Studying on own quantum mechanics I came across:

Preceeding text: A basic postulate of quantum mechanics tells us how to set up the operator corresponding to a given observable. Observables, $\Omega$, are represented by operators, $\hat\Omega$, built from the following position and momentum operators

$$\hat x=x\times \;,\qquad \hat p_x=\frac {\hbar}i\frac d{dx}.$$

How are they given? I think they are been postulated, but how? Also related question is why the eigenvalue of the operator corresponding to an observable is the value of the observable?


2 Answers 2


I suspect your text is taking $$\hat x=x\times \;,\qquad \text{ and } \qquad \hat p_x=\frac {\hbar}i\frac d{dx},$$ as postulates (it only holds in the Schrödinger Picture and with the Position Representation). And what it is saying is that it expects you to take any other observable $O$ and write it as a function of $t$, $x$, $p_x$, etcetera and replace every $x$ with a $\hat x$, and every $p_x$ with a $\hat p_x,$ etcetera to get an operator $\hat O$.

Obviously the order in which you write your observable $O$ as a function of $x$ and $p_x$ and such matters because you can get different operators $\hat O$ for different choices. And sometimes the operator you get won't even be self-adjoint or even hermitian. So really it only works for some observables and have to do it in a very specific way. And most very elementary texts don't want to get into it and think its OK to give a false sense of generality. Then you can feel good about yourself and imagine that you know how to make operators for anything you want and can get back to focusing on the next section of the book.

If you want to derive the results $$\hat x=x\times \;,\qquad \text{ and } \qquad \hat p_x=\frac {\hbar}i\frac d{dx},$$ then you have to make some other assumptions. But since it only holds in some pictures and some representations, maybe it isn't that important.

Then you asked a second question. Why is the eigenvalue of the operator corresponding to an observable the value of the observable? Your other result only held in the Schrödinger Picture and the Position Representation . Well, this result (about eigenvalues) doesn't hold for weak measurements. The whole idea is that a strong measurement, when repeated yields the same result. An eigenvector is very nice for that because before and after acting on an eigenvector you have (up to a scalar multiple) the same thing. So as long as you have different eigenvectors behave differently from each other but identically each time, then you can get that desirable (defining?) feature of a strong measurement. But many measurements don't send each eigenvector to someplace different. And they can't totally separate them because any dynamical development has to be linear, so if you have a whole 2d space of eigenvectors one you've sent two eigenvectors $v$ and $w$ that aren't both on the same line somewhere, everything else in the plane spanned by the two vectors is determined where it has to go, so ones that start out close to $v$ of those have to end up close to where $v$ went. These continuously close together nonzero eigenvectors have the same eigenvalues. So you are forced to some (some) eigenvectors with the same eigenvalue close together. One way to set up a strong measurement is to have it treat each eigenvector differently based solely on the different value of the eigenvalue. Then all the ones in the same eigenspace get operated on in the same way. Problem solved.

So one option is to just say that a measurement separates eigenvectors with different eigenvalues. For instance if a device deflects one eigenvector left and another one right then you can use multiple copies of the device and see that once deflected left the result is then always deflected left by the same machines. This reliability across reproduction at different times and places is a nice thing, and you can call that reliability the measurement outcome.

So in that sense it doesn't need a value per se. However if you actually described your particular machine to saw how quickly or how far it deflected something then you can relate it to a numerical eigenvalue. You could even have many detectors in a row and associate the value with the position of the detector that goes off. So there are senses where the numerical values make empirical sense.

Finally, if you assign the different outcomes the numerical value of the eigenvalue, then you can compute the expectation value of the operator (multiply each outcome by the frequency it happened and add them all up to get a sample average and compare that to a theoretical population average). if you do that, then the theoretical population average can be compared at the inner product of the original vector with the vector after if is acted on by the operator divided by the inner product of the vector with itself (that gives the eigenvalue for each eigenvector and a weighted average for the eigenvalues for a general vector so the relative weight of the linear combination is now related to the frequency weighing in the weighted average).

Obviously you can assign whatever labels you want to your outcomes, but if you assign the eigenvalues, then this simple computation "vector inner producted with the result after operator acts on vector divided by vector inner producted with itself" gives you the thing you wanted.


You can find derivation of these operators in most standard quantum mechanics textbooks. For your convenience, see https://en.wikipedia.org/wiki/Momentum_operator and https://en.wikipedia.org/wiki/Position_operator.

For the second question, Paul Dirac said in his classic The Principles of Quantum Mechanics:

A measurement always causes the system to jump into an eigenstate of the dynamical variable that is being measured, the eigenvalue of this eigenstate belongs to being equal to the result of the measurement.

The answer to this question does depend on interpretation of quantum mechanics. (See https://en.wikipedia.org/wiki/Measurement_in_quantum_mechanics)

  • $\begingroup$ (second part)is it a fact /postulate or proved/established theorem? $\endgroup$
    – RE60K
    Jul 29, 2014 at 10:16
  • 1
    $\begingroup$ @ADG No experiment can definitively tell us what happened when we weren't looking, only that it must be consistent with what we saw when we did look. And there are things called weak measurements that give results other than the eigenvalues. The poster already said their statement depends on interpretation, hence we must conclude it isn't really a science question. So not a theorem or a fact. I'll go as far as saying that people that talk about instantaneous quantum jumps want to falsely pretend they can do interactions faster than is possible. $\endgroup$
    – Timaeus
    May 2, 2015 at 21:16

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