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Suppose we are doing a measurement in a particular quantum field, i.e electron field. Are we looking for the probability of the electron to show up at that spot we are measuring or are we measuring the entire electron wavefunction (whatever that means)? Maybe it is checking every point in space?

I think mathematically we square the wavefunction to find the probability of electron at that spot or for every point in space.

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  • $\begingroup$ Quantum fields and wavefunctions are not the same thing. You are talking about wavefunctions. $\endgroup$
    – G. Smith
    Jun 30 '20 at 3:39
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Fields in quantum mechanics are not measurable quantities, and also wavefunctions are not measurable. What are measurable are interactions between elementary particles or composites of elementary particles, like atoms and nuclei.

Fields and wavefunctions are mathematical representations that allow a predictive modeling of these interactions . The predictions of quantum mechanics are probabilistic, therefore the probability of measuring an interaction at (x,y,z,t) is the only prediction that can be made and checked in the lab. The predictions of field theory are calculated using Feynman diagrams for the interactions.

For intuition, take this double slit experiment one photon a time.

dblslph

Single-photon camera recording of photons from a double slit illuminated by very weak laser light. Left to right: single frame, superposition of 200, 1’000, and 500’000 frames.

In QED the photon field is spread all over spacetime, and photons move by creation and annihilation operators as wavepackets in space. In this specific measurement experiment, "photon interacting with specific boundary conditions double slit" the individual photon footprint is a dot on the (x,y) plane of the screen at a distance z of the slit from the screen. That is the measurement of a mathematically modeled complicated wave function from the field of photons covering all space time.

The single photons seem randomly oriented on the screen. The prediction of the mathematics of QED is seen in the last screens where the interference pattern of the wavefunction representation can be seen. Same experiment exists for electrons. The (x,y) of the screen is a projection of the (x,y) at the double slit plane, where the quantum mechanical interaction happens and can be calculated by QED.

I think mathematically we square the wavefunction to find the probability of electron at that spot or for every point in space.

The wavefunction that is squared is the solution of the specific boundary conditions for the specific measurement and it will give a solution that fits the observed data. It is not the general wavefunctions making up the particle fields and the propagation of paricles with creation and annihilation operators. It is a specific one for the given experiment/observation , and field theory is used to get the Feynman diagrams of the interaction for the specific case. And it is the square of the specific to the interaction wave function that is measurable.

The success of the standard model in describing the elementary particle data supports the use of quantum field theory, but one has to keep in mind that it is only the complex conjugate squared of the wavefunction of the specific interaction under study that can be validated by experimental data, i.e. measured.

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  • $\begingroup$ Thanks, by interaction do you mean each green dot on the screen or the alternating light and dark bands in the example you used? $\endgroup$
    – user6760
    Jun 30 '20 at 5:09
  • $\begingroup$ in the specific experiment, the QM interaction happens of the photon/electron as it passes the slit with the atoms at the edge of the slits. That probability as a function of (x,y) is measured by the accumulating dots on the screen, The real x,y is in the space of the slits. It is the same as with bubble chamber experiments. The QM interaction is at the verices, the rest is the recording method. $\endgroup$
    – anna v
    Jun 30 '20 at 5:27
  • $\begingroup$ see physics.stackexchange.com/questions/423985/… $\endgroup$
    – anna v
    Jun 30 '20 at 5:33

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