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Say I have a non-Hermitian hamiltonian, such as one might have in an incomplete description of a system where the states are allowed to decay. Then probabilities are not conserved since magnitudes will decrease with time. In particular, say I have an eigenfunction $\mid \psi \rangle$ with $\langle \psi \mid \psi \rangle \sim e^{-kt}$, so that $\psi$ is normalized at $t=0$ (before any decay), but the probability of encountering $\psi$ decreases with increasing $t$.

My question is about the probability of finding $\psi$. If I pick a generic state $\mid \alpha \rangle$, I can calculate the probability that measuring $\psi$ gives $\mid \alpha \rangle$ by $\mid \langle \alpha \mid \psi \rangle \mid^2$. Now if I merely want to find my probability of finding $\mid \psi \rangle$ at all, then I would do the same as above but with $\alpha = \psi$, hence I compute $\mid \langle \psi \mid \psi \rangle \mid^2$. this feel very weird since now I am looking at $\mid \psi \mid^4$ as opposed to $\mid \psi \mid^2$, which is usually what we interpret at the probability.

What is going on here? (Note that I bring up a system with decay so that it genuinely matters what the exponent is. In a usual system, $\mid \psi \mid^2 = 1$ so it doesn't matter what power is chosen.)

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Suppose you have $\langle \psi(t) | \psi(0) \rangle = A e^{-k t}$, for some normalization constant $A$.

Then the probability for the system to be in state $|\psi \rangle$ at time $t$ is

\begin{equation} P_\psi(t) = |\langle \psi(t) | \psi(0) \rangle |^2 = |A|^2 e^{-2kt}. \end{equation}

I feel this answer is a little glib but I am not sure where you are stuck actually.

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  • $\begingroup$ Let me exapnd a bit, though I think you've answered my question. As a thought experiment, suppose I have $N>>1$ particles in state $\mid \psi \rangle$ at time $t = 0$, how many will be left at some later time $t = T$? I would compute this as $N \mid \psi(T) \mid^2 = N \langle \psi(T) \mid \psi(T) \rangle$. However, if we are measuring the probability that $\mid \psi \rangle$ is in state $\mid \alpha \rangle$ at time $T$, this is $\mid \langle \alpha \mid \psi(T) \rangle \mid^2$ but taking $\alpha = \psi(T)$ we get $\mid\langle \psi(T) \mid \psi(T) \rangle\mid^2$, which is not what we had. $\endgroup$
    – Eulerian
    Commented Oct 20, 2020 at 23:48
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    $\begingroup$ I think maybe your confusion might just be getting wrapped up in notation. First, forget about time, and note that $|\langle \alpha | \psi\rangle |^2 = |\psi(\alpha)|^2$ -- the "ket" is the function and the "bra" is the argument of the function. You don't have to think of it that way, but my point is that the bra and ket are different objects. In particular, I wouldn't think of $|\langle \psi | \psi \rangle|^2$ as being like a function $\psi(x)$ raised to the fourth power. Inserting a complete set of states, it is more like $\int d \alpha |\psi(\alpha)|^2$... $\endgroup$
    – Andrew
    Commented Oct 21, 2020 at 0:03
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    $\begingroup$ ...For the example you are considering, I think you have the right basic idea, but just need to be a bit more careful about the notation. Say that these particles are all bosons, and they all either live in a state with energy $E_0$ or else they decay. Then the probability to find a given particle in the $E_0$ state is $| \langle \psi(t) | \psi(0) \rangle |^2 = |\langle \psi(0) | e^{-i H t} | \psi(0)\rangle|^2 = e^{-2 k t} |\langle \psi(0) | \psi(0) \rangle|^2 = e^{-2 k t}$. Then the average number of particles remaining is $N e^{-2kt}$. $\endgroup$
    – Andrew
    Commented Oct 21, 2020 at 0:15
  • $\begingroup$ I see, that makes sense. Thank you for clarifying. $\endgroup$
    – Eulerian
    Commented Oct 21, 2020 at 2:34

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