce^{\frac{-x^2}{2}}e^{\frac{-iEt}{\bar{h}}}|^2 x dx$

$\displaystyle \int_{-\infty}^{\infty} {C}^{2}e^{-x^2}x \, dx$

$\displaystyle = 0$

Can anyone confirm if this is right?

I need help with the set-up for <p> please!

Are we meant to use the C I found in the first part for <x> and <p>? How should I attack the sketching part? I have the problem with there being too many variables.

This is correct.

Recall that the operator p is, in the position representation
$\displaystyle \hat{p} = -i \hbar \frac{d}{dx}$

So
$\displaystyle <p> = \int_{-\infty}^{\infty} \psi^* \left ( -i \hbar \frac{d}{dx} \right ) \psi ~dx$
Please note the position of the operator in the expression. Technically you got it wrong in the <x> calculation. However, since the $\displaystyle \hat{x}$ operator is simply "x" there was no problem with the order of the terms. The <p> calculation demonstrates that you do need to be careful of this in general.

For an arbitrary operator $\displaystyle \hat{A}$,
$\displaystyle <A> = \int_{-\infty}^{\infty} \psi^* \hat{A} \psi ~dx$

Also, once you have found a normalization constant you keep it. That is, after all, the meaning of "constant." :)

-Dan
• Mar 7th 2008, 06:39 PM
DiscreteW
Ah, yes! I foolishly forgot that.

I worked out <p> with your help:

$\displaystyle <\hat{p}> = \int_{-\infty}^{\infty} \psi^*\left ( -i \hbar \frac{d}{dx} \right ) \psi ~dx$

Thus, we have:

$\displaystyle \int_{-\infty}^{\infty}\left[Ce^{\frac{-x^2}{2}}e^{\frac{iEt}{\bar{h}}}\right]\left [ -i \hbar \frac{d}{dx} \right ] \left[Ce^{\frac{-x^2}{2}}e^{\frac{-iEt}{\bar{h}}}\right] dx$

Hence, we have:

$\displaystyle \int_{-\infty}^{\infty}\left[Ce^{\frac{-x^2}{2}}e^{\frac{iEt}{\bar{h}}}\right]\left [ -i \hbar\right] \left[ -Cx{e^{\frac{-x^2}{2}}}{e^{{\frac {-iEt}{h}}}}\right]~dx$

$\displaystyle \int_{-\infty}^{\infty} C^2\,i\hbar xe^{-x^2}~dx$

$\displaystyle \frac{-1}{2}\,i{C}^{2}h{e^{-{x}^{2}}}$

So for this we don't get a specific value it seems. How does that look?
• Mar 7th 2008, 06:45 PM
topsquark
Quote:

Originally Posted by DiscreteW
Ah, yes! I foolishly forgot that.

I worked out <p> with your help:

$\displaystyle <\hat{p}> = \int_{-\infty}^{\infty} \psi^*\left ( -i \hbar \frac{d}{dx} \right ) \psi ~dx$

Thus, we have:

$\displaystyle \int_{-\infty}^{\infty}\left[Ce^{\frac{-x^2}{2}}e^{\frac{iEt}{\bar{h}}}\right]\left [ -i \hbar \frac{d}{dx} \right ] \left[Ce^{\frac{-x^2}{2}}e^{\frac{-iEt}{\bar{h}}}\right] dx$

Hence, we have:

$\displaystyle \int_{-\infty}^{\infty}\left[Ce^{\frac{-x^2}{2}}e^{\frac{iEt}{\bar{h}}}\right]\left [ -i \hbar\right] \left[ -Cx{e^{\frac{-x^2}{2}}}{e^{{\frac {-iEt}{h}}}}\right]~dx$

$\displaystyle \int_{-\infty}^{\infty} C^2\,i\hbar xe^{-x^2}~dx$

Note two things:
1) <p> must be a real number. As this integral winds up with a complex coefficient, it had better be 0.

2) The integral you derived contains an odd integrand over even limits. Thus the integral has a value of 0. You didn't evaluate your result over the endpoints. :)

-Dan
• Mar 7th 2008, 06:50 PM
DiscreteW
Quote:

Originally Posted by topsquark
Note two things:
1) <p> must be a real number. As this integral winds up with a complex coefficient, it had better be 0.

2) The integral you derived contains an odd integrand over even limits. Thus the integral has a value of 0. You didn't evaluate your result over the endpoints. :)

-Dan

Hah, yup! Duh. I didn't. Sorry for wasting your time! I should have caught that. And yes, it is 0! Thanks again.
• Mar 7th 2008, 06:58 PM
topsquark
Quote:

Originally Posted by DiscreteW
Hah, yup! Duh. I didn't. Sorry for wasting your time! I should have caught that. And yes, it is 0! Thanks again.

You aren't wasting my time. Your questions are similar to what I had to learn at one point. I'm just happy to be able to help. :)

-Dan