Commutators involving components of angular momentum
Task number: 4572
Starting with the canonical commutation relations for position and momentum, work out the following commutators:
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a) \(\left [\hat L_z, \hat x \right]\), |
b) \(\left [\hat L_z, \hat y \right]\), |
c) \(\left [\hat L_z, \hat z \right]\), |
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d) \(\left [\hat L_z, \hat p_x \right]\), |
e) \(\left [\hat L_z, \hat p_y \right]\), |
f) \(\left [\hat L_z, \hat p_z \right]\), |
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g) \(\left [\hat L_z, \hat L_x \right]\), |
h) \(\left [\hat L_z, \hat L_y \right]\), |
i) \(\left [\hat L_z, \hat L_z \right]\). |
Based on the results from sections a) – i), first estimate the outcome and then verify by calculating the following commutators:
j) \(\left [\hat L_j, \hat x_k \right]\),
k) \(\left [\hat L_j, \hat p_k \right]\),
l) \(\left [\hat L_j, \hat L_k \right]\).
Note: In this task we use a dual notation, identifying the \(x\)-, \(y\)-, and \(z\)-components with the first, second, and third components, respectively: \(\hat x = \hat x_1, \, \hat y = \hat x_2, \, \hat z = \hat x_3\); \(\hat p_x = \hat p_1, \, \hat p_y = \hat p_2, \, \hat p_z = \hat p_3\); \(\hat L_x = \hat L_1, \, \hat L_y = \hat L_2, \, \hat L_z = \hat L_3\).
Hint 1
Recall the rules for evaluating commutators and for simplifying commutators of composite operators.
Hint 2
Recall how the components of the angular momentum operator may be expressed in terms of the position and momentum operators. Also recall the form of the canonical commutation relation.
Solution a) – c)
First, we express \(\hat L_z\) in terms of the position and momentum components. We then simplify the commutator of the composite operators
\[ \left [\hat L_z, \hat x \right ] = \left [\hat x \hat p_y - \hat y \hat p_x, \hat x \right] = \left [\hat x \hat p_y, \hat x \right] - \left [\hat y \hat p_x, \hat x \right] = \hat x \left [\hat p_y, \hat x \right] + \left [\hat x, \hat x \right] \hat p_y - \hat y \left [\hat p_x, \hat x \right] - \left [\hat y, \hat x \right] \hat p_x \, . \]We now use the fact that position operators commute with each other. Hence, the second and last terms vanish. Next, we apply the canonical commutation relation and simplify
\[ \left [\hat L_z, \hat x \right ] = \hat x \left (-i \hbar \delta_{21} \hat{\mathbb{E}} \right ) + 0 - \hat y \left (-i \hbar \delta_{11} \hat {\mathbb E} \right ) - 0 = i \hbar \hat y \, . \]The procedure for parts b) and c) is analogous. Therefore, we only provide a concise calculation here
\[ \left [\hat L_z, \hat y \right ] = \left [\hat x \hat p_y, \hat y \right] - \left [\hat y \hat p_x, \hat y \right] = \hat x \left [ \hat p_y, \hat y \right ] = -i \hbar \hat x \, , \] \[ \left [\hat L_z, \hat z \right ] = \left [\hat x \hat p_y, \hat z \right] - \left [\hat y \hat p_x, \hat z \right] = \hat x \cdot 0 - \hat y \cdot 0 = 0 \, . \]Solution d) – f)
The procedure in this section is analogous to that in the previous part. However, here we exploit the commutativity of momentum components rather than that of position operators. We provide a detailed calculation for case d)
\[ \left [\hat L_z, \hat p_x \right ] = \left [\hat x \hat p_y - \hat y \hat p_x, \hat p_x \right] = \left [\hat x \hat p_y, \hat p_x \right] - \left [\hat y \hat p_x, \hat p_x \right] = \] \[ = \hat x \left [\hat p_y, \hat p_x \right] + \left [\hat x, \hat p_x \right] \hat p_y - \hat y \left [\hat p_x, \hat p_x \right] - \left [\hat y, \hat p_x \right] \hat p_x = 0 + i \hbar \hat p_y - 0 - 0 = i \hbar \hat p_y \, . \]The procedure for parts e) and f) is analogous. Therefore, we only provide a concise calculation here
\[ \left [\hat L_z, \hat p_y \right ] = \left [\hat x \hat p_y, \hat p_y \right] - \left [\hat y \hat p_x, \hat p_y \right] = - \hat p_x \left [ \hat y, \hat p_y \right ] = -i \hbar \hat p_x \, , \] \[ \left [\hat L_z, \hat p_z \right ] = \left [\hat x \hat p_y, \hat p_z \right] - \left [\hat y \hat p_x, \hat p_z \right] = \hat p_y \cdot 0 - \hat p_x \cdot 0 = 0 \, . \]Solution g) – i)
In this section, we keep \(\hat L_z\) within the commutator and expand the angular momentum component on the right‑hand side. We then simplify the resulting expression into commutators of elementary operators
\[ \left [\hat L_z, \hat L_x \right ] = \left [\hat L_z, \hat y \hat p_z - \hat z \hat p_y \right] = \left [\hat L_z, \hat y \hat p_z \right] - \left [\hat L_z, \hat z \hat p_y \right] = \] \[ = \hat y \left [\hat L_z, \hat p_z \right] + \left [\hat L_z, \hat y \right] \hat p_z - \hat z \left [\hat L_z, \hat p_y \right] - \left [\hat L_z, \hat z \right] \hat p_y \, . \]Using the results from the previous sections, we can substitute for these commutators and simplify
\[ \left [\hat L_z, \hat L_x \right ] = 0 + \left (-i \hbar \hat x \right ) \hat p_z - \hat z \left (-i \hbar \hat p_x \right ) - 0 = i \hbar \left ({\hat z} {\hat p_x} - {\hat x} {\hat p_z} \right ) \, . \]Finally, we note that the expression in parentheses corresponds to the expanded form of the angular momentum component \(\hat L_y\). Hence, we obtain
\[ \left [\hat L_z, \hat L_x \right ] = i \hbar {\hat L_y} \, . \]The calculation for the \(\hat L_y\) component is analogous and is therefore presented only briefly
\[ \left [\hat L_z, \hat L_y \right ] = \left [\hat L_z, \hat z \hat p_x \right] - \left [\hat L_z, \hat x \hat p_z \right] = \hat z \left (i \hbar {\hat p_y} \right ) - \left (i \hbar \hat y \right ) \hat p_z = -i \hbar \hat L_x \, . \]The third commutator equals zero, since any operator commutes with itself
\[ \left [\hat L_z, \hat L_z \right ] = 0 \, . \]Hint – a useful relation
In calculating the commutator \(\left [\hat L_j, \hat L_k \right ]\), we make use of the relation connecting the Levi‑Civita symbol with the Kronecker delta. This relation reads
\[ \varepsilon_{abc} \varepsilon_{ade} = \delta_{bd} \delta_{ce} - \delta_{be} \delta_{cd} \, . \]Solution j) – l)
First, we are supposed to estimate the results. We will do this for the commutator of an angular momentum component and a position operator (part j). The other parts are analogous.
At first glance, it is evident that a factor of \(i \hbar\) appears in the commutators of an angular momentum component with a position operator. Moreover, we notice that for \(j = k\), the commutator equals zero. Next, we consider the sign of the commutator. For the case \(j = 3, \, k = 1\) the sign is \(+\), whereas for \(j = 3, \, k = 2\) it is \(-\). We also observe that for both combinations of distinct \(j\) and \(k\), the resulting position operator is the “missing” one, e.g. for \(j = 3, \, k = 1\) the position operator is \(\hat y\) and so on.
This leads to the insight that, in the general form, the Levi‑Civita symbol will appear. Combining these observations, we arrive at the estimate
\[ \left [\hat L_j, \hat x_k \right ] = i \hbar \varepsilon_{jkl} \hat x_l \, . \]We will now verify this estimate with an explicit calculation. We expand the angular momentum component and simplify the commutator of the composite operators
\[ \left [\hat L_j, \hat x_k \right ] = \left [\varepsilon_{jlm} \hat x_l \hat p_m, \hat x_k \right ] = \varepsilon_{jlm} \left (\hat x_l \left [\hat p_m, \hat x_k \right ] + \left [\hat x_l, \hat x_k \right ] \hat p_m \right ) \, . \]We can now substitute the values for these commutators. For the first term, we use the canonical commutation relation, and for the second term, we substitute zero, since position operators commute with each other. After substitution, we obtain
\[ \left [\hat L_j, \hat x_k \right ] = \varepsilon_{jlm} \left (\hat x_l \left (-i \hbar \delta_{mk} \right ) \right ) = \varepsilon_{jlm} \delta_{mk} \left (-i \hbar \hat x_l \right ) \, . \]We now sum over the index \(m\)
\[ \left [\hat L_j, \hat x_k \right ] = \varepsilon_{jlk} \left (-i \hbar \hat x_l \right ) = i \hbar \varepsilon_{jkl} \hat x_l \, . \]The procedure outlined above is also valid for parts k) and l), as all the commutators take the same form. The corresponding results are therefore
\[ \left [\hat L_j, \hat p_k \right ] = i \hbar \varepsilon_{jkl} \hat p_l \, , \] \[ \left [\hat L_j, \hat L_k \right ] = i \hbar \varepsilon_{jkl} \hat L_l \, . \]The verification for part k) is very similar, so we present it here in a more concise form
\[ \left [\hat L_j, \hat p_k \right ] = \varepsilon_{jlm} \left (\hat x_l \left [\hat p_m, \hat p_k \right ] + \left [\hat x_l, \hat p_k \right ] \hat p_m \right ) = \varepsilon_{jlm} \delta_{lk} \left (i \hbar \hat p_m \right ) = i \hbar \varepsilon_{jkm} \hat p_m \, . \]Here we have used the commutativity of the momentum components. Renaming the index \(m\) to \(l\) yields the final result
\[ \left [\hat L_j, \hat p_k \right ] = i \hbar \varepsilon_{jkm} \hat p_m = i \hbar \varepsilon_{jkl} \hat p_l \, . \]For the calculation in part l), we first expand the component of the angular momentum operator appearing on the right‑hand side of the commutator, so that we can apply previously obtained results
\[ \left [\hat L_j, \hat L_k \right ] = \varepsilon_{klm} \left (\hat x_l \left [\hat L_j, \hat p_m \right ] + \left [\hat L_j, \hat x_l \right ] \hat p_m \right ) \, . \]We now substitute the values of these commutators determined earlier. Next, we factor out \(i \hbar\) and perform a cyclic permutation of indices in the Levi‑Civita symbols so that the common index appears first
\[ \left [\hat L_j, \hat L_k \right ] = \varepsilon_{klm} \left \{\hat x_l \left (i \hbar \varepsilon_{jmn} \hat p_n \right ) + \left (i \hbar \varepsilon_{jlo} \hat x_o \right ) \hat p_m \right \} = i \hbar \left (\varepsilon_{mkl} \varepsilon_{mnj} \hat x_l \hat p_n + \varepsilon_{lmk} \varepsilon_{loj} \hat x_o \hat p_m \right ) \, . \]We now use the identity \(\varepsilon_{abc} \varepsilon_{ade} = \delta_{bd} \delta_{ce} - \delta_{be} \delta_{cd}\) and simplify
\[ \left [\hat L_j, \hat L_k \right ] = i \hbar \left \{ \left (\delta_{kn} \delta_{lj} - \delta_{kj} \delta_{ln} \right ) \hat x_l \hat p_n + \left (\delta_{mo} \delta_{kj} - \delta_{mj} \delta_{ko} \right ) \hat x_o \hat p_m \right \} = \] \[ = i \hbar \left (\delta_{kn} \delta_{lj} \hat x_l \hat p_n - \delta_{kj} \delta_{ln} \hat x_l \hat p_n + \delta_{mo} \delta_{kj} \hat x_o \hat p_m - \delta_{mj} \delta_{ko} \hat x_o \hat p_m \right ) \, . \]In the first and last terms, we sum over both Kronecker deltas. In the second term, we sum over the Kronecker delta \(\delta_{ln}\), and in the third term over \(\delta_{mo}\). This yields
\[ \left [\hat L_j, \hat L_k \right ] = i \hbar \left (\hat x_j \hat p_k - \delta_{kj} \hat x_n \hat p_n + \delta_{kj} \hat x_m \hat p_m - \hat x_k \hat p_j \right ) \, . \]Since the summation index in the second term is different from the indices of the Kronecker delta, we may rename it from \(n\) to \(m\). After this renaming, the second and third terms cancel each other, giving
\[ \left [\hat L_j, \hat L_k \right ] = i \hbar \left (\hat x_j \hat p_k - \delta_{kj} \hat x_m \hat p_m + \delta_{kj} \hat x_m \hat p_m - \hat x_k \hat p_j \right ) = i \hbar \left (\hat x_j \hat p_k - \hat x_k \hat p_j \right ) \, . \]The expression in parentheses has the same form as the component of a vector product that does not appear explicitly in the commutator. Denoting this component by the index \(l\), we obtain the final result in the form
\[ \left [\hat L_j, \hat L_k \right ] = i \hbar \varepsilon_{jkl} \hat L_l \, . \]Let us verify the calculation by substituting the indices \(j = z, \, k = x\)
\[ \left [\hat L_z, \hat L_x \right ] = i \hbar \varepsilon_{zxy} \hat L_y = i \hbar \hat L_y \, , \]which produces the same result as obtained in part g).
Answer
a) \(\left [\hat L_z, \hat x \right ] = i \hbar \hat y\)
b) \(\left [\hat L_z, \hat y \right ] = -i \hbar \hat x\)
c) \(\left [\hat L_z, \hat z \right ] = 0\)
d) \(\left [\hat L_z, \hat p_x \right ] = i \hbar \hat p_y\)
e) \(\left [\hat L_z, \hat p_y \right ] = -i \hbar \hat p_x\)
f) \(\left [\hat L_z, \hat p_z \right ] = 0\)
g) \(\left [\hat L_z, \hat L_x \right ] = i \hbar \hat L_y\)
h) \(\left [\hat L_z, \hat L_y \right ] = -i \hbar \hat L_x\)
i) \(\left [\hat L_z, \hat L_z \right ] = 0\)
j) \(\left [\hat L_j, \hat x_k \right ] = i \hbar \varepsilon_{jkl} \hat x_l\)
k) \(\left [\hat L_j, \hat p_k \right ] = i \hbar \varepsilon_{jkl} \hat p_l\)
l) \(\left [\hat L_j, \hat L_k \right ] = i \hbar \varepsilon_{jkl} \hat L_l\)
