# Can a projector serve as an operator for observable?

Suppose I have a two-level quantum system whose orthonormal basis is $$\{ |0\rangle,|1\rangle \}$$. Consider the projector onto the one-dimensional space spanned by $$|1\rangle$$: $$P_{1} = |1\rangle \langle 1|$$.

(viewpoint 1) Since $$P_{1}$$ is Hermitian, I think we can interpret $$P_{1}$$ as an "observable" whose eigenvalues are $$1$$ for $$|1\rangle$$ and $$0$$ for $$|0\rangle$$. If so, suppose then I prepare a state $$| \psi \rangle = | 0 \rangle$$. I wonder what will happen when we perform a "$$P_{1}$$ measurement" on this state $$| \psi \rangle = | 0 \rangle$$. From what I learn in quantum mechanics, the measurement will prepare for us the eigenstate of the observable, or more properly the quantum state will collapse to the eigenstate of the observable. Now, since we can interpret $$P_{1} = |0\rangle \cdot 0 \cdot \langle 0| + |1\rangle \cdot 1 \cdot \langle 1|$$, we might say that the measurement $$P_{1}$$ on $$| \psi \rangle = | 0 \rangle$$ will prepare for us still $$| 0 \rangle$$ (up to an overall phase) and we will obtain measurement result "$$0$$", and the probability for this to happen is actually $$1$$, because $$|0\rangle$$ is indeed just one eigenstate of $$P_{1}$$.

(viewpoint 2) However, I also sometimes heard people say that when we perform an measurement, we act the operator of the observable on the quantum state, which I am not sure whether it is a proper statement. This is because if we interpret $$P_{1}$$ as an observation, and act it on the state $$| \psi \rangle = | 0 \rangle$$, we will obtain zero because $$P_{1}| 0 \rangle=0$$. Then from this viewpoint the quantum state will just "disappear" after the measurement, which is weird. (I am considering a closed quantum system so I expect the probability to conserve)

Which of the viewpoints is correct? Thanks.

• also this Aug 16, 2021 at 20:15

However, I also sometimes heard people say that when we perform a measurement, we act the operator of the observable on the quantum state, which I am not sure whether is a proper statement.

Indeed, this is a completely incorrect statement. The correct statement is that the post-measurement state of an initial state $$\vert\psi\rangle$$ is described by (up to normalization) $$\mathbb{P}_i\vert\psi\rangle$$ with a probability $$\langle\psi \vert\mathbb{P}_i\vert\psi\rangle$$ where $$\mathbb{P}_i$$ is the projection operator corresponding to the $$i$$-th distinct eigenvalue of the operator being measured. See my answer here for a detailed discussion on the mathematical representation of the post-measurement state.

In your particular example, the projection operators corresponding to the eigenvalues of the operator being measured are $$O_1=\vert 1\rangle\langle1\vert$$ and $$O_2=\vert0\rangle\langle0\vert$$. So, the probability that the initial state $$\vert 0\rangle$$ will collapse to $$\vert 1\rangle$$ is given by $$\langle \psi\vert O_1\vert\psi\rangle=\langle 0\vert 1\rangle\langle1\vert0\rangle = 0$$ and the probability that it will collapse to $$\vert 0\rangle$$ is given by $$\langle \psi\vert O_2\vert\psi\rangle=\langle 0\vert 0\rangle\langle0\vert0\rangle = 1$$.

P.S.: It is a coincidence here that the projection operator corresponding to one of the eigenvalues of the operator being measured is the same as the operator being measured. This is simply due to the fact that the operator being measured itself is a projection operator. Due to this peculiarity of this example, if the initial state were to be $$\vert 1\rangle$$ then the post-measurement state would have been given by the action of the operator being measured on the initial state. However, I hope that it is clear that this would have been the case not because of a general rule but because of the fact that the operator being measured happens to be equal to one of its eigensubspace projectors.

• The one thing this is missing is that a complete description of a measurement has to have $\sum_i \mathbb{P}_i=\mathbb{I}$: if you start with state $|0\rangle$, you never get a "yes" when your observable is the projector $\mathbb{P}_1$, so the update rule for the state needs to use the projector $\mathbb{I}-\mathbb{P}_1$, which in this case happens to be $|0\rangle\langle 0|=\mathbb{I}-\mathbb{P}_1$. Aug 16, 2021 at 20:19
• @QuantumMechanic I don't understand your point. $\sum_i\mathbb{P}_i=1$ is manifestly true -- both in reality and in my description. I don't exactly understand what you mean by getting a "yes". Aug 16, 2021 at 20:41
• Yes it is all in your description, it just needs that extra step to connect to the original question. OP says a measurement is described by an observable $P_1$, while your answer talks about a single operator with two different eigenoperators. The point is that a projection operator as an observable defines a binary measurement (ie "yes/no") that is easier understood in terms of a set of projectors corresponding to different measurement results. Aug 17, 2021 at 0:08
• Otherwise there is no description of what happens when the initial state is $|0\rangle$ and you measure $P_1$ - does the state disappear? Does the measurement device explode? Does the measurement device give you an error code? Does the state remain the same? Aug 17, 2021 at 0:19
• @QuantumMechanic You're making the same mistake that OP is making. Measuring an operator does not mean acting on the state with the operator. There is a clear description of what happens when you measure $P_1$ on the initial state $\vert 0\rangle$. It collapses to $\vert 0\rangle$ with a probability $1$ because $\vert 0\rangle$ is an eigenstate of the operator being measured, namely, $P_1$. Aug 17, 2021 at 10:00

There are different ways to understand what "measuring an observable" means. In the following, let $$\rho$$ be some state, and $$A$$ some observable.

(Role of eigenspaces and eigenvalues of observables) Practically speaking, "measuring $$A$$" amounts to perform some kind of measurement which collapses $$\rho$$ into one of the eigenstates (or more generally, eigenspaces) of $$A$$. In other words, the eigenspaces of the observable describe the possible outcomes of the measurement. For example, if $$A=\lambda_1 P_1 + \lambda_2 P_2$$, with $$P_1,P_2$$ orthogonal projections, then "measuring $$A$$" amounts to "asking the quantum system" whether its state is in $$P_1$$ or in $$P_2$$. You can think of this as "forcing the system to decide" between one of these possibilities.

Of course, which outcome occurs at any given experimental run is probabilistic. For example, the probability of getting the outcome "$$1$$" when measuring $$A$$ will be $$\operatorname{Tr}(\rho P_1)$$.

It is worth noting that in the above discussion, the eigenvalues of the observable didn't have any role. The eigenvalues are only relevant when you consider the question:

Suppose I perform the measurement corresponding to the eigenspaces of $$A$$, and I note on a piece of paper the number $$\lambda_i$$ every time I find the outcome "$$i$$". Computing the average of the numbers I wrote down, in the limit of many measurements, what result do I get?

In other words, you can think of the eigenvalues as a way to ask a specific question about the outcomes, in (precisely) the same way that classical random variables ask a "specific question" about a probability distribution you are sampling from.

(Post-measurement states) Another thing worth stressing is that there is no notion of post-measurement states in the above discussion. Nor there is any need for it. When one talks about "the expectation value" of an observable $$A$$, this refers to the setting in which the state to be measured $$\rho$$ is reset before each measurement is performed. Whatever happens to the state after the measurement is irrelevant from this point of view. Sometimes it is meaningful to talk about post-measurement states (when you are doing a partial measurement, which only partially collapses the system), sometimes it's not. That won't affect the results of measuring the observable (regardless of what precisely you mean with the term).

(Action of a measurement on a state) Finally, let me remark that there is no obvious physical content in the result of $$A|\psi\rangle$$ for some state $$|\psi\rangle$$. This matrix product should only be understood, in this context, as an intermediate value you get in the process of computing the expectation value $$\langle A\rangle\equiv\langle\psi|A|\psi\rangle$$.

Granted, $$A|\psi\rangle$$ might be a meaningful state in some specific circumstances. In particular, if $$A$$ is unitary (on top of it being Hermitian), or more generally projective unitary, then $$A|\psi\rangle$$ can be considered as also a state (defining ket states as vectors modulo multiplication by a complex scalar). However, you are here considering $$A$$ as an evolution matrix, and this shouldn't be confused with what you do when $$A$$ describes a measurement.

For a more general way to describe measurements with their potential post-measurement states, see the general measurement formalism mentioned e.g. in the Wikipedia page.