One of the basic rules of Quantum mechanics is that after a measurement of an observable, the wavefunction is an eigenstate and any subsequent measurement will give the same result. This is not so in case of coordinate (x) measurement, because of the Heisenberg uncertainty or because the interaction used to specify (x) brings uncontrollable impulse to the particle. What is this mismatch due to?


[...] and any subsequent measurement will give the same result.

This is only the case if the operator commutes with the Hamiltonian of the system. Otherwise, time evolution will take the system out of the eigenstate and subsequent measurements will not give the same answer.


This can be seen easily from the Schrödinger equation $$ i\frac{\partial}{\partial t} |\psi\rangle = \hat{H} |\psi\rangle$$ which has the solution $$ |\psi(t)\rangle = e^{-i\hat{H}t} |\psi(0)\rangle.$$ So if at $t=0$ the system is in the eigenstate $|\psi(0)\rangle =|a\rangle$ of the operator $\hat{A}$ with eigenvalue $a$, it will for $t > 0$ be $$ |\psi(t)\rangle = e^{-i\hat{H}t} |a\rangle.$$ Applying the operator $\hat{A}$ again at time $t$ gives $$ \hat{A}|\psi(t)\rangle = \hat{A} e^{-i\hat{H}t} |a\rangle = e^{-i\hat{H}t} \hat{A} |a\rangle + [e^{i\hat{H}t},\hat{A}]|a\rangle = a |\psi(t)\rangle + [e^{i\hat{H}t}, \hat{A}]|a\rangle.$$ Clearly, if $[\hat{H},\hat{A}]$ = 0, the later application of $\hat{A}$ will give the same answer. But if not, this will generally not be the case.

  • $\begingroup$ The rule can be read here vergil.chemistry.gatech.edu/notes/quantrev/node20.html and it reads: Postulate 3. In any measurement of the observable associated with operator A, the only values that will ever be observed are the eigenvalues a, which satisfy the eigenvalue equation Aw=aw (w = wavefunction). If the system is in an eigenstate of A with eigenvalue a, then any measurement of the quantity A will yield a. There is nothing said about your proposal? If the system is in an eigenstate of $\hat{A}$ with eigenvalue $a$, then any measurement of the quantity $A$ will yield $a$ $\endgroup$ – Mercury Sep 8 '17 at 22:22
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    $\begingroup$ Yes, a measurement of $a$ at time $t$ means the system is in the eigenstate of $\hat{A}$ corresponding to the eigenvalue $a$ at time $t$. But that does not mean it will stay in this eigenstate for times $t_1 >t$. This is only the case if $\hat{A}$ commutes with the Hamiltonian. $\endgroup$ – noah Sep 9 '17 at 0:29
  • $\begingroup$ Thank you very much Noah! But would you be so kind to cite a book or a site where in the 3rd principle is explicitly stated that it is true only when [A,H]not=0. $\endgroup$ – Mercury Sep 10 '17 at 12:43
  • $\begingroup$ By the way [p,H] not=0 also. Does it mean that p changes for a single body (e.g. 1 Newton principle not true in QM)? $\endgroup$ – Mercury Sep 10 '17 at 18:40
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    $\begingroup$ I do not have a reference at hand, but you should be able to find one yourself. I also edited the answer, so it should be clear from that. How did you get $[\hat{p},\hat{H}] \neq 0$? The free particle Hamiltonian is $\hat{p}^2/2m$, which clearly commutes with $\hat{p}$. If you have a potential $V(\hat{x})$, this will not commute with $\hat{p}$, but is also equivalent to a force acting on the particle, thus changing the momentum. $\endgroup$ – noah Sep 10 '17 at 21:42

There is no such mismatch. If you ever manage to measure the position to an infinite precision, it will project it onto a position eigenstate $|x\rangle$; this is required by the uncertainty principle to have infinite momentum uncertainty, which among other things will cause the position uncertainty to begin growing immediately after the measurement, but does not introduce any inconsistencies into the theory.

Of course, in physical measurements we can only ever perform experiments with finite precision, which need to be modelled by the application of a suitable projector such as $$ \hat \Pi_{(x_0-\Delta x/2, x_0+\Delta x/2)} = \int_{x_0-\Delta x/2}^{x_0+\Delta x/2} |x\rangle \langle x| \: \mathrm dx $$ to the state. This will produce a state with a small-but-finite position uncertainty, and therefore with a large-but-finite momentum uncertainty, and all traces of unphysicality are removed from the situation.

It's also worth emphasizing that your understanding of the measurement postulate,

the wavefunction is an eigenstate and any subsequent measurement will give the same result

is rather incomplete. If one performs a measurement of an observable $A$ and obtains the result $A=a$, then subsequent measurements of $A$ performed after a negligible time interval will also yield $A=a$. However, if the second measurement happens after some finite time interval $\Delta t$ and $A$ is not preserved by the time evolution (or, in technical language, $A$ does not commute with (is incompatible with) the hamiltonian) then that ceases to be true.

For the case of the position of a massive particle, the observable $A=x$ is incompatible with the hamiltonian $H=p^2/2m$, so if you let $H$ act for any length of time at all (i.e. if you propagate for any finite time) on a position eigenstate, or indeed after projecting with finite precision with $\hat \Pi_{(x_0-\Delta x/2, x_0+\Delta x/2)} $ as above, then the "uncontrollable impulse" (or, with the finite-precision measurement, the finite-but-large impulse) brought on by the measurement will start to act and the state will spread out.

However, as above, if the two measurements are performed in quick succession, i.e. if there is not enough time for the hamiltonian to act, then the first measurement result will be preserved.

  • $\begingroup$ I cant understand you. You say there is no mismatch but in fact you repeat exactly what the mismatch is: the postulate in the case of x measurement is not true! $\endgroup$ – Mercury Sep 10 '17 at 19:02

The postulate as stated is correct, and you are also correct. Here's what you're saying:

If I measure $x$ and get into an eigenstate of $\hat{X}$, then there is huge momentum uncertainty. So if I come back ten seconds later and measure $x$ again, it will probably be in a totally different place!

Here is what the postulate is saying:

If I measure $x$ and get into an eigenstate of $\hat{X}$, and then immediately (no waiting) measure $x$ again, I'll get the same result.

There is no contradiction here. It's true that a particle with a well-defined position has a huge momentum uncertainty, but that only means that in the future the wavefunction will spread out. The postulate isn't saying anything about what happens in the future, it's talking about what would happen if you measure two times in a row with no elapsed time in between.


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