In Feynman's volume III, chapter 7, he describes a particle which is in a superposition of the energy states $E_1$ and $E_2$ as

$$ e^{-i(E_1/\hbar)t} |E_1\rangle + e^{-i(E_2/\hbar)t} |E_2\rangle $$

See formula $(7.3)$.

If I understand that correctly, what does he mean with interferences in the following statement:

And if we have some combination of the two, we will have an interference.

I know that the superposition of two states allows for interferences, but what would those be in the case of a particle at rest? What experiment could be performed to show the interference?

EDIT: To make it clear what I mean by "a particle at rest". I quote Feynman:

Suppose we have an atom — or an electron, or any particle — which at rest would have a definite energy $E_0$

Feynman explains the most simple case of a quantum mechanical particle: a particle with a definite momentum $p=0$, where the uncertainty principle says that the position uncertainty $\Delta q$ must be infinite because $\Delta p=0$, so the particle's position amplitude is everywhere the same:

$$ a e^{-i(E_0/\hbar)t} $$

  • $\begingroup$ It means that the amplitude (modulus square) of the state is not just the same of the two single amplitudes. There is an additional contribution proportional to $\cos(E_1-E_2)$. $\endgroup$
    – gented
    Commented Oct 7, 2015 at 11:12

1 Answer 1


The interferences come out when you look at the expectation value of some operator.

If $$|\Psi(t)\rangle = e^{-i(E_1/\hbar)t} |E_1\rangle + e^{-i(E_2/\hbar)t} |E_2\rangle $$ and we want to look at the expectation value of some Hermitian operator $O$, then we get

\begin{eqnarray}\langle \Psi(t) |O|\Psi(t)\rangle =& \langle E_1 |O|E_1\rangle + \langle E_2 |O|E_2\rangle + e^{i \omega_{12} t} \langle E_1 |O|E_2\rangle + e^{-i \omega_{12} t} \langle E_2 |O|E_1\rangle \\ =& \langle E_1 |O|E_1\rangle + \langle E_2 |O|E_2\rangle + 2 \cos(\omega_{12}t) \Re \langle E_2 |O|E_1\rangle + 2 \sin(\omega_{12}t) \Im \langle E_2 |O|E_1\rangle \end{eqnarray}

where $\omega_{12} = (E_1 - E_2)/\hbar$. That last term is the interference contribution.

After reading this section from the Feynman lectures, I think I can explain as follows. A particle can be described as at rest if its momentum $\langle \hat{p} \rangle$ is zero and the uncertainty in momentum is also zero. This means, as you mention, the uncertainty in position is infinite.

But it could still be in a superposition state if the energy eigenstates $|E_1\rangle$ and $|E_2\rangle$ describe its internal energy - like the electronic state of an atom, or the particle's spin, or the particle's angular momentum. So what we mean specifically when we say "nonstationary state" is that it is not in a single eigenstate of the Hamiltonian.

See the answers here for a nice discussion.

  • $\begingroup$ See edit for clarification. $\endgroup$
    – Bass
    Commented Oct 7, 2015 at 15:48
  • $\begingroup$ Does Feynman's definition contradict the thing you're saying about nonstationary states? This might well be, he is simplifying many things to explain the important things as simple as possible. $\endgroup$
    – Bass
    Commented Oct 8, 2015 at 9:20
  • 1
    $\begingroup$ So I've always heard that physicists absolutely love Feynman's lectures, but that they are horrible textbooks to learn from the first time around. Looking at the page you are referencing now on Google Books.... $\endgroup$
    – Jason B.
    Commented Oct 8, 2015 at 9:24

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