I am currently studying for an exam in Quantum Mechanics and came across a solution to a problem I have trouble with understanding.

The Problem:

A Particle sits in an infinite potential well described by

\begin{align} V(x) &= 0, & 0 \leq x \leq L \\ V(x) &= \infty, & \text{otherwise} \end{align}

We know that the energies are given by $E_n = \dfrac{n^2 \pi^2 \hbar^2}{2 m L^2}$ and $\Psi(x) = A_n \sin(n \pi x /L)$.

At time $t_0$ the potential well is suddenly doubled in size, such that the potential is now

\begin{align} V(x) &= 0, & 0 \leq x \leq 2L \\ V(x) &= \infty, & \text{otherwise} \end{align}

So the energies are now given by $\tilde{E}_n = \dfrac{n^2 \pi^2 \hbar^2}{2 \cdot 4 m L^2}$ and $\tilde{\Psi}(x) = \tilde{A}_n \sin(n \pi x /2L)$.

  1. If the particle is in the ground state of the potential well before the change, what is the probability to find the particle in the ground state of the new potential after the change?

This is absolutely clear to me. We find a non vanishing probability as a result. But now it gets tricky:

  1. What is the expectation value of the energy of the particle directly after the change? How does the expectation value of the energy evolve in time?

The solution suggests that the expectation value of the energy does not evolve in time, which is clear to me, since the Hamiltonian is time independent and thus energy is conserved. But it also suggests that the expectation value does not change after we double the width of the potential wall which I understand from the argument of energy conservation but not in terms of quantum mechanics. If the probability that the particle is in the state $\tilde{\Psi}$ does not vanish the particle could have the energy $\tilde{E}_n$ which is lower than $E_n$ and this would mean that the expectation value of energy could change (with a given probability).

What am I missing here, where is my mistake? Any help is appreciated!


2 Answers 2


The expectation value of the energy stays the same after the doubling of size but it doesn't mean that the spectrum is the same. For a normalized $\psi$, the expectation value of the energy is simply $$ \int_{-L}^{+L}dx\,\psi^* \left( -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x) \right) \psi $$ because the integral may be reduced to the interval as the wave function vanishes outside the interval. Now, immediately when you double the size of the well, the value of $\psi(x)$ remains the same so it still vanishes outside the interval $(-L,L)$ and makes the integrand vanish as well (even though the second derivative could refuse to vanish). That's why the integral above may still be rewritten as $$ \int_{-2L}^{+2L}dx\,\psi^* \left( -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x) \right) \psi $$ without any change. It's the expectation value of the new Hamiltonian. Note that $V(x)=0$ wherever $\psi(x)\neq 0$ so the potential term may be omitted.

You're right that there is some probability that in the larger well, the particle sits at a lower-than-the-initial-energy-eigenvalue value of energy. However, there is some probability that the energy is raised as well – the wave packet is unnecessarily squeezed in a small part of the well which adds more kinetic energy than the minimum possible one. These positive and negative changes cancel in the expectation value of the energy: the calculation above showed that it stayed constant.

The expectation value of the energy stays constant when the particle evolves according to the larger-well Hamiltonian, too.

The probabilities for each energy eigenvalue are constant for all $t<0$ and then for $t>0$ but there is a discontinuity at $t=0$. However, as the simple calculation above shows, in the expectation value of the energy itself, the change of the spectrum etc. at $t=0$ cancels when it comes to the expectation value of the energy.

  • $\begingroup$ Thanks for your detail response and explanation, this makes it clear! $\endgroup$
    – user20486
    Feb 1, 2013 at 13:05
  • $\begingroup$ What is the value for the Hamiltonian after the change? I understand that the values in $E_{n}$ change to $\tilde E_{n}$, but shouldn't that change the value of the Hamiltonian? (using $\langle H \rangle = \Sigma |c_{n}|^{2}E_{n}$) $\endgroup$
    – user44816
    Feb 3, 2015 at 16:18
  • $\begingroup$ Hi @user44816, it's the first point of this calculation (and first sentence in my answer) that the expectation value (not just value!!!) of the Hamiltonian stays constant after the change because the two Hamiltonians only differ by the potential energy at places where the particle was 100% guaranteed to be absent. A general change of the Hamiltonian changes its expectation value in a general state by a general amount - but in this particular problem, my argument clearly shows that the amount is zero. $\endgroup$ Feb 4, 2015 at 6:36

Under sudden perturbation the state does not change, but the basis does. This state gets expanded in the new basis whose coefficients evolve correspondingly. Normally it is covered in chapters with the time-dependent perturbation theory $\hat{V} = \hat{V}(t)$.

If the potential is time-dependent, the energy is not conserved in general case. In your case the energy from certain becomes uncertain.

  • $\begingroup$ You surely have some spirit of the right answer but you can't call this change a "perturbation" because it's not small in any sense. The change (doubling) shouldn't be described by any perturbation theory. - For general time-dependent potentials, the energy isn't conserved but the point here is that it is not the general case. Because $V=0$ wherever $\psi\neq 0$, the expectation value of the energy actually is conserved but you haven't really explained why it is - despite the fact, as you admit, for general changes of the Hamiltonian, it isn't conserved. $\endgroup$ Feb 1, 2013 at 12:56
  • $\begingroup$ @LubošMotl: I never mentioned an expectation value, but the eigenvalue, to tell the truth. The initial state is an eigenstate of the old Hamiltonian and it ceases to be such of the new Hamiltonian. You mistake an expectation value and an eigenvalue, unfortunately. $\endgroup$ Feb 1, 2013 at 14:08

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