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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 $|A\rangle$?

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    $\begingroup$ In general nothing. If that is an arbitrary state it doesn't mean anything physical to act on it with an observable. $\endgroup$
    – Charlie
    Jul 6, 2021 at 18:32
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    $\begingroup$ Ah no, absolutrly not. The eigenvalues of the operators dictate the possible outcomes of measurement. $\endgroup$
    – Charlie
    Jul 6, 2021 at 18:37
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    $\begingroup$ Apologies my "in general nothing" comment was directed at the last line of your question, not the title. $\endgroup$
    – Charlie
    Jul 6, 2021 at 18:38
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    $\begingroup$ As I said in general you dont get anything interesting, if $|A\rangle$ is an eigenstate of $L$ on the other hand the action of $L$ is to multiply the state by the associated eigenvalue which is the value you would get if you measured $L$ and found the system in the state $|A\rangle$. $\endgroup$
    – Charlie
    Jul 6, 2021 at 18:43
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    $\begingroup$ Yes, basically. Note that a general spin state doesn't have a spin associated to it, only the spin eigenstates do $\endgroup$
    – Charlie
    Jul 6, 2021 at 19:09

2 Answers 2

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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).


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 $$P_2|\psi\rangle = \pmatrix{0 \\ i \\ 1/\sqrt{2}}$$


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.

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  • $\begingroup$ While I like everything in this answer, I feel as though it was not Susskind's main intended point - I read it with the emphasis on "some kind of operation that apparatus does to the system" - thoughts? $\endgroup$ Jul 6, 2021 at 20:27
  • $\begingroup$ @QuantumMechanic Some additional context likely would have clarified this. On p. 81 of this book (which I found via Google), the paragraph partially quoted by the OP includes the line "It is often thought that measuring an observable is the same as operating with the corresponding operator on the state," and "[...] if the state of the system before we do the measurement is $|\psi\rangle$, it is not correct to say that measurement of $\mathbf L$ changes the state to $\mathbf L|\psi\rangle$." $\endgroup$
    – J. Murray
    Jul 6, 2021 at 20:35
  • $\begingroup$ I see, your answer what I got from reading the full context $\endgroup$ Jul 6, 2021 at 20:49
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    $\begingroup$ @alpal The point you're missing is that $|\psi\rangle=\pmatrix{1/\sqrt{2} \\ 0}$ and $\sqrt{2}|\psi\rangle = \pmatrix{1\\0}$ represent precisely the same state. They are different vectors, but the state of a system is only defined up to multiplication by some nonzero constant. It is convenient to represent the state of the system with normalized vectors for which $\langle \psi|\psi\rangle=1$ simply because the denominator of the $\mathrm{Prob}$ expressions is equal to $1$, but this is not necessary (which I demonstrated by explicitly using a non-normalized vector for the initial state). $\endgroup$
    – J. Murray
    Jul 7, 2021 at 2:26
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    $\begingroup$ In other words, having obtained $P_1|r\rangle = \pmatrix{1/\sqrt{2}\\0}$, you are free to re-normalize by multiplying this state by $\sqrt{2}$ to obtain $\pmatrix{1\\0}$ before proceeding with any further calculations. This is a convenient step, but it is by no means a necessary one. $\endgroup$
    – J. Murray
    Jul 7, 2021 at 2:28
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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.

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  • $\begingroup$ Mechanics Thank you very much for your thorough and advanced answer. Unfortunately I can not choose more than one answer as the accepted answer otherwise I would have chosen yours as well, although yours is way more advanced and it will take me some time to digest and figure it out. $\endgroup$
    – al pal
    Jul 6, 2021 at 20:29
  • $\begingroup$ No problem - in the current context of the book, the answer you accepted is correct - mine is what you will learn about when you start thinking about more than one quantum system interacting with each other. $\endgroup$ Jul 6, 2021 at 20:50
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    $\begingroup$ Yes I understand. In fact I had started thinking about the relation between measurement and entanglement and thought when system A measures system B, it has to somehow get entangled with it. Now you provided some material I can use to study such. $\endgroup$
    – al pal
    Jul 6, 2021 at 21:47

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