# Measurement of a State Not in the Eigenbasis of the Operator

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?

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:

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

• Does this correspond to an observable physically, or is it a Hermitian operator that has no physical analgoue? I mean this in the sense that we can measure position and so it gets an operator and the same with momentum etc. Commented Apr 6, 2019 at 12:09
• This isn't really correct - you've chosen a particular (and rather arbitrary) assignment of eigenvalues to each eigenstate of your measurement operator. I think the correct question that the OP in asking is "What operators correspond to measurements that can yield the state $| 0 \rangle$ as an outcome?" and the answer is "Any self-adjoint operator that has $|0\rangle$ as an eigenvector." The operator's action on the subspace orthogonal to $|0\rangle$ is irrelevant. Commented Apr 6, 2019 at 14:18
• @user258521 How easily this operator $|0\rangle \langle 0 |$ could be physically implemented depends completely the the actual physical system that you're describing. For example, if it's a spin-1/2 particle with $|0\rangle$ and $|1\rangle$ representing spin-up and spin-down along the z-axis respectively, then this operator equals $\frac{1}{2} \hat{I} + \hat{S}^z$. You could measure in this basis via e.g. a Stern-Gerlach apparatus. Commented Apr 6, 2019 at 14:24
• tparker; No, I've not " chosen a particular (and rather arbitrary) assignment of eigenvalues to each eigenstate of your measurement operator". If you do the multiplication you'll find that the $|0\rangle$ is certainly an eigenvector of $|0\rangle\langle 0|$ and that its eigenvalue is 1. Commented Apr 6, 2019 at 20:58
• @CarlBrannen Yes, of course, but it seems rather arbitrary to assign the eigenvalue $1$ to the eigenstate $|0\rangle$ and the eigenvalue $0$ to the eigenstate $|1\rangle$. Your proposed operator certainly works, but I just wanted to clarify that it's not the only operator that works, as your answer seems to imply. I think the issue is that the OP asked multiple questions in his/her post, and we answered slightly different ones. Commented Apr 10, 2019 at 12:41

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.

• I was looking more for an analogue to the fact that position is an observable in experiments and so we can represent it by an operator. As @CarlBrannen pointed out, the projection operator satisfies this so I think my confusion lies in what an observable is in finite dimensional Hilbert spaces?? Commented Apr 6, 2019 at 12:39
• @user258521 Sounds like you should post another question Commented Apr 6, 2019 at 12:40

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 $$|\langle 1 | \psi(t) \rangle|^2$$.

• Great comments ! That has definitely helped. The difference in your first point is quite subtle but I shall keep that in mind. With regards to the second point, I believed it was an assumption that measurements only give eigenvalues of the operator. So measuring in the basis you have given would correspond to measuring some self-adjoint operator that has eigenvector $|0 \rangle$? Commented Apr 6, 2019 at 14:59
• @user258521 Yes, exactly. You should think of the eigenvalues as "labeling" the eigenvectors. Their actual numerical values are actually often not terribly conceptually important. The physical process of making a measurement "tells" you an eigenvalue, but what you really care about is the eigenvector that it corresponds to, because that tells you about the physical state of the system. Commented Apr 6, 2019 at 15:02
• @user258521 This idea is made rigorous in the blog post blog.jessriedel.com/2016/11/12/…, but I warn you that it's quite advanced and (I mean this entirely respectfully) probably above your current level of understanding. Regarding the first point in my answer, you can look up the "Kochen-Specker theorem" for the mathematically precise version, but again I warn you that it's quite advanced for a beginner. Commented Apr 6, 2019 at 15:05
• I think I was struggling to identify what the operator for $|0 \rangle$ could be but the spin example clarifies this. I suppose it wouldn't be much of a leap to show that all self-adjoint operators on finite dimensional spaces correspond to 'some' observable, but not necessarily for infinite dimensional spaces? Commented Apr 6, 2019 at 15:09
• @user258521 That's actually a rather subtle question, that people still argue about. Everyone agrees that every observable corresponds to a self-adjoint operator (well, up to a few subtle caveats that are beyond this discussion), but whether every self-adjoint operator corresponds to a physical observable is more complicated. The problem is basically that many self-adjoint operators correspond to crazy nonlocal quantities like "the product of the position of this electron here times the spin of that proton the Andromeda galaxy". Any actual experiment to measure this quantity would be so ... Commented Apr 6, 2019 at 15:15