2
$\begingroup$

Suppose we have some Hamiltonian $H$ with at least one normalized eigenstate $v$ with real eigenvalue $\lambda$.

The time evolution operator is given by $$ U(t,0) = e^{- i \frac{H}{\hbar}t} \ .$$ Now let $\psi_0 = v$ and let $\psi(t) = U(t, 0) \psi_0$. Now if my computation is correct we have \begin{align} \left| \left< \psi(t) | \psi_0 \right> \right|^2 & = \left| \left< \psi_0 | \psi(t) \right> \right|^2 = \left| \left< \psi_0 | U(t,0)\psi_0 \right> \right|^2 = \left| \left< \psi_0 | e^{- i \frac{\lambda}{\hbar}t} \psi_0 \right> \right|^2 \\ & = \left| e^{ i \frac{\lambda}{\hbar}t} \left< \psi_0 | \psi_0 \right> \right|^2 = \left|\left< \psi_0 | \psi_0 \right> \right|^2 = 1 \ . \end{align} Now in the context of my exercise this makes no sense as I am asked what the "survival probability" of some state $\psi(t)$ is and the exercise in question asked me to calculate the above probability, but I seem to get that the probability is $1$ for all times, so the interpretation would be that the state remains the same for all time, or something like that.

I have a feeling I am misunderstanding something in the above calculation.

$\endgroup$
  • 3
    $\begingroup$ So why would that be wrong? How did you get to eigenstates as relevant quantities in the first place? $\endgroup$ – Emilio Pisanty Oct 4 '16 at 18:58
  • $\begingroup$ In the context of the exercise I am dealing with this result makes no sense. The physical example that this situation corresponds to is that we have a neutrino that is initially produced as an electron neutrino, where the electron neutrino state corresponds to an eigenstate. Now we are asked to calculate the survival probability of this electron neutrino as a function of distance, but the result I get above basically states that the electron neutrino is an electron neutrino always. $\endgroup$ – Kayle of the Creeks Oct 4 '16 at 19:23
  • 3
    $\begingroup$ Yeah, that sounds about right. Neutrinos will remain neutrinos. Electron neutrinos will oscillate to other flavours because they're not eigenstates of the propagation - they're the useful basis for the weak interaction, but they're not eigenstates of the mass operator. If you prepare in an eigenstate of the governing hamiltonian, it will remain that way. $\endgroup$ – Emilio Pisanty Oct 4 '16 at 19:32
2
$\begingroup$

Your observation is totally correct and nothing is wrong with that.

Eigenstates of the (time independant) Hamiltonian are called stationary states. Such system will stay in a state who's amplitude is constant. Only the relative phase will change:

$$ \psi(t) = U(t,0) \psi_0 = e^{-iHt/\hbar}\psi_o = e^{-i\lambda t/\hbar}\psi_0 $$

So a system with definite energy will not change it's energy, commonly known as energy conservation.

Also, if a system is in a constant superposition of states $\psi_0,~\psi_1,~\psi_2,\dots$ with different energy eigenvalues $\lambda_0,~\lambda_1,~\lambda_2,\dots$, the system's probability to be in one of those states is constant:

$$ \left| \langle\psi_0|U(t,0)\sum_i\alpha_i|\psi_i\rangle \right|^2 = |\alpha_i|^2 \left| \langle\psi_0|\psi_0\rangle \right|^2 = const. $$

However, the propablity to stay in that superposition may oscillate:

$$ \left| \left(\sum_j\alpha_j^*\langle\psi_j|\right)U(t,0)\sum_i\alpha_i|\psi_i\rangle \right|^2 = \left| \sum_i |\alpha_i|^2 e^{-i\lambda_i t/\hbar}\right|^2 \ne const. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.