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 :)

  • 2
    $\begingroup$ 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 $x\phi_i^*\phi_j$. $\endgroup$
    – Will
    Jun 6, 2013 at 15:27
  • $\begingroup$ Ok, that is probably one of the flaws then :) $\endgroup$ Jun 6, 2013 at 15:36

1 Answer 1


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

  • $\begingroup$ 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... $\endgroup$ Jun 6, 2013 at 19:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.