I'm working with a particle in a box. The potential is $V(x) = \begin{cases} 0 & 0 \leq x \leq a \\ \infty & \text{otherwise} \end{cases}$, where $a > 0$ a real constant. We're told the system is prepared in the state $\Psi(x,0) = \frac{\sqrt{2}}{a}\left( c_1\sin{\frac{\pi x}{a}} + c_2\sin{\frac{2\pi x}{a}} \right)$, where $c_1, c_2$ are complex constants. I've therefore calculated the wavefunction at later times, it comes to $\Psi(x,t) = \frac{\sqrt{2}}{a}\left( c_1\sin{\frac{\pi x}{a}} + c_2\sin{\frac{2\pi x}{a}} \right) \exp{\left(\frac{-iE_n t}{\hbar}\right)}$. I now want to calculate the expected value of position at time $t$, that is, $\langle \hat{x} \rangle_\Psi$, and I want to show it is oscillatory. To do this I wrote

$$\langle \hat{x} \rangle_\Psi = \int_0^a \frac{\sqrt{2}}{a}\left( c_1\sin{\frac{\pi x}{a}} + c_2\sin{\frac{2\pi x}{a}} \right) x \frac{\sqrt{2}}{a}\left( c_1\sin{\frac{\pi x}{a}} + c_2\sin{\frac{2\pi x}{a}} \right) dx$$

$$ = \frac{c_1^2 a}{2} - \frac{32c_1c_2a}{9\pi^2} + \frac{c_2^2 a}{2}$$

This result is not oscillatory and it is actually constant, so it seems obviously wrong. What went wrong in my calculation? What's the right way to proceed?

  • $\begingroup$ The initial wavefunction is a superposition of two linearly independent states. The time evolution of each sine function is different, which will lead to time-dependent interference terms in the position expectation value. $\endgroup$ Commented May 31, 2023 at 18:38
  • $\begingroup$ Ah thank you, I now realise that I calculated $\langle \hat{x} \rangle_\Psi$ incorrectly, we get a sum of two complex exponentials so they won't cancel, you are write, there will be interference terms @QuantumFieldMedalist $\endgroup$ Commented May 31, 2023 at 18:55
  • $\begingroup$ The general state of a particle in a box is a sum of an infinite number of possible states. You can find a superposition of these states that represents a somewhat localized classical particle but you need quite a few of these states with higher energy and you have to add them with the right phases. $\endgroup$ Commented May 31, 2023 at 19:09

2 Answers 2


You have the wrong $\Psi(x,t)$. If $\Psi(x,0)=\sum_n c_n \psi_n(x)$, then $$ \Psi(x,t)=\sum_n c_n \psi_n(x)e^{-iE_n t/\hbar} $$ with the factor $e^{-iE_n t/\hbar}$ different for different $\psi_n(x)$ since they have different energies.

In your specific case you have a single exponential $\exp(-i E_nt/\hbar)$ and your $\psi(x,0)$ is a sum of solutions with different $E_n$'s; to see that your approach of multiplying by a single exponential is conceptually wrong ask yourself which one value of $E_n$ (out of all of them) are you suggesting to choose to multiply $\psi(x,0)$?


You may want to write your initial wave function as:

$$ |\psi\rangle = c_1|1\rangle + c_2|2\rangle $$

where $H|n\rangle = E_n|n\rangle$

and I am using bra-ket notation for brevity. Also:

$$ |n, t\rangle = e^{-iE_n t}|n\rangle $$

where I have slightly abused the notation to include time-dependence, and set $\hbar=1$.


$$\bar x(t) = \langle \psi|x|\psi\rangle $$

One problem with your formula is that you failed to take the complex conjugate of the "bra", so that:

$$\bar x(t) = (c_1^*e^{iE_1 t}|\langle 1|+ c_2^*e^{iE_2 t}\langle 2|)x(c_1e^{-iE_1 t}|1\rangle + c_2e^{-iE_2 t}|2\rangle) $$

$$\bar x(t) = ||c_1||^2\langle 1|x|1\rangle + ||c_2||^2\langle 2|x|2\rangle + c_1^*c_2 e^{i(E_1-E_2)t}\langle 1|x|2\rangle + c_2^*c_1 e^{-i(E_1-E_2)t}\langle 2|x|1\rangle $$

Since $||c_1||^2 + ||c_2||^2 = 1$:

$$\bar x(t) = \frac a 2 + c_1^*c_2 e^{i(E_1-E_2)t}\langle 1|x|2\rangle + c_2^*c_1 e^{-i(E_1-E_2)t}\langle 2|x|1\rangle $$

which can be further simplified.


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