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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?

Purui
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4 Answers4

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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.

Community
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Ruslan
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  • Am I understanding you correctly, that te wave function has truly collapsed into a single eigenstate, and that the density matrix just represents our classical uncertainty given the limited resolution of our measurement?

    Or, put another way, if we measure a particle's momentum, the position is truly delocalized across all of space? This seems unphysical and inconsistent with what we experience day-to-day. This explanation is also inconsistent with https://physics.stackexchange.com/questions/301223/after-measuring-momentum-it-seems-like-the-particles-position-could-be-literal .

     – Dragonsheep Mar 06 '20 at 08:18
  • @Dragonsheep this is not inconsistent once you consider Heisenberg uncertainty principle and compare day-to-day measurement errors to the quantum uncertainties. The latter ones will be dwarfed as to be irrelevant in daily life. – Ruslan Mar 06 '20 at 08:39
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The wave function is "reduced", meaning that there is a reduction in size of the continuous range of states (positions) that have non-zero probability. However, it never goes to being a single eigenstate, due to the quantum uncertainty of the probe used in the measurement or of the measurement apparatus, itself. That uncertainty can never go to exactly zero.

Position measurements are of the unfortunate breed that do not refine with successive measurements, due to the back action of the position measurement on the momentum, which, in turn, affects the position that was just measured. That is, measuring the position demolishes the position that was just measured. On the other hand, measurements of momentum can be non-demolition, so that successive measurements further reduce the size of the range of non-zero probability momentum states.

David
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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.

Aung
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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 06:10
  • @Anug, I refined my second question a little bit. Can you take a look? Thanks. – Purui Feb 12 '14 at 07:02
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  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.

Ján Lalinský
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