0
$\begingroup$

Say we know the probability amplitudes of finding a particle at a state with well defined energy. Does this alone provide complete description of the particle's state?

$\endgroup$
  • $\begingroup$ No, energy eigenvalues can be degenerate. For free particles this is due to spin polarizations. $\endgroup$ – Slereah Dec 21 '17 at 12:17
  • $\begingroup$ @Slereah May I also know that, if it is degenerate, then how do I calculate the probability of finding the state in a specific energy E (since it may have different values of spin)? How will the system evolve after it (will it's spin be randomised)? $\endgroup$ – user148792 Dec 21 '17 at 12:19
0
$\begingroup$

Ignore spin for a moment. Lets look at a free electron in $3D$ with Hamiltonian

$$\hat{H}=-\frac{\hbar^{2}}{2m}\nabla^{2}$$

The corresponding eigenfunctions are

$$\psi_{\boldsymbol{k}}\left(\boldsymbol{r}\right)=\frac{1}{\left(2\pi\right)^{\frac{3}{2}}}e^{i\boldsymbol{k}\cdot\boldsymbol{r}}$$

with energies $E_{\boldsymbol{k}}=\frac{\hbar^{2}\left|\boldsymbol{k}\right|^{2}}{2m}$. As you can see, all the values of $\boldsymbol{k}$ that satisfy

$$\left|\boldsymbol{k}\right|=\frac{\sqrt{2mE}}{\hbar}$$

lead to the same energy $E_{\boldsymbol{k}}=E$ (i.e. the energy level $E\neq 0$ is degenerate). Therefore, if you know that your particle is in a state with a definite energy $E$, then its wavefunction is

$$\psi\left(\boldsymbol{r},t\right)=\sum_{\boldsymbol{k}\in S^{2}\left(\frac{\sqrt{2mE}}{\hbar}\right)}c_{\boldsymbol{k}}\psi_{\boldsymbol{k}}\left(\boldsymbol{r}\right)e^{-i\frac{E}{\hbar}t}$$

where $\sum_{\boldsymbol{k}\in S^{2}\left(\frac{\sqrt{2mE}}{\hbar}\right)}\left|c_{\boldsymbol{k}}\right|^{2}=1$ is the normalization condition and $S^{2}\left(R\right)$ denotes a sphere of radius $R$. Hence there are infinite number of possible wavefunctions for your particle, and you can't determine the specific combination based on energy alone. If you add spin, things become even more degenerate.

$\endgroup$
  • $\begingroup$ If the energy eigenstates are degenerate, then if I've measure the energy of the system, which state does the wave function collapse into? $\endgroup$ – user148792 Dec 21 '17 at 13:42
  • $\begingroup$ @delickcrow123 It remains in this state since it has a definite value of the energy $E$. $\endgroup$ – eranreches Dec 21 '17 at 13:43
  • $\begingroup$ What about the situation in which the wave function is a linear combination of several states of energy E1 and several states of energy E2? Given that I measured the system to have energy, say, E1, then which of the "several" states will the system be in (or will it be in a linear combination of the "several" states of E1 it was originally in)? $\endgroup$ – user148792 Dec 21 '17 at 13:57
  • 1
    $\begingroup$ @delickcrow123 What you wrote in the parenthesis is correct. The key point here is that there is nothing special about $\psi_{\boldsymbol{r}}$ as eigenstates. Every normalized linear combination of degenerate eigenstates is also an eigenstate with the same energy. $\endgroup$ – eranreches Dec 21 '17 at 14:02

Your Answer

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