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The expectation value of a physical observable in a given state is just a single real scalar quantity. On the other hand, the probability distribution of the eigenvalues of the observable is a set of $N$ numbers, where $N$ is the dimensionality of the Hilbert space. My question is the following: is it possible to measure only the expectation value of an observable without having to measure the entire probability distribution of its eigenvalues?

P.S: I am not looking for answers that make any additional assumptions, or approximations, or that apply only in specific cases.

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  • $\begingroup$ Yes, but only if $N=1$ or the state of the system is an eigenvector of the observable. Otherwise, no. $\endgroup$ – AccidentalFourierTransform Jan 6 '18 at 12:15
  • $\begingroup$ I'm guessing you mean if it is possible to do it more efficiently that what's possible by collecting samples and doing statistics? Otherwise the answer is trivially yes, just use a black box that takes the samples and only gives you the average as output, thus basically allowing you to "measure only the expectation value" without having access to the samples/prob distribution. $\endgroup$ – glS Jan 7 '18 at 10:01
  • $\begingroup$ You guessed it right. $\endgroup$ – Girish Jan 7 '18 at 17:58
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Measuring directly expectation values by a single measurement can be done for macroscopic observables only, and is indeed routinely done for these.

The values recorded for any macroscopic measurements are, according to nonequilibrium statistical mechanics, the ensemble expectations of macroscopic observables such as mass densities, currents or energy densities.

Apart from this, the only situation where single measurements give expectation values without approximation is when an observable is measured in an eigenstate.

In particular, with the caveat requiring not to ''make any additional assumptions, or approximations, or that apply only in specific cases'', specified in the bounty announcement (you should move this to your - in its present form not clearly phrased - question, since it makes an essential difference), the answer of your question is negative.

There cannot be a proof in the absence of a rigorous definition of what a measurement is. Such a definition does not exist. But there is no hint at all in either experiment or theory that it could be possible.

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  • $\begingroup$ Yes, that is right. It works here because the macroscopic system can itself be understood as an ensemble of microscopic systems in an identical state. Consequently, the macroscopic measurement directly yields the expectation value of the observable over the ensemble. However, I am interested in whether fundamentally it is possible to get the expectation value directly while still performing microscopic measurements, and not macroscopic ones. $\endgroup$ – Girish Jan 6 '18 at 11:48
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    $\begingroup$ @Girish: Apart from this, the only situation single measurements give expectation values is when an observable is measured in an eigenstate. $\endgroup$ – Arnold Neumaier Jan 6 '18 at 12:02
  • $\begingroup$ Can it be proved that the answer is negative in general? $\endgroup$ – Girish Jan 7 '18 at 17:57
  • $\begingroup$ @Girish: There cannot be a proof in the absence of a rigorous definition of what a measurement is. Such a definition does not exist. But there is no hint at all in either experiment or theory that it could be possible. $\endgroup$ – Arnold Neumaier Jan 8 '18 at 10:53
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I'm not sure how general of an answer you want, but here's an explicit proof it can't be done for standard measurements of Hermitian operators. Consider an arbitrary operator $A$ and suppose we want to measure its expected value by measuring some other operator $B$.

Now, no matter what state we're in, measuring $B$ should always give us a definite value, the expectation value. This is only possible if every vector is an eigenvector of $B$, with the eigenvalue (the corresponding measured value) equal to $\langle A \rangle$. That is, we want $$B | v \rangle = \langle v | A | v \rangle |v \rangle.$$ We know for sure that if $A$ has a definite value, then $B$ should also have that same definite value, $$A | v \rangle = \lambda |v \rangle \quad \rightarrow \quad B |v \rangle = \lambda |v \rangle.$$ But since $A$ is Hermitian, it has a complete basis of eigenvectors, so this constraint alone completely determines $B$; it must be equal to $A$. So this 'expected value' operator can't exist.

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  • $\begingroup$ Your argument assumes a von Neumann measurement, which applies only for very simple systems. But there are more general measurement schemes in microscopic settings, e.g., POVMs. Moreover, the microscopic setting does not apply to measurements of macroscopic quantities. $\endgroup$ – Arnold Neumaier Jan 7 '18 at 11:24
  • $\begingroup$ Measuring an operator and acting the operator on a state are two distinct operations. For instance, when the position $\hat{X}$ of a state $\int c(x) |x\rangle \mathrm{d}x$ is measured, the system jumps into the state $ |x\rangle $ with probability density $|c(x)|^2$ . On the other hand, when you simply act the operator we get $\hat{X}\int c(x) |x\rangle \mathrm{d}x= \int c(x)x |x\rangle \mathrm{d}x$ , which is still a superposition state. $\endgroup$ – Girish Jan 7 '18 at 17:44
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    $\begingroup$ @Girish I added a bit more explanation of what I was doing. $\endgroup$ – knzhou Jan 8 '18 at 1:54
  • $\begingroup$ I am probably being silly, but what about $B=\langle A\rangle \boldsymbol 1$, with $\boldsymbol 1$ the identity operator? $\endgroup$ – AccidentalFourierTransform Jan 8 '18 at 11:11
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    $\begingroup$ @AccidentalFourierTransform The right-hand side depends on the state but the left-hand side can’t. $\endgroup$ – knzhou Jan 8 '18 at 11:29
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Expectation values are defined

In general, the expectation value for any observable quantity is found by putting the quantum mechanical operator for that observable in the integral of the wavefunction over space:

expectvalu

Seeing a measurment as an eigenvalue

When applied to a general operator Q, it can take the form

eigenvalue

means one is picking an instance under the first integral.

You ask:

is it possible to measure only the expectation value of an observable without having to measure the entire probability distribution of its eigenvalues?

In the case of wavefunctions of potential problems, as for example the solutions of the hydrogen atom, measuring the energy spectrum has very high probability of falling on one of the energy eigenvalues of the solution, because the wavefunctions give very high probability on the eigenvalues of the energy. Thus measuring an energy level identifies the wavefunction of origin without the need of accumulating a probability distribution.

In summary if the wavefunction contains in some variable discrete eigenvalues, measuring those identifies the wavefunction too.

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We can 'measure' the expectation value of an operator by performing repeated measurements of it, and taking the mean. (I am using quotes here because it is more precise to say that we are estimating the expectation value.)

So, suppose we would like to estimate the expectation value $\left< \psi \right| \hat{A} \left| \psi \right>$ of an operator $\hat{A}$ in a state $\left|\psi\right>$. We can repeatedly do the following: prepare the system in the state $\left|\psi\right>$, and measure $\hat{A}$. If we do this $n$ times, we will obtain $n$ measurement results $A_1,\dots,A_n$. Here each $A_k$ is a real number -- the result of measurement $k$ of the operator $\hat{A}$. The mean of the measurements is an estimate of the expectation value. For large $n$, the law of large numbers tells us that we should expect \begin{align} \left< \psi \right| \hat{A} \left| \psi \right> \approx \frac{1}{n} \sum_{k=1}^n A_k \,. \end{align}

So yes, one can estimate the expectation value without measuring the full distribution.

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    $\begingroup$ Thanks for the response. However, the procedure you have described actually yields the probability distribution. One can choose to not infer the distribution, but operationally this procedure is no better than the standard von Neumann measurement in the observable's eigen-basis. $\endgroup$ – Girish Jan 4 '18 at 8:38
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    $\begingroup$ Only if we take $n$ very large (assuming $N$ is large to begin with). In practice we take $n$ not very large and get a reasonable estimate of the expectation but not of the whole distribution. (Of course we get some information about the distribution, just not enough to reconstruct the whole thing.) This is what happens in real experiments. $\endgroup$ – Guy Gur-Ari Jan 4 '18 at 13:52

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