Measurement of a State Not in the Eigenbasis of the Operator

Question

Suppose I have a two dimensional Hilbert space $\{ |0 \rangle,|1\rangle \}$ with these states being orthonormal. Now suppose I have the Hamiltonian $H=|1\rangle \langle 0|+|0\rangle \langle 1| .$

It is clear that the (normalised) eigenstates of this are:

$|\phi_0\rangle=\frac{1}{\sqrt{2}}\Big(|0\rangle+|1\rangle\Big)$ and $|\phi_1\rangle=\frac{1}{\sqrt{2}}\Big(|0\rangle-|1\rangle\Big)$ with eigenvalues $+1$ and $-1$ respectively.

Now what I am confused about is calculating probabilities: I know that given the state $| \psi(t)\rangle$ of the system, if I want to calculate the probability of this state being in an arbitrary state $|\phi\rangle$, then I just need to calculate their overlap $\big($ie. $\mathcal{P}(\phi)=|\langle \phi | \psi(t) \rangle |^2\big).$ Suppose I chose the state to be $|0\rangle$, then I would need to find its overlap.

What is the actual grounding for this calculation $\dots$ From the principles of QM the only outcomes from measurement are eigenvalues of the operator in question, and the probability is given by the overlap of the eigenstate with the time evolving state of the system. So I should only be able to find the probabilities of being in states $|\phi_0 \rangle$ or $|\phi_1 \rangle$ by this axiom (obviously not true though!)

So how can I find the probability of being in state $|0\rangle$ if it isn't an eigenstate of the Hamiltonian operator. Is there an operator corresponding to the original basis states which we can measure (though this would be the identity)? What is the intuition behind this definition so that I can some appreciation as to why it makes sense to compute the overlap of states?

Answer 1

Perhaps the confusion here is "what is an operator for the measurement whose eigenvector is $|0\rangle$?"

The usual operator is the pure density matrix or projection operator $|0\rangle\langle 0|$ for which $|0\rangle$ has eigenvalue 1. The other eigenvector is $|1\rangle$ which has eigenvalue 0.

If you want to you can write $|0\rangle$ in the $\phi_j$ basis by using:

$|0\rangle = (|\phi_0\rangle + |\phi_1\rangle)/\sqrt{2}$

Answer 2

So how can I find the probability of being in state $|0\rangle$ if it isn't an eigenstate of the Hamiltonian operator

You don't need to do anything special. $\langle 0|\psi\rangle$ tells you "how much of $|0\rangle$ is in $|\psi\rangle$". So just compute $|\langle 0|\psi\rangle|^2$.

Of course this is easier if $|\psi\rangle$ is already expressed in the $\{|0\rangle,|1\rangle\}$ basis, but it is not required.

Answer 3

I think there are two issues that are messing you up.

The first is your claim that $|\langle \phi | \psi(t) \rangle|^2$ represents "the probability of this state being in an arbitrary state |𝜙⟩". Whether or not this claim is actually true is something that philosophers have debated for a century, but I would posit that it's not a useful statement for a beginner to believe. What that the quantity $|\langle \phi | \psi(t) \rangle|^2$ really represents is the probability that a measurement made in a basis that includes the state vector $|\phi\rangle$ will yield that state - no more, no less.

The second is your implicit assumption that you are only "allowed" to measure in the eigenbasis of the Hamiltonian. That's not true - you can in principle measure in any orthonormal basis you'd like. In particular, you're certainly free to measure the system in the $\{ |0\rangle, |1\rangle\}$ basis. If you do, then the relevant outcome probabilities are $|\langle 0 | \psi(t) \rangle|^2$ and $|\langle 1 | \psi(t) \rangle|^2$.