What do we have to perform physically that is equivalent to applying those quantum mechanical operators on a state $|\psi\rangle$?

Edit: I have removed the part I was asking regarding measurement because it takes us away from the real question.

  • $\begingroup$ In practice I don't think one usually does measure it in that way. Although quantum mechanics of course is perfectly correct, there is not always a straightforward correspondence between the basic theory and what is done in practice. Generally, the momentum of a charged particle is done by measuring it radius of curvature in a known magnetic field. If you want to prepare a particle in a particular momentum state this can be done by having two parallel plates with a small hole in each of them so the particle has to be traveling in a straight line between the holes to get out. $\endgroup$ – Virgo Nov 1 '14 at 22:10
  • $\begingroup$ What is the closest we can get between applying mathematical operations and physical operations? $\endgroup$ – Harshfi6 Nov 2 '14 at 5:08
  • $\begingroup$ As stressed by @AlfredCentauri your question is somewhat confused. A measurement does not correspond to operating on a state with the associated quantities operator. $\endgroup$ – Virgo Nov 3 '14 at 3:07

Perhaps I misunderstand your question but I would like to make clear that operating on a state with say, the momentum operator is not meant to be the equivalent of measuring the momentum of the system in that state.

Consider, for example, a state that is a superposition of two momentum eigenstates:

$$|\psi\rangle = \frac{1}{\sqrt{2}}\left(\,|p_1\rangle + |p_2\rangle \,\right)$$

If we operate on this state with the momentum operator, we get a different state:

$$\hat p |\psi\rangle = \frac{1}{\sqrt{2}}\left(\,p_1|p_1\rangle + p_2|p_2\rangle\, \right)$$

But note that $\hat p |\psi\rangle$ is a superposition of momentum eigenstates, i.e., operating on the state with the momentum operator did not 'collapse' the state to one or the other momentum eigenstate.

However, if we measure the momentum of the system in this state, we will measure either $p_1$ or $p_2$ and, further, the state of the system, immediately after the measurement, will be the associated eigenstate.

A measurement always causes the system to jump into an eigenstate of the dynamical variable that is being measured, the eigenvalue this eigenstate belongs to being equal to the result of the measurement.

P.A.M Dirac in "The Principles of Quantum Mechanics"

Thus, the 'momentum measurement operator' (whatever that is) is not the momentum operator.

Put another way, the result of operating on the state with the momentum operator is determined by the state; the result of the operation is certain.

However, the result of measuring the momentum of the system in this state is not determined. The result will be either $p_1$ or $p_2$ but which value will be measured is not determined by the state.

  • $\begingroup$ So, you mean we can't do anything physically, to convert the eigenstate $ |\psi\rangle $ to $\hat p |\psi\rangle$. Only measurement is possible physically. $\endgroup$ – Harshfi6 Nov 2 '14 at 13:20
  • $\begingroup$ @harshfi6, I don't understand the first sentence in your comment and neither $|\psi\rangle$ or $\hat p|\psi\rangle$ are eigenstates. What I mean, in short, is that the mathematical description of physical measurement is far different and far more complex than the operators that correspond to observables. $\endgroup$ – Alfred Centauri Nov 2 '14 at 13:25
  • $\begingroup$ i was asking if it possible to operate $\hat p$ to $|\psi\rangle$ physically. $\endgroup$ – Harshfi6 Nov 2 '14 at 13:44
  • $\begingroup$ @harshfi6, it's not clear to me what you're asking. $\hat p$ and $|\psi\rangle$ are abstract mathematical objects. Are you asking what physical manipulation of the system corresponds to operating with $\hat p$ on $|\psi\rangle$? $\endgroup$ – Alfred Centauri Nov 2 '14 at 14:12
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    $\begingroup$ @harshfi6 if you read Alfred's answer carefully, you will see the answer to your question is there... $\endgroup$ – Phonon Nov 5 '14 at 2:57

The whole artillery of Quantum Mechanics, operators and all, is a mathematical description of the result of measurements.

The contact of the mathematics with measurements comes from the postulates:

1) To every observable there corresponds an operator

2) The square of the wave function for the specific system gives the probability of finding the system at (x,y,z) at time t.

From these, an operator operating on the state function of the system under consideration will give a distribution of the probability of finding a specific value of that observable with a single measurement. For example the momentum operator operating on the wave function will give the probability distribution for finding the particle with momentum p. These type of distributions have been checked against measurements and that is what we mean that quantum mechanics is a validated theory.

So an operator can only give distributions to be checked against data, it does not operate on the data, but on the mathematical functions.


Here are two examples where these things are measured in practice.


In photoelectron spectroscopy, electrons are knocked out of atoms or molecules by ultraviolet radiation - photons with 10 or more eV of energy.

The energy of the electrons emitted $E_e$ is given by $E_e = h\nu - \text{binding energy of electron}$, where $h\nu$ is the energy of the photon.

The energy of the electron - or photoelectron - is measured with a hemispherical energy selector (or parallel plate or similar) and an electron detector.

A spectrum of electron energies is the result of the measurment.


The momentum of electrons in an atom or molecule can be measured with e,2e spectroscopy. Electrons are fired at an atom or molecule and events are detected where an electron is knocked out by the incident electron - both the incident and scattered electron must be detect 'in coincidence'. When they are detected in coincidence it is possible to work backwards to determine the original momentum of the electron knocked out of the atom or molecule.

Applying operators

The dipole moment operator is applied in absorption or emission of a photon

In Raman scattering the polarizability operator is applied to the wavefunction

...but I don't think this is exactly what you had in mind.

  • $\begingroup$ OK, So we use experimental techniques, but do we have nothing to perform physically that is equivalent to applying those quantum mechanical operators? I have edited the question, a little for that. $\endgroup$ – Harshfi6 Nov 2 '14 at 5:05
  • $\begingroup$ @harshfi6 - expeimental techniques are not perfect and suffer from problems such as non-perfect resolution, which blurs answers, however, by measuring experimental data we can get good evidence to support QM calculations. The best agreement I have seen between QM calculations of $\psi$ and experiment are in very high resolution spectroscopy measurements, but I'm no expert in that field and don't know a good reference - the examples in the answer above are an indication of how experimentalist can get close to measureing $E$ and $p$ for %\psi$ wavefunctions $\endgroup$ – tom Nov 2 '14 at 9:42
  • $\begingroup$ @harshfi6 added something about operators, but I am not sure you are going to like it... hope it is helpful. $\endgroup$ – tom Nov 2 '14 at 21:59

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