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I'm looking for some help to a question.

I'm working in the infinite square well, and I have the wavefunction:

$$\psi(x,t=0)=A\left( i\sqrt{2}\phi_{1}+\sqrt{3}\phi_{2} \right).$$

For every time t, the wavefunction is:

$$\psi(x,t)=A\left( i\sqrt{2}\phi_{1}e^{-iE_{1}t/\hbar}+\sqrt{3}\phi_{2}e^{-iE_{2}t/\hbar} \right).$$

Now, I'm asked to calculate the expectation value of the particles position $\left\langle x\right\rangle (t)$.

My guess was to just do it like this:

$$\left\langle x \right \rangle(t) = \int_{-\infty}^{\infty}x|\Psi(x,t)|^{2}dx,$$

and then use last of my two wavefunctions, and use the Kronecker Delta argument to remove the terms that got $i \neq j$.

But, that doesn't seem to work. The answer should be:

$$\langle x \rangle (t) =\frac{32\sqrt{6}a}{45\pi^{2}}\sin((E_{1}-E_{2})t/\hbar).$$

But if I do what I said, I will get something like 1 I think.

So I'm guessing I'm going all wrong about, so I was hoping someone could give me a hint :)

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Careful with your comment about using the Kronecker Delta to remove terms for $i\neq j$ - this is the case when integrating $\phi_i^*\phi_j$, as the wavefunctions are orthogonal, but not, in general, integrating $x\phi_i^*\phi_j$. – Will Jun 6 '13 at 15:27
Ok, that is probably one of the flaws then :) – Denver Dang Jun 6 '13 at 15:36
up vote 1 down vote accepted

Since you're only asking for a hint I will point out what I think you're missing:

The integral for the expectation value will yield 4 terms $$\langle x \rangle (t) = \int_{-\infty}^\infty \psi(x,t)^* x \psi(x,t) dx $$ $$= |A|^2 \int_{\infty}^{\infty} \left( -i\sqrt{2}\phi_1^* e^{+iE_1t/\hbar} + \sqrt{3} \phi_2^* e^{+iE_2t/\hbar}\right)x\left(i\sqrt{2}\phi_1e^{-iE_1t/\hbar} + \sqrt{3}\phi_2 e^{-iE_2t/\hbar}\right)dx$$ $$ = |A|^2 \int_{-\infty}^{\infty}\left(2\phi_1^*x\phi_1 + 3\phi_2^*x\phi_2 + i\sqrt{6}\left( \phi_2x\phi_1^*e^{i(E_2-E_1)t/\hbar} - \phi_1^* x \phi_2 e^{-i(E_2-E_1)t/\hbar}\right)\right)dx$$

The expressions $\phi_1^*\phi_1$ and $\phi_2^*\phi_2$ are even functions in position space, when standing alone. Remember that if you integrate over an even function multiplied by an odd function (e.g. $x$) it evaluates to zero:

$$\int_{-\infty}^{\infty} \phi_1^* \phi_1 dx = 1$$ However $$\int_{-\infty}^{\infty} \phi_1^* x \phi_1 dx = 0$$ Hence, you actually lose the first two terms (not the 'cross' terms like you mentioned).

On the other hand, for your cross terms, you have $$\int_{-\infty}^{\infty} \phi_1^* \phi_2 dx = 0$$ However, this is your mistake (which Will pointed out in his comment) $$\int_{-\infty}^{\infty} \phi_1^* x \phi_2 dx \neq 0$$ After you remove the appropriate terms you are left with the following expression to evaluate.

$$ = |A|^2 \int_{-\infty}^{\infty}\left(i\sqrt{6}\left( \phi_2x\phi_1^*e^{i(E_2-E_1)t/\hbar} - \phi_1^* x \phi_2 e^{-i(E_2-E_1)t/\hbar}\right)\right)dx$$

And remember that $$ \sin(\theta) = \frac{e^{i\theta} - e^{-i\theta}}{2i}$$

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That makes sense. Haven't even thought about that. Thank you :) Now I just need the wierd factor in front of $\sin$. But maybe I'll figure that out... – Denver Dang Jun 6 '13 at 19:50

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