What is the correspondence between operators and measurement in quantum mechanics?

Question

Leonard Susskind, in his "Quantum Mechanics: The Theoretical Minimum" says the following

The correspondence between operators and measurement is fundamental in quantum mechanics. It is also very easy to misunderstand.......When an operator acts on a state vector it produces a new state-vector.......The measurement is some kind of operation that apparatus does to the system but that operation is no way the same as operating on the state vector with operator $L$.......If the state of the system before measurement is $|A\rangle$, it is not correct to say that measurement of $A$ changes the state to $L|A\rangle$

If it is so then, when and what triggers the $L$ operator to be applied to state $|A\rangle$?

Answer 1

To simplify things, consider the Hilbert space $\mathcal H = \mathbb C^3$. Vectors in this space consist of triples of complex numbers, and linear operators take the form of $3\times 3$ matrices. For a physical interpretation, this would be the appropriate space for modeling a spin-1 particle which is fixed in place.

Consider the operator $\hat A$ given by $$\hat A = \pmatrix{6 & 0 & 0 \\ 0 & -8 & 0 \\ 0 & 0 & -8}$$ Note that I can decompose this matrix as follows:

$$\hat A = \pmatrix{6 & 0 & 0 \\ 0 & -8 & 0 \\ 0 & 0 & -8}= 6 \pmatrix{1 & 0 & 0 \\0&0&0\\0&0&0} + (-8) \pmatrix{0 &0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1}\equiv \lambda_1 \hat P_1 + \lambda_2 \hat P_2$$ where $\lambda_1=6$, $\lambda_2 = -8$, and $\hat P_1$ and $\hat P_2$ are the corresponding matrices. The set $\{\lambda_1,\lambda_2\}$ is the set of eigenvalues of $\hat A$, and $\hat P_1$ and $\hat P_2$ are the orthogonal projection operators onto the corresponding eigenspaces. Note also that $\sum_i \hat P_i = \mathbb I$, where $\mathbb I$ is the $3\times 3$ identity matrix, and that $\hat P_i$ are all hermitian (trivially, in this case).

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If $\hat A$ is the operator corresponding to some physical observable, then a measurement of that observable must return a result in the set $\{\lambda_1,\lambda_2\}$ - in this case, either $6$ or $-8$. Let's say the system is in the state $$|\psi\rangle = \pmatrix{1/\sqrt{2}\\ i \\ 1/\sqrt{2}}$$

which, for the sake of generality, I haven't bothered to normalize. The probability of obtaining $\lambda_1=6$ as the result of a measurement is given by the following:

$$\mathrm{Prob(\hat A,\lambda_1)} = \frac{\langle \psi |\hat P_1 | \psi\rangle}{\langle \psi|\psi\rangle} = \frac{1/2}{2} = \frac{1}{4}$$ while $$\mathrm{Prob(\hat A,\lambda_2)} = \frac{\langle \psi |\hat P_2 | \psi\rangle}{\langle \psi|\psi\rangle} = \frac{3/2}{2} = \frac{3}{4}$$

The fact that these probabilities add to 1 is no accident; as you can clearly see,

$$\sum_i \mathrm{Prob(\hat A,\lambda_i)} = \sum_i\frac{\langle \psi |\hat P_i | \psi\rangle}{\langle \psi|\psi\rangle} = \frac{\langle \psi |\sum_i\hat P_i | \psi\rangle}{\langle \psi|\psi\rangle}=\frac{\langle\psi|\psi\rangle}{\langle\psi|\psi\rangle} = 1$$

The state of the system after the measurement depends on what the measurement result was. If we obtain $\lambda_1=6$ as a result, then the post-measurement state is given by $$P_1|\psi\rangle = \pmatrix{1/\sqrt{2}\\0\\0}$$ whereas if we obtain $\lambda_2=-8$, the post-measurement state will be $$P_2|\psi\rangle = \pmatrix{0 \\ i \\ 1/\sqrt{2}}$$

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In summary, the recipe goes as follows. Given an operator $\hat A$ which represents a physical observable, the eigenvalues of $\hat A$ (more generally, the spectrum, but nevermind that now) correspond to the possible measurement results. If we decompose $\hat A = \sum_i \lambda_i \hat P_i$ such that each $\lambda_i$ is distinct, then the hermitian projection operators $\hat P_i$ tell us (i) how to compute the probability of measuring $\lambda_i$ given some initial state $|\psi\rangle$, and (ii) what the post-measurement state of the system will be if our measurement yields $\lambda_i$ as a result.

In the example I've given here, the eigenvalues $\lambda_i$ and corresponding projection operators $\hat P_i$ were fairly obvious. It is a reasonable to ask under what circumstances it is possible to decompose $\hat A$ in this way - as the sum of projection operators scaled by eigenvalues - and it turns out that this is possible if and only if $\hat A$ is normal, meaning that $[\hat A,\hat A^\dagger]=0$ where $\dagger$ means conjugate transpose.

If we also require that the eigenvalues $\lambda_i$ are real (which is a reasonable requirement for physical observables such as energy or angular momentum), then we must have that $\hat A=\hat A^\dagger$ - i.e. $\hat A$ is hermitian. This is why hermitian operators are of central importance in quantum theory.

Finally, I should note that for infinite-dimensional Hilbert spaces like $L^2(\mathbb R)$, this nice simple picture gets very complicated and technical because operators can have continuous spectra (rather than just a discrete set of eigenvalues). This leads to an enormous increase in technical sophistication and a very deep and interesting theory of measurement which generalizes the one I've presented here. That being said, not too much changes in spirit from the much simpler finite-dimensional case.

Answer 2

My reading (without context) of Susskind's statement is that a measurement requires an interaction between two systems, so describing a measurement as the evolution of a single system may miss some of the details.

For example, one system can be in the state $|A\rangle_{\mathrm{system}}$ and the other can be the measurement device in some state $|B\rangle_{\mathrm{m.\,device}}$. The desired, evolution equation is $$|A\rangle_{\mathrm{system}}\otimes|B\rangle_{\mathrm{m.\,device}}\to L|A\rangle_{\mathrm{system}}\otimes |B\rangle_{\mathrm{m.\,device}},$$ which must come about through an interaction between the two systems. One way of achieving that is via an interaction with an interaction Hamiltonian of the form $$H=J\otimes M\quad \Rightarrow\quad e^{-iHt/\hbar}=e^{-i\frac{t}{\hbar}J\otimes M}.$$ In that case, the joint system evolves under the Schrödinger equation as $$|A\rangle_{\mathrm{system}}\otimes|B\rangle_{\mathrm{m.\,device}}\to e^{-i\frac{t}{\hbar}J\otimes M}\left(|A\rangle_{\mathrm{system}}\otimes |B\rangle_{\mathrm{m.\,device}}\right).$$ Depending on the state of the measurement device, a different thing will happen to the system state! In general, the two systems will become entangled, and so the first system may seem to be in a mixed state when you inspect it on its own.

We can expand the measurement device's state in the eigenbasis of the operator $M$, with eigenvalues $\lambda_m$ and eigenstates $|\phi_m\rangle$ by using some normalized set of coefficients $b_m$: $$|B\rangle_{\mathrm{m.\,device}}=\sum_{m}b_m |\phi_m\rangle_{\mathrm{m.\,device}}.$$ Then, the overall interaction will look like \begin{align}|A\rangle_{\mathrm{system}}\otimes\sum_{m}b_m |\phi_m\rangle_{\mathrm{m.\,device}}\to &\sum_{m}b_m e^{-i\frac{t}{\hbar}J\otimes M}\left(|A\rangle_{\mathrm{system}}\otimes |\phi_m\rangle_{\mathrm{m.\,device}}\right) \\ &=\sum_{m}b_m e^{-i\frac{t}{\hbar}J\lambda_m}\left(|A\rangle_{\mathrm{system}}\otimes |\phi_m\rangle_{\mathrm{m.\,device}}\right) \\ &=\sum_{m}b_m e^{-i\frac{t}{\hbar}J\lambda_m}|A\rangle_{\mathrm{system}}\otimes |\phi_m\rangle_{\mathrm{m.\,device}} .\end{align}

How do we get a single operator from this? If the measurement device is measured to be in a particular state $|\phi_n\rangle$, we know immediately that the overall system has "collapsed" to the state $$ |\Psi_{\mathrm{final}}\rangle\propto b_n e^{-i\frac{t}{\hbar}J\lambda_n}|A\rangle_{\mathrm{system}}\otimes |\phi_n\rangle_{\mathrm{m.\,device}}. $$ This looks like the system underwent the desired transformation with $$L=\exp\left(-i\frac{t}{\hbar}J\lambda_n\right).$$

We required an interaction for the measurement to actually effect a change on the system.