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Do linear and angular momentum operators commute? If I use the canonical commutation relations I get that they commute. Say, for $x$-component,

$[p_x, L_x] = p_x y p_z - y p_z p_x - p_x z p_y + z p_y p_x = y[p_x, p_z] - z [p_x, p_y] = 0$

However, e.g., Akhiezer in his textbook claims without proof that they don't commute. Am I getting wrong somewhere?

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  • $\begingroup$ You are misreading the General statement to a special particular case where the nontrivial result collapses. L rotates p, so try $[p_x,L_y]$, instead. $\endgroup$ – Cosmas Zachos May 4 '18 at 16:10
  • $\begingroup$ Aha, so physically, if I know $L_z$, I can measure $\{p_x, p_y\}$ simultaneously, but not $p_z$, right? $\endgroup$ – MsTais May 4 '18 at 16:16
  • $\begingroup$ The other way around. If you know $L_z$, $p_z$ commutes with it, so you may measure these two simultaneously. $\endgroup$ – Cosmas Zachos May 4 '18 at 16:18
  • $\begingroup$ Oh, yeah, sorry... Right! $\endgroup$ – MsTais May 4 '18 at 16:40
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They must have non-trivial commutation relations, since all vector operators have certain commutation relations with the angular momentum operators, due to the fact, that they generate rotations and vectors transform under rotation in a specific fashion.

The relations can also be derived directly for the momentum: \begin{align*} [p_i, L_j] &= \varepsilon_{jlm}[p_i, x_lp_m] = \varepsilon_{jlm} \big(x_l [p_i, p_m] + [p_i, x_l]p_m \big) = -\varepsilon_{jlm} i\hbar \delta_{il} p_m = i\hbar\varepsilon_{ijm}p_m. \end{align*}

However, the term $[L_x, p_x]$ you compute is indeed zero since $\varepsilon_{xxj} = 0$ for all $j$, but $[L_y, p_x]$ and $[L_z, p_x]$ are not.

On the general statement

The operator for a spacial rotation around the axis $\vec \varphi$ by the angle given by its absolute value in quantum mechanics is given by $U = e^{-\frac i \hbar \vec \varphi \cdot \vec L}$ (this corresponds to the way $T = e^{-\frac i \hbar \vec a \cdot \vec p}$ implements spatial translations on the states). The corresponding rotation matrix is${}^1$ $A = e^{\vec \varphi \times}$ and the the components of the expectation values of vector operators $\vec v$ in all states (and therefore the components of vector operators) must transform according to${}^2$: \begin{align*} U \vec v U^\dagger = A\vec v. \end{align*} The operators $U$ and $A$ work in different ways here, the operator $A$ transforms between the components of the vector, so the r.h.s. reads $A_{ij}v_j$ in components, on the left hand-side the operator $U$ is a scalar, in the sense that $U$ acts on each component of $\vec v$ independently, that is $v_i$ is transformed to some linear combination of the components of $\vec v$.

Now we look at the component $i$ and use the formula${}^3$ $e^{-B}Ae^B = e^{[B, \cdot]}A$ to expand the left hand side of the transformation formula and expand the exponential on the right hand side: \begin{align*} U^\dagger v_i U &= \sum_{n=0}^\infty \frac{i^n}{\hbar^n n!} [\varphi_m L_m, \cdot]^n v_i = \sum_{n=0}^\infty \frac{(\vec \varphi \times)^n_{ij} v_j}{n!} = \big(e^{\varphi \times}\vec v \big)_i. \end{align*} By comparing coefficients (in terms of powers of components of $\varphi$) on the left and right-hand side, we arrive at: \begin{align*} (i/\hbar)^n[\varphi_mL_m, \cdot]^n v_i &= (\vec \varphi \times)^n_{ij}v_j. \end{align*} Taking $n = 1$ gives: \begin{align*} (i/\hbar)\varphi_m[L_m, v_i] &= \varepsilon_{ikj} \varphi_k v_j & [L_m, v_i] &= -i\hbar \varepsilon_{imj} v_j & [L_m, v_i] &= i\hbar\varepsilon_{mij} v_j \end{align*} (the second equation follows by comparing coefficients, note that $\vec \varphi$ can be chosen arbitrarily). It is left as an exercise for the reader to show that this commutator derived from the first order term fulfils the equation in all orders.

This discussion can in fact be extended to tensor operators of any order, including scalars (all scalars commute with the angular momentum components, since $U^\dagger s U = s$).


${}^1$This notation considers $\vec\varphi \times$ as the linear operator that maps a vector $\vec v$ to $\vec\varphi \times \vec v$, in components this linear operator is given by the matrix $(\vec \varphi \times)_{ij} = \epsilon_{ikj} \varphi_k$.

${}^2$Usually the orbital angular momentum is derived, the other way around, by prescribing it in terms of the transformation behaviour and the corresponding preserved quantity in case of a rotation symmetry

${}^3$The notation $[A, \cdot]$ denotes the super-operator $[A, \cdot] \colon B \mapsto [A, B]$, this means $[A, \cdot]^n = \underbrace{[A, [A, \cdots [A, B]\cdots]]}_{\text{$n$ commutators}}$.

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  • $\begingroup$ Can you provide some references? Thanks in advance! $\endgroup$ – MsTais May 4 '18 at 19:21
  • $\begingroup$ Did you mean U is the rotation operator around the axis $z$? $\endgroup$ – MsTais May 4 '18 at 19:23
  • $\begingroup$ I wrote it down from memory/re-derived the missing parts. Franz Schwabl: Quantum Mechanics does derive the general rule (although I think the book uses infinitesimal transformations to derive the commutator) and I only have access to the German edition, so I can't give a precise hint. I'll edit to clarify what $U$ and $A$ and $\varphi$ represent. $\endgroup$ – Sebastian Riese May 4 '18 at 20:06

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