I have seen Zetilli's QM book deals with $E>V$ and $E< V$ (tunnelling) in case of the potential wells deliberately avoiding the E=V case, so I thought maybe something is intriguing about this and made this up.

Suppose the total energy of the particle is equal to its potential energy.Then its kinetic energy should be zero, (speaking non-relativistically). But Kinetic energy operator is $\hat{T}=\hat{p}^2/2m$ (where $\hat{p}=-i\hbar\frac{\partial}{\partial x}$), So clearly since Kinetic energy is 0 here, momentum eigenvalue will also vanish.

Now, Putting $E=V$ in time-independent Schrodinger equation (1D) we get, $$\frac{\partial^2\psi}{\partial x^2}=\frac{2m(E-V)}{\hbar^2}\psi\implies\frac{d^2\psi}{d x^2}=0\implies\psi=Ax+B$$ where $A$ and $B$ are arbitrary constants. Since, the wave function must vanish at $\pm\infty$, $A=0$,hence the wave function equals a constant=$B$ and is not normalizable. So, a particle with no momentum(or kinetic energy), gives a physically unrealizable wave function!

Does this imply $E=V$ is a restricted critical case or momentum cant be zero in quantum mechanics or did i just go wrong somewhere?


2 Answers 2


You are not wrong, but it is worth noting that the same thing is true of any momentum eigenstate (or closely related unbound eigenstate of a Hamiltonian with a potential well in it). Explicitly $$ -i\hbar\frac{\partial}{\partial x} \psi(x) = p\,\psi(x) $$ then $$ \psi = A e^{i \frac{p}{\hbar}x} $$ which is not normalisable either. This means that we can never truly realise a momentum eigenstate, but we can still use them as a basis for physically realisable states usng the rigged Hilbert space formalism.

So yes we cannot realise a state with exactly zero momentum, but this is not a special property of the zero momentum state.

  • $\begingroup$ Does that imply that a quantum particle can never be at rest(wrt lab frame)? $\endgroup$ Mar 10, 2020 at 14:05
  • 4
    $\begingroup$ It depends what you mean by at rest. I can never really put the particle into a state where momentum has no uncertainty, and so I cannot put it a state where a measurement of momentum is guaranteed to give zero. I can, however, find a frame where the expectation of the particle's momentum is zero (for a massive particle) and Ehrenfest's theorem tells us that in this case the expectation of the particles position is constant. Which of these notions of a rest frame is relevant depends on what you want to do with it. $\endgroup$ Mar 10, 2020 at 15:46
  • $\begingroup$ I got it now.Thank you. $\endgroup$ Mar 10, 2020 at 16:01

I would like to add two points to the accepted answer:

  • If you use the periodic boundary conditions trick to normalize the momentum eigenstates, then all the momentum eigenstates become normalizable, including the zero momentum eigenstate.

  • The discussion of tunneling in QM books is usually within the quasi-classical approximation, which breaks down when the difference $V - E$ is small. Thus, one usually treats the cases when the particle energy is well below or well above the barrier.


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.