1
$\begingroup$

What is the random variable that belongs to the expectation value of momentum in quantum mechanics?

Or in general: Is there any way we can define the expectation values that occur in quantum mechanics via the definition of expectation values by using random variables?

Or more precisely: What are the random variables in quantum mechanics?

For people who are interested in getting some background on what I am asking:

  1. Expected Values of Operators in Quantum Mechanics
  2. http://en.wikipedia.org/wiki/Expectation_value_%28quantum_mechanics%29#Example_in_configuration_space
$\endgroup$
  • $\begingroup$ You get not only the expectation value but the whole probability distribution of any measurable from the wave function, what more are you asking for? $\endgroup$ – Abdulh Khazzak Gustav ElFakiri Nov 18 '13 at 19:07
  • $\begingroup$ What is the randomn variable that belongs to 'let's say: the expectation value of the momentum?' $\endgroup$ – user66906 Nov 18 '13 at 19:47
  • $\begingroup$ The measured momentum $\endgroup$ – Abdulh Khazzak Gustav ElFakiri Nov 18 '13 at 19:51
  • $\begingroup$ could you write down the actual map please? $\endgroup$ – user66906 Nov 18 '13 at 19:57
  • $\begingroup$ You set up your experiment, then you rpess a button to perform experiment, a computer probably records the momentum and you read off the screen? $\endgroup$ – Abdulh Khazzak Gustav ElFakiri Nov 18 '13 at 19:58
1
$\begingroup$

As the link you posted from another question explain, in the position representation $\psi(x)$, the eigenvectors with eigenvalues $\hbar k$ of the momentum operator $\hat{p}=\frac{\hbar}{i}\frac{\partial}{\partial x}$ are functions of the form $\psi_k(x)=e^{ikx}$. If we want the probability distribution for the measurement of some observable $O$ in a given state, we need to perform the spectral decomposition $O=\int k dP_k$, where $P_k(\cdot)=\psi_k(\psi_k,\cdot)$ is the projector in the eingespace of eigenvalue $k$ (I assumed that the eigenvalue is non-degenerate for simplicity, but the same also holds for the general case). Then the probability distribution will be given by $p(k)=(\psi, P_k \psi)$. So, for the case of the momentum operator, we have the probability distribution $$p(k)=(\psi,\psi_k)(\psi_k,\psi)=|(\psi_k,\psi)|^2=\left|\int e^{-ikx}\psi(x)dx\right|^2=|\tilde{\psi}(k)|^2,$$ where $\tilde{\psi}(k)$ is the Fourier transform of $\psi(x)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy