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Text books say that when you measure a particle's position, its wave function collapses to one eigenstate, which is a delta function at that location. I'm confused here.

  1. A measurement always have limited accuracy. Does the wave function collapse to exactly a eigenstate no matter what accuracy I have?

  2. When a particle is in an eigenstate of position, I can represent the state in momentum basis, and calculate it's expected value (average) of kinetic energy. This gives me infinity. Can a particle ever be in such a state?

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3 Answers 3

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  1. No, it doesn't collapse to an eigenstate. Collapse to an eigenstate is a picture of an ideal measurement. In general the final state will not be describable by a wave function, because it's not a pure state, it is instead a mixed state. See this question, which is about inexact measurements.

  2. Position eigenstate in position representation is $\langle x_{}|x_0\rangle=\delta(x-x_0)$. This gives the following in the momentum representation: $\langle p_{}|x_0\rangle=e^{\frac{i}{\hbar}px}$. For this function probability density is constant, thus its expectation value is undefined (one can't find a center of infinite line). Similarly, for free particle expectation value of energy will also be undefined. This is because such state is an abstraction, a useful mathematical tool. Of course, such states can't be prepared in real experiment, but one can come very close to it, e.g. shoot an electron at a tiny slit and observe state of the electron at the very exit of that slit.

As to finding expectation value of energy in position eigenstate, first mistake which you make using the formula $\overline E=\langle x|\hat H|x\rangle$ is forgetting to normalize the eigenvector. But position operator has continuous spectrum, which makes all its eigenvectors unnormalizable (i.e. if you try to normalize them, you'll get null vector, which is meaningless as a state). Thus you can't directly find expectation value of energy in position eigenstate.

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It is due to the nature of quantum mechanics.

In classical regime, the lowest possible energy is zero. But in QM, the lowest state(ground state) still has energy. Quantum nature is wave-nature. In CM, you can pinpoint a location of an object but in QM, it is a distribution (probability density). What you can only do is find the smallest distribution not a specific point of an object.

For your question, It is due to wave nature: Uncertainty Principle says that$ \Delta x \: \Delta p \ge \frac{\hbar}{2}$. So when you try to measure a location , $\Delta x$ (deviation in position) becomes zero and to satisfy this relation, your $\Delta p$ has to be really large (like a delta function). So as soon as you do the measurement, you just destroy the wave function.After you do the exact measurement on something in QM, you just destroy that "something" and you cannot say anything about it such as its energy.

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Thanks for the answers. What if I calculate <x|H|x>, where x is an eigenstate of position? Does that make any sense? –  Purui Feb 12 '14 at 6:10
@Anug, I refined my second question a little bit. Can you take a look? Thanks. –  Purui Feb 12 '14 at 7:02
  1. No. Position operator does not have normalizable eigen-functions ($\delta(x-x_0)$ is not normalizable). The closest thing one can do in this formalism is to contract the wave function to some sharp peak with non-zero width and finite height, based on the accuracy of the measurement.

  2. With continuous space, particle cannot be in an "eigenstate of position", because there is no such thing there. On a discrete set of admissible positions this would be possible, but the whole physics and formalism would be very different.

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