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In class, it was mentioned that beginning students often mix up the application of an operator to a state with taking a measurement of that state, and that this is not the case. What's more, the TA said that it has nothing to do with taking a measurement. This confuses me.

For this, you can make all the assumptions one would ordinarily make in an intro class (the operator is Hermitian and unitary, for example).

So let's say the operator is $$\sigma_x$$ and the state is $$|0\rangle + |1\rangle$$. What is $$\sigma_x\left[|0\rangle + |1\rangle\right]$$. If it has nothing to do with measurement, why do we calculate it at all? Is it just an intermediary step to calculating the expectation value of the observable, $$\langle\psi|\sigma_x|\psi\rangle$$?

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Both quantum gates and measurements can be represented as matrices. Every quantum gate corresponds to a unitary matrix, and every measurement corresponds to a Hermitian matrix. Some matrices happen to be both unitary and Hermitian, which is probably the source of your confusion. But if a matrix $U$ is both unitary and Hermitian, the process of applying the transformation $U$ to a state $\lvert\psi\rangle$ is not the same as performing the corresponding measurement on $\lvert\psi\rangle$.

Here are some differences between these two actions, described in terms of your example, the Pauli $X$ matrix $U=\sigma_x$, which happens to be both unitary and Hermitian. One interesting fact about this operator is that it inverts itself - that is, given any state $\lvert\psi\rangle$, applying the operator $\sigma_x$ twice to this state will give back the original state, i.e. $\sigma_x^2 \lvert\psi\rangle = \lvert\psi\rangle$ or $\sigma_x^2 = I$. More generally, any quantum gate describes a bijection on the space of qubits $\mathbb C^2$, so that it may not be its own inverse, but it will always have an inverse.

On the other hand, consider the process of performing the measurement described by the matrix $\sigma_x$, which makes sense because it is Hermitian. This matrix has eigenvectors $\lvert +\rangle$ and $\lvert -\rangle$ with respective eigenvalues $+1,-1$. This means that when we measure an unknown state $\lvert\psi\rangle$ in this basis, the resulting state will always be either $\lvert +\rangle$ or $\lvert -\rangle$, and the eigenvalue, which represents the information we receive from this measurement, tells us which of these two states the system is now in. But since $\lvert\psi\rangle$ could have been in any of infinitely many different states to begin with, after performing this measurement, we cannot return the system to its original state. Also, depending on the original state $\lvert\psi\rangle$ of the system, the outcome of this measurement may be inherently probabilistic. Another thing to keep in mind is that we could just as easily represent the same measurement by a different matrix that isn't unitary, as long as it has the same basis of eigenvectors. For instance, the matrix $$\frac{I+\sigma_x}{2} = \begin{bmatrix}\tfrac{1}{2} & \tfrac{1}{2} \\ \tfrac{1}{2} & \tfrac{1}{2}\end{bmatrix}$$ has precisely the same eigenvectors, $\lvert +\rangle$ and $\lvert -\rangle$, this time with respective eigenvalues $1$ and $0$. This measurement is effectively the same as the measurement described by $\sigma_x$, except that in this case an observed eigenvalue of $0$ (rather than $-1$) indicates that the system has ended up in state $\lvert -\rangle$. But notice that this matrix is not even unitary, and therefore does not correspond to any quantum gate. So the fact that $\sigma_x$ describes both a unitary operator and a measurement is kind of a fluke.

To sum up, applying a quantum gate can be described as a bijective linear transformation on the state space. Applying a measurement cannot be described by a linear transformation, and in fact cannot really be described by a function at all, since its outcome is probabilistic.

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  • $\begingroup$ Thank you so much for this thorough answer! $\endgroup$ Mar 14, 2023 at 9:32

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