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If $$P_{+} = |+\rangle\langle+|=\frac{1}{2}(|0\rangle\langle0|+|0\rangle\langle1|+|1\rangle\langle0| +|1\rangle\langle1|)$$ and $$P_{-} = |-\rangle\langle-|=\frac{1}{2}(|0\rangle\langle0|-|0\rangle\langle1|-|1\rangle\langle0| +|1\rangle\langle1|),$$ then we can choose $\lambda_{+}=1$ and $\lambda_{-}=-1$, so that $\begin {bmatrix}0&1\\ 1&0\end{bmatrix}$ is a Hermitian operator for a single qubit measurement in the Hadamard basis?

My confusion is about what this even means. Surely a measurement in the Hadamard basis simply involves the application of the associated projectors $P_{+}$ and $P_{-}$ to whatever state you possess, with $$\frac{P_{i}|\psi\rangle}{\sqrt{{\rm tr}(P_{i}|\psi\rangle\langle\psi|P_{i})}}$$ giving the new state and $\langle\psi|P_{i}|\psi\rangle$ giving the associated probability of obtaining said state. What does the above operator even do? How is it even applied in the role of measurement?

I just don't see what use the above operator has, beyond maybe making it clear that $|+\rangle\to|+\rangle$ and $|-\rangle\to-|-\rangle$

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  • $\begingroup$ I assume you mean $P_-=|-\rangle \langle -|$, in which case you are exactly right. The operator you mention is the Pauli $X$ operator and it is not only a hermitian, it's the important unitary that maps $|0\rangle \mapsto |1\rangle$. Could you elaborate on what you are confused about? $\endgroup$ Commented Jul 20, 2020 at 16:27
  • $\begingroup$ Yeah, changed that in an edit. I am aware it is the pauli X operator, but what does that mean from the perspective of getting the operator in another basis? What's the point? It gives you no immediate eigenvalues or associated projectors with which to perform a measurement with. Also, did you mean to say it maps $|+\rangle\to|+\rangle$? It maps $|0\rangle\to|1\rangle$ $\endgroup$ Commented Jul 20, 2020 at 16:31

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The reason for this viewpoint on measurement is primarily historical. Physicists often think of measurements in terms of observables (aka hermitian operators). The way this is related to the more mathematical notion of projective measurements (a type of POVM measurement) is by thinking about the spectral decomposition of a hermitian operator $$M=\sum_i \lambda_i P_i,$$ where the $\lambda_i$'s are the real eigenvalues of $M$ (historically the eigenvalues were relevant measurable quantities in some physical experiment, which is the reason why this viewpoint exists) and each $P_i$ is the orthogonal projection onto the corresponding eigenspace.

Now, measuring the observable and obtaining the eigenvalue $\lambda_i$ is analogous to measuring the $i$th outcome of the projective measurement (this projective measurement is exactly the one defined by $\sum_i P_i=I$). Another common viewpoint is to define the expected value of an observable acting on a state by the formula $$E(\lambda_i)=tr(M|\psi\rangle \langle \psi|),$$ where $E(\lambda_i)$ is called the expected value of obtaining the eigenvalue $\lambda_i$.

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You can think of an observable as giving you a more coarse-grained type of information than a projective measurement.

A projective measurement $\{P_i\}$ amounts to asking in which of these states am I going to find the system? The answer to this question is of course probabilistic in general, and thus in practice you would find sometimes the state corresponding to $P_1$, sometimes the state corresponding to $P_2$, etc.

Suppose now you have an observable written as $A=\sum_k \lambda_k P_k$. Compute the expectation value of this observable amounts to doing the above projective measurement $\{P_i\}$, then attaching the value $\lambda_i$ to the $i$-th outcome, and then calculating the average of the values thus obtained. You can easily tell from this that the amount of information you get from $A$ is smaller than that you get from $\{P_i\}$. Indeed, if you know the probabilities of finding the system in some basis, you can calculate the expectation value of any observable diagonal in that basis.

In other words, observables and projective measurements answer different questions. If you want to know the probability of finding some state, you use a projective measurement. But you might only be interested in some average quantity (say, the average position of some particle), in which case it is convenient to describe things in terms of observables.

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  • $\begingroup$ I understand the difference between the two. What I don't understand is what is the point of the above process I listed? Why bother changing the basis of the observable associated to the hadamard states? What is the point? This new operator doesn't even give any eigenvalues, and calculating them of course just brings you back to before you changed basis. Why is this a measurement observable in the hadamard basis when it isn't even spectrally decomposed? $\endgroup$ Commented Jul 20, 2020 at 18:00
  • $\begingroup$ Is the reason that doing so provides a matrix to obtain the expectation value if the basis in question is not that of the eigenvectors that span the hilbert space of the original observable? $\endgroup$ Commented Jul 20, 2020 at 18:07
  • $\begingroup$ the point of using an observable depends on what you are doing. You don't mention any specific application so I'm not sure how to answer that. If you want to measure an expectation value you can use an observable, if you want to know outcome probabilities then you don't need it (although you can always also describe projective measurements in the observable formalism if you want to) $\endgroup$ Commented Jul 20, 2020 at 18:44
  • $\begingroup$ They don't give any specific application. All they state is that they construct a hermitian operator corresponding to measurement of a single qubit in the Hadamard basis. They then expand out the projectors in the computational basis, multiply them by their eigenvalue and then sum them. What I am guessing they mean is that they have taken the observable and changed it's basis representation into the computational basis, and that the use of this operator with any state in the computational basis gives the expectation value. The wording is just... strange :/ $\endgroup$ Commented Jul 20, 2020 at 19:14
  • $\begingroup$ i don't know who "they" are $\endgroup$ Commented Jul 20, 2020 at 19:49

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