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If $E$ (energy) and $P$ (momentum) only commute in constant potential, how could we have an $E$-$K$ diagram in a solid material?

$[E,p] \neq 0$. Then we cannot prepare electrons whose $E$ and $P$ are both very specified. However, in solid state material, we often have $E$-$K$ diagram, where every point is a pair of specified $E$ and $P$. The potential in a solid is definitely not a constant, so can somebody explain this?

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In the band diagrams you refer to, $K$ is not actually momentum; instead, it is something called crystal momentum (also often 'quasimomentum'; for more details and its relationship to real momentum see this, this, this and this questions, among many others on this site).

In short, $\hat H$ does not commute with $\hat p$, but it does commute with a restricted subset of operators which are functions of $\hat p$. In particular, it can be seen that the operator $$ \hat U(a) = e^{i\hat p a/\hbar} $$ produces a translation by $a$ in position space by the displacement $a$. If the hamiltonian has the form $$ \hat H = \frac{1}{2m} \hat p^2 + V(\hat x) $$ where $V(x+a)=V(x)$ is a periodic potential, then generically $\hat H$ will not commute with arbitrary displacements, i.e. $[\hat H, e^{i\hat p \tilde a/\hbar}] \neq 0$ for arbitrary $\tilde a$, but it will commute with $e^{i\hat p \tilde a/\hbar}$ when $\tilde a = a$, i.e. when the displacement length matches the period of the potential.

Thus, we have $$ [\hat H, e^{i\hat p \tilde a/\hbar}] = 0, $$ which means that $\hat H$ and $e^{i\hat p \tilde a/\hbar}$ admit a common eigenbasis. Moreover, $\hat U(a) = e^{i\hat p a/\hbar}$ is a unitary operator, which means that all of its eigenvalues must have unit modulus and it is always possible to write them as $u = e^{ik a/\hbar}$, where $k$ is only ever defined up to a multiple of $2\pi\hbar/a$; to remove this ambiguity, we restrict ourselves only to $k$'s in the interval $(-\pi\hbar/a, \pi \hbar/a]$.

This $k$ is the crystal momentum that's used in the band diagrams you're bothered by. While it does have a nontrivial relationship with physical momentum, it is not the same thing.

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This is not an answer to your question, but rather a response to the premise you wrote, to clear up a misconception: Energy and momentum can commute in a non-constant potential, it just depends what the potential is. You'd have to check it explicitly for a given potential by acting the momentum operator on the potential on a test function.

In general, in any closed system (so no external potential), momentum is conserved, which means we must have $$[H,p_i]=0$$

Why should they commute? A conserved quantity commutes with $H$ so that its time evolution is unaffected, thus it is conserved: $$P(t)=e^{-itH} P e^{itH}=P e^{-itH} e^{itH}=P$$

Edit: In the comments it was suggested that the claim above was false. So as an important example, below is a proof of this for the coulomb potential of two charge particles of mass $m$. Note that the conserved quantity is the summed (total) momentum of both particles, $\vec{p}_1+\vec{p}_2$. The individual momenta are not conserved and do not commute with $H$.

$$[H,p]$$ $$=[p_1^2/2m +p_2^2/2m+V(x_1,x_2), p_1 +p_2]$$

The first two terms of the hamiltonian commute with the total momentum so they vanish. $$=[V(x_1,x_2), p_1 + p_2]$$ $$=[V(x_1,x_2), p_1]+[V(x_1,x_2),p_2]$$

Consider the first commutator, for $p_1$. Use the notation $\partial_{x1}\equiv \frac{\partial}{\partial x_1}$.

$$[V(x_1,x_2), p_1]$$ $$=-i\hbar[V(x_1,x_2), \partial_{x1}]$$

where $V(x_1, x_2)=\frac{-k}{|x_2-x_1|}$

Let's find this commutator. Consider on some test function $\psi(x_1, x_2)$

$$\partial_{x1} (V(x_1, x_2) \psi)=V(x_1, x_2) \partial_{x1}\psi + \partial_{x1}V(x_1, x_2) \psi$$ $$\implies V(x_1, x_2) \partial_{x1}\psi-\partial_{x1} (V(x_1, x_2) \psi)=[V,\partial_{x1}]\psi=-\partial_{x1}V(x_1, x_2) \psi$$

The derivative of the potential is

$$\partial_{x1}V(x_1, x_2)=\frac{k(x_2-x_1)}{|x_2-x_1|^3}$$

Therefore $[V(x_1, x_2),\partial_{x1}]=\frac{k(x_2-x_1)}{|x_2-x_1|^3}$. This gives

$$[V(x_1, x_2),p_1]=-i\hbar\frac{k(x_2-x_1)}{|x_2-x_1|^3}$$ $$[V(x_1, x_2),p_2]=+i\hbar\frac{k(x_2-x_1)}{|x_2-x_1|^3}$$

..where the second commutator came from exchanging $x_1 \leftrightarrow x_2$. These commutators are negatives of each other, so adding them gives $$[V(x_1, x_2),p_1+p_2]=[V(x_1, x_2),p]=0$$ $$\implies [H,p]=0$$

This is a proof in one spacial dimension. The same proof works for the Coulomb potential in three spacial dimensions, where the generalized result is $[H,p_i]=0$ for all three components of the momentum.

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    $\begingroup$ Why is this dead-wrong answer getting upvoted? The claim that "Energy and momentum can commute in a non-constant potential" is patently false, and has nothing to do with the actual answer to the question at hand. $\endgroup$ – Emilio Pisanty Sep 30 '19 at 16:51
  • $\begingroup$ Added the proof. I believe this answer is actually helpful to OP, or else I would delete it. $\endgroup$ – doublefelix Sep 30 '19 at 18:25
  • $\begingroup$ That's a remarkable exercise in semantic gymnastics. Yes, if you go to a two-dimensional configuration space and you use a potential which is constant in one direction but not another, then one component of momentum will be conserved and the other one won't. But no, that has nothing to do with what any reasonable person would interpret your original claim to mean. This answer continues to be wrong, and cures no misconceptions exhibited by OP. (The OP is explicitly in 1D; if you need to invoke more degrees of freedom, then you're way out of scope.) $\endgroup$ – Emilio Pisanty Sep 30 '19 at 18:51
  • $\begingroup$ I think you misread my answer. All three components of the combined momentum operator for two particles, $\vec{p}=\vec{p}_1+\vec{p}_2$ are conserved. The potential is also not constant in any direction, it is the standard coulomb potential. I wrote the "..." because all 3 directions were implicit. In case that wasn't clear I'll edit away that shorthand in the definition of V. I did it in 3 dimensions but the proof is exactly the same (actually simpler) in 1d. $\endgroup$ – doublefelix Sep 30 '19 at 19:29
  • $\begingroup$ No, I didn't misread your answer, but you misread my comment. If you want to insist on proliferating dimensions from 1 to 3 per particle, then you've increased the dimension of the configuration space to six instead of two; from those six dimensions, the potential is constant along three orthogonal directions. And again: if your answer does not fit in a single dimension, it does not fit in this thread, it does not correct any "misconception" exhibited by OP, and it does not help OP or anyone else, or indeed do anything other than add confusion where none was needed. $\endgroup$ – Emilio Pisanty Sep 30 '19 at 19:51

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