Calculating <x> and <p2> for Wavefunctions

[knut-o]

Homework Statement

1. I got the wavefunctions:$$\psi _0=(\frac{m\omega }{\pi\hbar})^{\frac{1}{4}}\cdot e^{-\frac{m\omega}{2\hbar}\cdot x^2}$$,
and $$\psi _1=(\frac{m\omega }{\pi\hbar})^{\frac{1}{4}}\cdot \sqrt{\frac{2m\omega }{\hbar}}\cdot e^{-\frac{m\omega}{2\hbar}\cdot x^2}$$.
Also recomended by the task given to introduce two new variables:
$$\xi =\sqrt{\frac{m\omega }{\hbar}}\cdot x\\ \alpha=(\frac{m\omega}{\pi\hbar})^{\frac{1}{4}}$$
Now, I am tolk to find <x>, <p>, <x²> and <p²>.

Homework Equations

I am informed how to find <x> and <p>:
$$<x>=\int _{-\inf} ^{\inf} x|\psi(x,t)|^2dx=\inf\psi*(x)\cdot \psi dx$$
$$<p>=-i\hbar \int \psi* \cdot\frac{\partial\psi}{\partial x}dx$$
I also wonder what the * stands for, it's not a normal multiplication-sign is it?

The Attempt at a Solution

What I am mostly curious about, is how do I find <x²> and <p²>?
I have also found:
$$\psi _0=\alpha\cdot e^{-\frac{\xi ^2}{2}}\\ \psi _1=\alpha\cdot\xi\cdot e^{-\frac{\xi ^2}{2}}$$.
Do I, when I calculate
$$<x>=\int _{-\inf} ^{\inf} x|\psi(x,t)|^2dx=\inf\psi*(x)\cdot \psi dx$$ get insterted for x $$x=\xi \cdot\sqrt{\frac{\hbar}{m\omega}}$$ and $$\frac{d\xi}{dx}=\sqrt{\frac{m\omega}{\hbar}}\Rightarrow dx=d\xi\cdot\sqrt{\frac{\hbar}{m\omega}$$? Giving even more variables to work with in thei ntegratian/calculation?

And to find <x^²>, do I simply just square the function standing inside there, giving me [/tex]|\psi |^4[/tex] and the function I calculate for <p> and just square it?

I am so not getting this thing..

 

[kuruman]

The * stands for "complex conjugate". Your wavefunctions are both real, so in this case you can just ignore it.

For the integrals, start from

$$<x^2>=\int^{\infty}_{-\infty}\psi(x)x^2\psi(x)dx$$

$$<p^2>=\int^{\infty}_{-\infty}\psi(x) \left( -\hbar^2\frac{\partial^2 \psi(x)}{\partial x^2}\right)dx$$

It will probably be easier for you to assemble the integrals in terms of the given functions using x as your variable, then replace x with ξ before you integrate.

 

[knut-o]

Oh, no wonder I had some severe mistakes in my calculations. I calculated
$$<x^2>=\int x\psi (x,t)^4dx\\ <p^2>=\int (\psi\cdot \frac{\partial\psi}{\partial x})^2dx$$

But let us do some calculations, or rather. I got osme answers that I don't know if are correct or not. For <x²> I got $$\frac{\hbar}{m\omega}$$ without any clue knowing wether or not it's correct. I also get 0 for both <x> and <p>, which I like, but don't see any physical understanding in. Cause after I find <p²> and <x²>, I am told to check them up to the uncertainty principle, so I am guessing that I am going to end up with some of them to cancel each other out.
But I am kind of stuck at <p²>.
$$-\hbar^2\frac{\partial^2 \psi(x)}{\partial x^2}=2\sqrt{m\omega \hbar ^3}(e^{-\sqrt{\frac{m\omega}{\hbar}}x^2}-2\sqrt{\frac{m\omega}{\hbar}}x^2e^{-\sqrt{\frac{m\omega}{\hbar}}x^2})$$
Now, I am not looking forward to putting that into the integral for finding <p²>. Especially not since I got to do it with the wavefunction of state 1 (this is state 0) aswell. Does it look right to you guys?

 

[kuruman]

But let us do some calculations, or rather. I got osme answers that I don't know if are correct or not. For <x²> I got $$\frac{\hbar}{m\omega}$$ without any clue knowing wether or not it's correct.

One clue is this: Is your expression dimensionally correct?

I also get 0 for both <x> and <p>, which I like, but don't see any physical understanding in.

Look at the two integrands, they are both odd functions in x. What happens when you integrate an odd function over symmetric limits?

Cause after I find <p²> and <x²>, I am told to check them up to the uncertainty principle, so I am guessing that I am going to end up with some of them to cancel each other out.

Don't guess. Just do the integrals correctly and see if anything cancels out.

But I am kind of stuck at <p²>.
$$-\hbar^2\frac{\partial^2 \psi(x)}{\partial x^2}=2\sqrt{m\omega \hbar ^3}(e^{-\sqrt{\frac{m\omega}{\hbar}}x^2}-2\sqrt{\frac{m\omega}{\hbar}}x^2e^{-\sqrt{\frac{m\omega}{\hbar}}x^2})$$
Now, I am not looking forward to putting that into the integral for finding <p²>. Especially not since I got to do it with the wavefunction of state 1 (this is state 0) aswell. Does it look right to you guys?

Replace x with the parameter ξ before you integrate. This should make your life simpler. Also, how did you get those radicals in the exponential terms? The original wavefunction doesn't have them.

 

[knut-o]

One clue is this: Is your expression dimensionally correct?

I assume I am going to end up with meter, and I don't do that now, end up wit Js² or something..

Look at the two integrands, they are both odd functions in x. What happens when you integrate an odd function over symmetric limits?

Odd function? Odd function over symetric limits? I must admit I am not quite sure about this.

Don't guess. Just do the integrals correctly and see if anything cancels out.

Part of the problem is doing it correcltly since I most likely going to screw it up when it comes to the algebraic calculations.

Replace x with the parameter ξ before you integrate. This should make your life simpler. Also, how did you get those radicals in the exponential terms? The original wavefunction doesn't have them.

And you're absolutly right, I don't know what I was thinking when I tried to calcuate it here. Guess I need a night of sleep on it before I continue tomorrow.

 

[knut-o]

Now, I've done some real attempts. Now, it's hard to write the many integrals in latex.
The integrals still goes from -inf to inf.
For psi 0:
$$<x^2>=\sqrt{\frac{\hbar ^3}{(m\omega)^3}}\alpha ^2\int \xi ^2e^{-\xi ^2}=\frac{\hbar}{2m\omega}$$
$$<p^2>=\alpha ^2\hbar\sqrt{m\omega}(\int e^{-\xi ^2}d\xi -\sqrt{\frac{m\omega}{\hbar}}\int\xi e^{-\xi^2}d\xi =\hbar m\omega$$

Now, this, put into the uncertainty principle $$\sigma _x \sigma _p=\sqrt{<x^2><p^2>}=\frac{\hbar}{\sqrt{2}}$$ more or equal to $$\frac{\hbar}{2}$$ which is true :D :D .

On psi 1 I end up with a total of $$\sqrt{\frac{3}{8}}\hbar$$ is more or equal to $$\frac{\hbar}{2}$$ Which is also true :)


Now I got to calculate the kinetic and potential energy for each of them, phew, more integration to come. But that I think I can handle now, now that I control the tool of integration :) .


But, another task is to construct psi 2, which I got the formula to do:
$$\psi _n=\frac{1}{\sqrt{n!}}(a_+)^n\psi _0$$ . It's that a₊ⁿ I struggle with. $$a_+=\frac{1}{\sqrt{2\hbar m\omega}}(ip+m\omega x)$$, but squaring this, and with operators and stuff. In another example, where they construct psi 1 (which I also used for the earlier tasks) it seems like they insert $$-\hbar\frac{d}{dx}+m\omega w$$, which squared would be $$\hbar ^2 \frac{d^2}{dx^2} -2\hbar m\omega +mwx^2$$, now I derived the x and the deriving operator, under the assumption that $$a_+=\frac{1}{\sqrt{2\hbar m\omega}}(ip+m\omega x)=-\hbar\frac{d}{dx}+m\omega w$$ but I am really unsure here.
I am also told to sketch it, so do I am unsure for psi 2 that I get (-a+bx²)e^{-cx^2} where a,b,c is just random constants that would be sat to 1 when I would sketch it, since like psi 0 = e^{-cx^2}, psi 1 = xe^{-cx^2}, would psi 2 be (-a+bx²)e^{-cx^2} or bx²e^{-cx^2}?

Then I am told to check ortogonality, which is where the oddness and evenness of functions come in (think I understood what it means), $$\int x^{2n-1}\cdot e^{-x^2}dx=0$$ for n=1,2,3... ? I also had a comment on my last assigment that the constants actually played a role for $$\int\psi _0\cdot\psi _2 dx$$, but that part I haven't really given a go so far.

 

[vela]

I think you wrote the creation operator down incorrectly. It should be

$$\hat{a}_+=\frac{1}{\sqrt{2\hbar m\omega}}(m\omega\hat{x}-i\hat{p})$$

To express it in the x-basis, you use the substitutions

$$\hat{x} \rightarrow x$$

$$\hat{p} \rightarrow \frac{\hbar}{i}\frac{d}{dx}$$

to get

$$\hat{a}_+=\frac{1}{\sqrt{2\hbar m\omega}}(m\omega x-\hbar\frac{d}{dx})$$

You had a "w" in a₊, which I assume was a typo and was supposed to be x.

When you square a₊, remember that $\hat{x}$ and $\hat{p}$ don't commute, so the cross terms don't combine the way you did it. Instead of squaring a₊ and then applying the resulting operator to $\psi_0$, try applying a₊ twice. It makes the steps a bit simpler.

 

[knut-o]

Instead of squaring a₊ and then applying the resulting operator to $\psi_0$, try applying a₊ twice. It makes the steps a bit simpler.

Does that mean I multiply $$\psi _1$$ with $$a_+$$? And sorting out the constants, and deriving psi _1 where it's needed for x, or simply multiply it with mwx ?
So I get something along the lines of: Where $$B1=(\frac{m\omega}{\pi\hbar})^{\frac{1}{4}}*\frac{1}{\hbar}$$
$$\psi _2=a_+\psi _1=B1(m\omega x -\hbar\frac{d}{dx})\cdot e^{-\frac{m\omega}{2\hbar}x^2}=B1(m\omega xe^{-\frac{m\omega}{2\hbar}x^2} - \hbar\frac{de^{-\frac{m\omega}{2\hbar}x^2}}{dx})?$$
If so, psi 2 is fairly easy to calculate, and I think the ortoghonality should be fairly easy to calculate.


Arf, this is not so easy subject as I thought it would become.

 

[vela]

Does that mean I multiply $$\psi _1$$ with $$a_+$$?

Yes.

And sorting out the constants, and deriving psi _1 where it's needed for x, or simply multiply it with mwx ?

I have no idea what that means.

 

[vela]

So I get something along the lines of: Where $$B1=(\frac{m\omega}{\pi\hbar})^{\frac{1}{4}}*\frac{1}{\hbar}$$
$$\psi _2=a_+\psi _1=B1(m\omega x -\hbar\frac{d}{dx})\cdot e^{-\frac{m\omega}{2\hbar}x^2}=B1(m\omega xe^{-\frac{m\omega}{2\hbar}x^2} - \hbar\frac{de^{-\frac{m\omega}{2\hbar}x^2}}{dx})?$$
If so, psi 2 is fairly easy to calculate, and I think the ortoghonality should be fairly easy to calculate.


Arf, this is not so easy subject as I thought it would become.

You have the wrong wavefunction for $\psi_1$. You're using $\psi_0$ in your calculation.

 

[vela]

I was looking over your original post, and I noticed two minor errors. The first one I think is just a typo. You left out a factor of x in $\psi_1$. The second one was when you converted $\psi_1$ to the dimensionless variable $\xi$, there's a factor of $\sqrt{2}$ missing.

 

[knut-o]

$$\psi _2=a_+\psi _1=B1(m\omega x -\hbar\frac{d}{dx})\cdot e^{-\frac{m\omega}{2\hbar}x^2}=B1(m\omega x^2e^{-\frac{m\omega}{2\hbar}x^2} - \hbar\frac{dxe^{-\frac{m\omega}{2\hbar}x^2}}{dx})?$$
If I do this derivation, sort out the constants etc, I will end up with $$\psi _2$$?
Which should give me something along the lines of $$\psi _2=B1(m\omega x^2e^{-\frac{m\omega}{2\hbar}x^2}}+\hbar e^{-\frac{m\omega}{2\hbar}x^2}} +m\omega xe^{-\frac{m\omega}{2\hbar}x^2}})=2B1m\omega (2x^2-1)e^{-\frac{m\omega}{2\hbar}x^2}}$$?

Starting to run a little out of time here, got to hand it in before 1200 tomorrow.


Edit: Now it's just weird, and the latex is failing me.. So $$\psi _2=2B1m\omega (2x^2-1)e^{-\frac{m\omega}{2\hbar}x^2}}$$?
[/tex]
Ok, now I don't understand anything from this forum and the latex-code it is currently doing. Is my psi 2 correct or like far off?

 

[vela]

Well, you said

$$\psi_n(x)=\frac{1}{\sqrt{n!}}(a_+)^n\psi_0(x)$$

right? So

$$\psi_2(x)=\frac{1}{\sqrt{2!}}(a_+)^2\psi_0(x) =\frac{1}{\sqrt{2}}a_+\left[\frac{1}{\sqrt{1!}}a_+\psi_0(x)\right] =\frac{1}{\sqrt{2}}a_+\psi_1(x)$$

Your textbook should list the first few wavefunctions for the harmonic oscillator, so you can check your answer. That said, I don't see anything obviously wrong with your answer.