Let $\Psi_1$ and $\Psi_2$ be widely separated narrow packets and $\Psi(x,0) = \Psi_1(x,0) +\Psi_2(x,0)$. I want to know how the overlap integral $\gamma(t)$ changes over time: $$\gamma(t) =\int_{-\infty}^\infty \Psi_1^*(x,t)\Psi_2(x,t)dx $$ At time equal to zero the value of $|\gamma(0)|$ is very small. As the packets evolve and spread, what will happen to the value of $|\gamma(t)|$?

And here is my solution :

Differentiating $\gamma(t)$ with respect to $t$, \begin{align*} \frac{d\gamma}{dt} =& \int_{-\infty}^\infty \left( \frac{\partial \Psi_1^*}{\partial t}\Psi_2 + \Psi_1^*\frac{\partial \Psi_2}{\partial t} \right) dx. \end{align*} Also, by Schrodinger equation,

\begin{align*} \frac{\partial \Psi_1^*}{\partial t} =& -\frac{1}{i\hbar}\left[ - \frac{\hbar^2}{2m}\frac{\partial^2\Psi_1^*}{\partial x^2} + V\Psi_1^* \right] \\ \frac{\partial \Psi_2}{\partial t} =& \frac{1}{i\hbar}\left[ -\frac{\hbar^2}{2m}\frac{\partial^2\Psi_2}{\partial x^2} + V\Psi_2 \right] \end{align*} Thus the integrand of $d\gamma /dt$ is $$ \frac{1}{i\hbar}\frac{\partial}{\partial x} \left[ -\frac{\hbar^2}{2m}\left( \Psi_1^* \frac{\partial \Psi_2}{\partial x}-\frac{\partial \Psi_1^*}{\partial x} \Psi_2 \right) \right]. $$ Hence, the integral is $$ \frac{1}{i\hbar}\left[ -\frac{\hbar^2}{2m}\left( \Psi_1^* \frac{\partial \Psi_2}{\partial x}-\frac{\partial \Psi_1^*}{\partial x} \Psi_2 \right) \right] $$ evaluated at $x = \infty , -\infty$. By boundary condition this integral vanishes, thus $d\gamma/dt=0$. This means that $\gamma(t)$ doesn't change with respect to $t$. This is the end of my solution.

However, why doesn't it changes with respect to time? I think it should change over time when two wave packets overlaps, and vanishes when they are totally separated. What is wrong with my solution?

For reference, this is problem set from Prof. Barton Zwiebach of MITOCW Physics 8.04.

  • 1
    $\begingroup$ I suppose the phrase "widely separated" is meant to imply this, but it's not obvious how to justify that $\psi_{1,2}$ sayisfy the SE individually and not just their sum. $\endgroup$
    – jacob1729
    Commented Jan 13, 2021 at 17:35

1 Answer 1


Unitary evolution implies time-independent inner products of evolving states

You are correct that the overlaps are independent of time. This is a basic postulate of quantum mechanics: time evolution is unitary. An initial state $|\psi_0\rangle$ evolves in time as $|\psi(t)\rangle = \hat{U}(t)|\psi_0\rangle$ where $\hat{U}(t)$ is a unitary operator. As a consequence, any two evolving states $|\psi(t)\rangle=\hat{U}(t)|\psi_{0}\rangle$ and $|\phi(t)\rangle=\hat{U}(t)|\phi_{0}\rangle$ will have a time-independent inner product: $$ \langle\psi(t)|\phi(t)\rangle = \langle \psi_0|\hat{U}^\dagger(t)\hat{U}(t)|\phi_0\rangle $$ $$\langle\psi(t)|\phi(t)\rangle = \langle \psi_0|\phi_0\rangle$$ To emphasize, this comes from unitarity: $\hat{U}^\dagger\hat{U}=1$. Your proof is just like this but explicitly uses the Schrodinger equation.

Unitary evolution does not imply time-independent measurement probabilities

But this does not mean that our measurement probabilities are time-independent. Suppose we prepare a system in a state $|\psi_0\rangle$ at $t=0$ and we want to measure some obserservable $\hat{M}$ at a later time $t$. In particular, suppose we are interested in the probability of finding a particular measurement outcome ($\hat{M}$ eigenvalue) $m$ corresponding to a single $\hat{M}$ eigenstate $|m\rangle$. Note that $|m\rangle$ is not evolving. It is simply the state we are interested in measuring. The probability of getting measurement outcome $m$ at time $t$ is $$ p(m, t) = |\langle m|\psi(t)\rangle|^2 $$ $$ = |\langle m | \hat{U}(t)|\psi_0\rangle|^2 $$ Since there is no $\hat{U}^\dagger$ to cancel the $\hat{U}$, this probability will clearly depend on time in general.

  • $\begingroup$ Thank you for your kind answer. Everything is little confusing but it's getting better. Thank you a lot. $\endgroup$ Commented Jan 22, 2021 at 16:13

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