# Quantum harmonic oscillator problem

1. Jan 9, 2008

### natugnaro

1. The problem statement, all variables and given/known data

Is there any way to find $$<\varphi_{n}(x)|x|\varphi_{m}(x)|>$$ (where phi_n(x) , phi_m(x) are eigenfunction of harmonic oscillator) without doing integral ?

2. Relevant equations

perhaps orthonormality of hermite polynomials ?

$$\int^{+\infty}_{-\infty}H_{n}(x)H_{m}(x)e^{-x^{2}}dx=\delta_{nm}(Pi)^{1/2}2^{n}n!$$

3. The attempt at a solution

Actually i need to find <x> from known $$\psi(x,t)$$.
$$<x>=<\psi(x,t)|x|\psi(x,t)|>$$

This gives me a lot of $$<\varphi_{n}(x)|x|\varphi_{m}(x)|>$$ terms, some of them cancel out (for m=n) ,
but at the end I'm left with three terms thath I have to calculate ($$<\varphi_{1}(x)|x|\varphi_{2}(x)>,<\varphi_{2}(x)|x|\varphi_{3}(x)>, <\varphi_{3}(x)|x|\varphi_{1}(x)>$$
and that also looks like a lot of work.

final result is <x>=0

(This is problem from Schaums QM (supplementary prob.))

2. Jan 9, 2008

3. Jan 9, 2008

### Rainbow Child

There is a recurrence relation, which is satisfied by the Hermite polynomials

$$H_{n+1}(x)=2\,x\,H_{n}(x)-2\,n\,H_{n-1}(x)\Rightarrow x\,H_n=\frac{1}{2}\left(H_{n+1}(x)+2\,n\,H_{n-1}(x)\right)$$

Thus you do not have to do the integrals, i.e.

$$\int^{+\infty}_{-\infty}H_{1}(x)\,x\,H_{2}(x)e^{-x^{2}}dx=\int^{+\infty}_{-\infty}H_{1}(x)\left(\frac{H_{3}(x)+4\,H_{1}(x)}{2}\right)e^{-x^{2}}dx=2\,\int^{+\infty}_{-\infty}H_{1}(x)\,H_{1}(x)e^{-x^{2}}dx=4\,\sqrt{\pi}$$

Similar expressions hold for the other integrals.

4. Jan 9, 2008

5. Jan 9, 2008

### malawi_glenn

ok, it is very powerful and it is used very often in QM (more than just about harm osc). Send me a private message and I'll teach you if you want ;)

6. Jan 9, 2008

### natugnaro

Thank's Rainbow Child, that is really useful.
Using this recurrence relation for hermite polynomials and $$<\varphi_{n}(x)|x|\varphi_{m}(x)>=<\varphi_{m}(x)|x|\varphi_{n}(x)>*$$
I was able to reduce <x> to the form:
$$\frac{2}{7}(4\sqrt{\pi}A_{1}A_{2}e^{i\omega t}+4\sqrt{\pi}A_{1}A_{2}e^{-i\omega t}+24\sqrt{\pi}A_{2}A_{3}e^{-i\omega t}+24\sqrt{\pi}A_{2}A_{3}e^{i\omega t})$$
Where An's are normalization constants, but I don't see how this leads to <x>=0 ?
Even if I expand, just Sin(wt) terms will cancel out, but not Cos(wt).
(It is problem 5.15 from Schaum's QM)

Last edited: Jan 10, 2008
7. Jan 9, 2008

### malawi_glenn

\pi and \cdot

if you want to learn real QM, try ladder operators (A)

8. Jan 9, 2008

### natugnaro

If I don't solve the problem this way, I'll send you a message again.

9. Jan 10, 2008

### Rainbow Child

Your answer is correct (up to constants, I didn't carried out the whole calculations). The only case where $<x>=0$ would be if we had eigenfunctions of harmonic oscillator of the form

$$\varphi_{n}(x),\varphi_{m}(x)$$

with $$n\neq m\pm 1$$

Only in this case the integrals

$$\int_{-\infy}^{+\infty}H_n(x)\,x\,H_m(x)\,e^{-x^2}\,d\,x=\int_{-\infty}^{+\infty}H_n(x)\frac{1}{2}\left(H_{m+1}(x)+2\,m\,H_{m-1}(x)\right)\,d\,x=\frac{1}{2}\int_{-\infty}^{+\infty}H_n(x)H_{m+1}(x)\,d\,x+m\,\int_{-\infty}^{+\infty}H_n(x)\,H_{m-1}(x)\,d\,x$$

would vanish. Thus there must be a typo error in the book!

10. Jan 13, 2008

### natugnaro

I didn't get the constants right up there.
I did the thing again and find

$$<x>=\frac{2(2+\sqrt{3})}{7}Cos(\omega t)\sqrt{\frac{\hbar}{m\omega}}$$

At last I've tried also ladder operators (thanks malawi_glenn).
$$x=\sqrt{\frac{\hbar}{2m\omega}}(a+a^{+})$$

then using
$$a|k>=\sqrt{k}|k-1>$$
$$a^{+}|k>=\sqrt{k+1}|k+1>$$

arrive at
$$<n|x|k>=\sqrt{\frac{\hbar}{2m\omega}}(\sqrt{k}<n|k-1>+\sqrt{k+1}<n|k+1>)$$
where n and k are eigenfunctions of H.O.
From here it's easy, I just use the dot product.
There is a solved problem in Schaums QM for this.

I got the same result for <x> using ladder op. ,that's nice .
Ok, I'm gonna mark this thread as solved even though the solution in the book is <x>=0.

11. Jan 13, 2008

### malawi_glenn

$$a|n>=\sqrt{n}|n-1>$$
$$a^{\agger}|n>=\sqrt{n+1}|n+1>$$

$$<n|x|n> =\sqrt{\frac{\hbar}{2m\omega}}(\sqrt{n}<n|n-1>+\sqrt{n+1}<n|n+1>) = 0$$

Last edited: Jan 13, 2008
12. Jan 13, 2008

### natugnaro

Ok, but I have <1|x|2> and <2|x|3>, so I can't have zero, right ?

13. Jan 13, 2008

### malawi_glenn

no of course not, but the case <n|x|n> = <x> = 0, I thought you said that you did not understand why you did not get it.

The thing you have: <n|x|k> is correct, just plug in your values for n and k and you will get the correct answer.

14. Jan 13, 2008

### natugnaro

Ok, no problem .
As Rainbow Child said "Thus there must be a typo error in the book!"

15. Jan 13, 2008

### malawi_glenn

ok, but <x> always refer to be taken between same two states.

<x> and <1|x|2> and <2|x|3> ?

16. Jan 14, 2008

### natugnaro

Here is the text from the book:
The wave function of a harmonic oscillator at time t=0 is
$$\psi(x,0)=\sqrt{2}A\phi_{1}+\frac{1}{\sqrt{2}}A\phi_{2}+A\phi_{3}$$
Where phi_n is stationary eigenfunction of the harmonic oscillator for the nth state,
and A is normalization constant.
a) compute A.
b) find psi(x,t) for all values of t.
c) calculate the average values of <E> at times t=0, t=Pi/w, t=2Pi/w.
d) find expectation values <x> and <p> for t>=0.

a) done that , got result as in the book
b) done that, got result as in the book
c) skipped that
d) calculated just <x>, <x>=2(2+sqrt(3))/7 * sqrt(h/(2Pi*mw))*Cos(wt)

Solutions from the book:
a) $$A=\sqrt{2/7}$$

b) $$\psi(x,t)=\sqrt{2/7}(\sqrt{2}\phi_{1}e^{\frac{-3i \omega t}{2}}+\frac{1}{\sqrt{2}}\phi_{2}e^{\frac{-5i \omega t}{2}}+\phi_{3}e^{\frac{-7i \omega t}{2}})$$

d) <x>=0 , <p>=0

17. Jan 14, 2008

### malawi_glenn

Great, always post the original question in the first post so no confusion aries :)

You now also know how to write p as a combination of a and a(dagger).

shall see if can get the right values for <x>

18. Jan 14, 2008

### natugnaro

Yes, I should have done that