# Repeating a measurement vs uncertainty

The wikipedia says on measurement in quantum mechanics that:

Repeating the same measurement without any evolution of the quantum state will lead to the same result.

On the other hand, doesn't uncertainty (in momentum) entail that I can't expect to measure the same position of a particle twice?

EDIT - I found the following excerpt by Feynman in the second chapter of the first volume of his lectures; it seems related to the question and possibly at odds with some of the answers:

What keeps the electrons from simply falling in? This principle: If they were in the nucleus, we would know their position precisely, and the uncertainty principle would then require that they have a very large (but uncertain) momentum, i.e., a very large kinetic energy. With this energy they would break away from the nucleus. They make a compromise: they leave themselves a little room for this uncertainty and then jiggle with a certain amount of minimum motion in accordance with this rule. (Remember that when a crystal is cooled to absolute zero, we said that the atoms do not stop moving, they still jiggle. Why? If they stopped moving, we would know where they were and that they had zero motion, and that is against the uncertainty principle. We cannot know where they are and how fast they are moving, so they must be continually wiggling in there!)

## 3 Answers

The Projection Postulate states that if we have an observable $O$ with discrete spectrum $\{\lambda_i\}$, after a measure in a system resulting in the eigenvalue $\lambda_a$, the initial system $|\psi_i\rangle$ is reducted to the state $$|\psi_f\rangle=\frac{P_a|\psi_i\rangle}{\|P_a|\psi_i\rangle\|}$$ where $P_a$ is the projection on the eigenspace of the eigenvalue $\lambda_a$. Any repeated measurement of $O$ in this state without time evolution will yield the same result $\lambda_a$, since $$O|\psi_f\rangle=\frac{OP_a|\psi_i\rangle}{\|P_a|\psi_i\rangle\|}=\frac{\lambda_aP_a|\psi_i\rangle}{\|P_a|\psi_i\rangle\|}=\lambda_a|\psi_f\rangle$$ i.e., $|\psi_f\rangle$ is an eigenvector of $O$ with eigenvalue $\lambda_a$.

In the case of continuous spectrum, there is some complications, since the eigenvectors are not normalizable and hence are not acceptable states. But as the measurement of an exact value of position, for example, cannot be made with arbitrary precision, what is really being measured is the projection over some region of the spectrum, like the dimension of the detector. In that case, say the particle is measured in the region $[a,b]$. By the projection postulate, the final state will be (in position representation, ignoring the normalization): $$\psi_f(x)=P_{[a,b]}\psi_i(x)=\chi_{[a,b]}\psi_i(x)$$ where $\chi_{[a,b]}$ is the characteristic function of $[a,b]$. That is, the final wavefunction will be the restriction of the initial one to the interval $[a,b]$. Realize that we still have an uncertainty in the momentum as expected, but if the system does not evolve with time, any other measurement of position, will still be restricted to $[a,b]$, since the final wavefunction support is in that region.

• but, as we increase the accuracy of measurement of position, we also increase the uncertainty of momentum, and therefore shouldn't we expect it to involve a change in the position of the particle, to the extent that it may not be restricted to the interval [a,b]? – nir Nov 1 '14 at 18:07
• I guess you are thinking too classically. While you are correct in your statement that the uncertainty of the momentum will be increased as we increase the precision, we are talking about successive measurements without time evolution. That being said, while it's true that if we let the system evolve, we would have a probability of finding the particle outside $[a,b]$, that can't happen if there's no time for the system to evolve. – Mateus Sampaio Nov 3 '14 at 23:13

What you are missing is the context of that statement. In the sentences that precede it, you are told that the the system has been prepared, and then a measurement made, resulting in the collapse of the wave function into an eigenstate. The measured property is the eigenvalue of that eigenstate.

Any further measurements of that property will result in the same eigenvalue. The system has collapsed into that eigenstate, and further measurements of that property will not cause the system to move from that state.

The complete uncertainty that you have in momentum guarantees that you can know the position exactly.

• But, isn't momentum related to a change in position? if the uncertainty in momentum entails it may be non zero, then how can we expect the same measurement of position again? – nir Nov 1 '14 at 18:03
• Your notion of momentum might be a bit too classical. Momentum does imply a change in position. But if the particle has a definite position, the uncertainty in the momentum is unknown. Perhaps a better way to look at it is to recognize that the particle is in a state that is a superposition all possible momenta, positive and negative. The average momentum is zero. – garyp Nov 1 '14 at 18:43

Wait, are you sure you have understood the uncertainty principle correctly? It doesn't rule out the possibility of measuring the same value of position twice. The uncertainty principle only relates how much of an uncertainty in measured momentum you're gonna get if you know the position to a known precision.

But it is one of the important results of quantum mechanics, that measurement will collapse a wavefunction into one of its eigenstates, so that further measurements will yield the same values.

• I am sure that I still don't understand it correctly, since I have only recently begun studying this (on my own). – nir Nov 1 '14 at 18:08
• If you're getting confused, go back to the fundamental statement. $\Delta x \Delta y \geq \frac{h}{4\pi}$ and remember that $\Delta A$ is the standard deviation in the measurement of A, defined as $\sqrt{<A^2> - <A>^2}$ where $<A>$ and $<A^2>$ are the expected values of $A$ and $A^2$ respectively – quarkleptonboson Nov 1 '14 at 19:16
• doesn't this mean that you cannot have the same exact measurement of position a second time, since that would entail momentum is zero, in contradiction to the uncertainty principle? – nir Nov 1 '14 at 19:42
• oh right, i should repeat my emphasis on the fact that once measurement has been made, the eigenfunction collapses. When this happens, you lose the wave behavior of the particle, resulting in that particle having a definite position and momentum. (The uncertainty principle depends on wave behavior) – quarkleptonboson Nov 3 '14 at 17:42
• Heisenberg: "It is impossible to determine accurately both the position and the direction and speed of a particle at the same instant" - en.wikipedia.org/wiki/Uncertainty_principle#History – nir Nov 3 '14 at 20:29