Consider an electron in an infinite square well. The expectation values of momentum and angular momentum are all zero for energy eigenstates. An electron transition is accompanied by the emission or absorption of photons. And we know the momentum of a photon is hk and the angluar momentum (spin) is 1.

The momentum and angular momentum should be conserved in the transition process. Does that mean all electron transitions between energy levels are prohibited in such system?

If such transitions are prohibited, how do we explain light-emitting quantum dots?


Thanks for the answers and comments below. Now I realize a quantum dot is more like an atom than a hollow box. But if we consider the theoretical ideal square well, there seems to be no simultaneous eigenstates of both energy and momentum - at least I can't figure it out.

Consider a 2D quare well. The energey eigenstates can be shown as:

Energy eigenstates in a 2D Box

Apparently they are not momentum eigenstates. Since these states form a complete set, we can set a momentum eigenstate to be:

$|k_x\rangle = \sum_{n_x,n_y} C_{n_x,n_y} |n_x,n_y\rangle$

where C's are constants, and

$P_x |k_x\rangle = \hbar k_x |k_x\rangle$

Keep going on:

$P_x |k_x\rangle = -i\hbar\nabla_x \sum_{n_x,n_y} C_{n_x,n_y} |n_x,n_y\rangle = -i\hbar\nabla_x \left( C_{1,1} \sin(k_{1}x)\sin(k_{1}y)+\cdots\right)$ $= -i\hbar\left( C_{1,1} k_1 \cos(k_1 x)\sin(k_1 y)+\cdots\right)$

$=???\; \hbar k_x \left( C_{1,1} \sin(k_{1}x)\sin(k_{1}y)+\cdots\right) = \hbar k_x |k_x\rangle$

I don't know any way to transform cos*sin's to sin*sin's without messing them up.

Is there anything wrong with my calculation?

  • 2
    $\begingroup$ A quantum dot (3D infinite square well) has states with nonzero l and m. $\endgroup$ – kennytm Dec 9 '10 at 6:41
  • $\begingroup$ To elaborate on KennyTM's note: energy levels in multi-dimensional system can be hugely degenerated (because energy is no longer the only quantum number characterizing the system). For each such degenerated level you can choose more quantum numbers (this is not unique. It might be momentum, or angular momentum, or some function of the two; and for both you also have freedom in direction). If you choose angular momentum then it will in general have to distinguish between some of the degenerated states (so that it is a good quantum number). $\endgroup$ – Marek Dec 9 '10 at 12:43
  • $\begingroup$ To make the above a little bit more technical: if you found some basis of the Hilbert space on which some operator had only zero eigenvalues (which you claim for energy eigenstates and angular momentum operator) then it would imply that the operator itself is a zero operator. It should be obvious that this is not quite true. $\endgroup$ – Marek Dec 9 '10 at 12:46
  • $\begingroup$ So, what are the simultaneous eigenstates of enegey and momentum/angular momentum for such system? It's not spherical symmetric so that I don't see any easy connection to common n,l,m states in an atom. $\endgroup$ – skywaddler Dec 10 '10 at 5:07
  • $\begingroup$ skywaddler: I suppose that a particle in a box has cannot have a defined momentum due to the uncertainty principle $\Delta p \ge \hbar/2\Delta x$ $\endgroup$ – gigacyan Dec 10 '10 at 12:38

In an ideal 3-D square well the electron's movement along the three axes is independent. This is not true for a real quantum dot which has a spherical shape (close to it) and is thus more similar to hydrogen atom. As soon as there is a coupling between movement along $x$,$y$ and $z$ the electron will start orbiting around the quantum dot and will posess an angular momentum.

Now, a linear conjugated molecule can be considered as a 1-D square well. Linear carbon chains have been discovered in the interstellar space and also produced in the lab. To interact with a photon such molecule has to have a transition dipole moment which means that the electron cloud should move. If we don't consider $p$-orbitals (that would add a second dimension), only a non-symmetric molecule like $C_6N$ can have a nonzero transition moment and such molecules are in fact very strong light absorbers.

| cite | improve this answer | |

If you're thinking about a 1-D square well, it doesn't really make sense to worry about angular momentum, since angular momentum necessarily requires multiple dimensions. When it comes to the linear momentum, there may be some subtlety regarding the meaning of expectation values. The expectation value of momentum ($\langle p \rangle $) is zero for a square-well state, but that does not mean that the particle is not moving. The expectation value of momentum squared ($\langle p^2 \rangle$) is non-zero. The conceptual explanation of this is that the particle is always in motion, but is equally likely to be found moving in either direction. Thus, when you average the momentum of many different realizations to get the expectation value, the momentum averages to zero, but when you remove the directional dependence by squaring the momentum, you get a non-zero value that corresponds to the speed of the particle.

This might allow enough wiggle room to let a photon absorption/emission conserve both energy and momentum, in that the momentum squared increases as you increase the energy state (as it must, the kinetic energy being $\frac{p^2}{2m}$). I'm not sure the numbers really work out, though.

I'll also note that when we talk about transitions in atoms at the same sort of level as used in talking about the 1-D infinite square well, we generally ignore the photon momentum-- we talk about electrons moving from one orbital to another, but the selection rules we find are based only on conservation of energy and conservation of angular momentum. The linear momentum carried by the photon is transferred to the atom as a whole, which is the basis for laser cooling, but this is generally neglected at the intro quantum physics level. So, it's not unprecedented for linear momentum to be ignored at this sort of crude level of approximation.

| cite | improve this answer | |

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.