5
$\begingroup$

This operator identity showed up in a course I was taking, and it was given without proof.

$$[L^2,[L^2,\vec{r}]]=2 \hbar ^2 \{ L^2, \vec{r}\}$$

The curly brackets denote the anticommutator, $AB+BA$. The $\vec{r}$ operator is the position operator. The $L^2$ operator is given by:

$$L^2 = -\hbar ^2 \left( \frac{1}{\sin \theta} {\partial\over\partial\theta} (\sin \theta {\partial\over\partial\theta}) + \frac{1}{\sin^2 \theta} {\partial^2\over\partial\phi^2}\right)$$

Is there a way of proving this identity without tediously expanding all the commutators? I've been trying to find one but was unable to.

$\endgroup$
  • $\begingroup$ To clarify, is your notation here such that $r=\sqrt{\mathbf x\cdot\mathbf x}$? $\endgroup$ – joshphysics Jan 25 '14 at 19:46
  • $\begingroup$ Is the other option that $r$ represents $x$, $y$, or $z$? $\endgroup$ – BMS Jan 25 '14 at 19:57
  • $\begingroup$ $r=\sqrt{\vec x \cdot \vec x}$ would no make sense, because $[L^2,r]=0]$. So whats $r$? $\endgroup$ – pressure Jan 25 '14 at 20:21
  • $\begingroup$ @pressure Agreed. Hence my attempt at getting clarification. $\endgroup$ – joshphysics Jan 25 '14 at 20:24
  • 1
    $\begingroup$ It is definitely $\vec r$, the identity is correct with that. It doesn't matter which coordinates you use; the identity is correct in all coordinates. $\endgroup$ – Luboš Motl Jan 26 '14 at 9:44
8
$\begingroup$

The symbol $r$ in the identity represents (and will represent in the text below) the whole three-component vector of operators $\hat{\vec r} = (\hat x, \hat y, \hat z)$.

The simple way I found to prove the identity is to verify that all matrix elements of both sides match. Let's calculate the matrix elements of the operators $LHS,RHS$ between $$\langle j,m,a| LHS| k,n,b\rangle$$ and similarly for the right hand side. Here, $j,m$ and $k,n$ are the usual total angular momenta (which I will assume to be integers, just the orbital angular momentum case) and the $z$-component and $a,b$ represent the other quantum numbers that won't matter.

The advantage is that $\vec L$ combine to $L^2$ almost everywhere. The left hand side operator is $$ L^2 L^2 r - 2 L^2 r L^2 + r L^2 L^2 $$ so the matrix element (because $L^2$ acts either on the bra or ket vector in a simple way) is the same as the matrix element of $$ \hbar^4 r[ j(j+1)j(j+1) - 2j(j+1)k(k+1) + k(k+1)k(k+1)] $$ The coefficient in the parenthesis is equal to a complete square, $$ \hbar^4 r [j(j+1)-k(k+1)]^2 $$ Note that $\hbar^4 r$ is in all terms. The right hand side has the same matrix elements as the operator $$ 2\hbar^4 r [j(j+1) + k(k+1)] $$ They don't look "obviously" equal: one is quartic, one is quadratic. But we must realize that the operators on both sides are $j=1$ vector operators, from the $\vec r$ factor, so they only change the angular momentum by zero or $\pm 1$.

So it is enough to compare the expressions for these three choices; for higher changes of $j$, the matrix elements on both sides clearly vanish (and are therefore equal). For $j=k$, the matrix element vanishes because of parity: $r$ carries the negative parity while the parities $(-1)^l$ are $(-1)^j$ or $(-1)^k$ for the bra/ket vectors.

For $j=k+1$, the LHS is $$\hbar^4 r (k+1)^2 (k+2 - k)^2 = 4\hbar^2 r (k+1)^2 $$ while the RHS is $$2\hbar^4 r[(k+1)(k+2)+k(k+1)]= 4\hbar^4 r(k+1)^2$$ so it works. The same verification applies to the case $k=j+1$, too, just $j,k$ are interchanged.

There are many other ways to calculate or verify the identity but I found this one easiest. Note that I am not assuming any coordinates; the abstract calculation above works in any coordinates.

$\endgroup$
1
$\begingroup$

I) For the record, here is the operator calculation that OP wants to avoid. The benefit of the calculation is that the operators are not sandwich with any bra/ket representation, and hence we do not have to worry about whether the bra/ket representation is faithful. Let us put $\hbar=1$ for simplicity. The starting point is the CCR

$$ [x^i, p_j]~=~\mathrm{i}\delta^i_j .\tag{1}$$

The CCR (1) ensures that the definition of the orbital angular momentum operator $$ L_i~:=~\varepsilon_{ijk}~ x^jp_k~\stackrel{(1)}{=}~-\varepsilon_{ijk}~ p_jx^k \tag{2}$$ is Hermitian and that it does not suffer from operator ordering ambiguities. In particular, the position and the angular momentum are mutually perpendicular as operators $$ x^i L_i~\stackrel{(2)}{=}~0~\stackrel{(2)}{=}~L_i x^i \tag{3}.$$ Einstein's summation convention is implicitly assumed everywhere in this answer.

II) Now let us calculate the LHS of OP's identity.

$$ [L_i,x^j]~=~\mathrm{i}\varepsilon_{ijk}~ x^k \tag{4}$$

$$ \Downarrow $$

$$[L^2,x^j]~=~\{L_i ,[L_i,x^j]\} ~\stackrel{(4)}{=}~\mathrm{i} \varepsilon_{ijk}~\{L_i ,x^k\} \tag{5}$$

$$ \Downarrow $$

$$[L^2,[L^2,x^j]]~\stackrel{(5)}{=}~ \mathrm{i} \varepsilon_{ijk}~\{L_i ,[L^2,x^k]\} ~\stackrel{(5)}{=}~ \varepsilon_{ijk}\varepsilon_{k\ell n}\{L_i , \{L_{\ell} ,x^n\} \}$$ $$~=~ \left(\delta_{i\ell}\delta_{jn} -\delta_{j\ell}\delta_{in} \right)\{L_i , \{L_{\ell} ,x^n\} \} ~=~ \{L_i , \{L_i ,x^j\} \} -A_j, \tag{6}$$ where $$A_j~:=~ \{L_i , \{L_j ,x^i\} \} ~\stackrel{(3)}{=}~ [L_i , [L_j ,x^i] ] ~\stackrel{(4)}{=}~ \mathrm{i}\varepsilon_{ijk}[L_i ,x^k ] ~\stackrel{(4)}{=}~ \varepsilon_{jik}\varepsilon_{ik\ell}~x^{\ell}~=~2x^j .\tag{7}$$

III) On the other hand, the RHS yields $$ 2\{L^2, x^j\}~=~L_i \left( \{ L_i, x^j\} + [ L_i, x^j] \right)+\left( \{ x^j, L_i \} + [ x^j, L_i ] \right) L_i ~=~ \{L_i , \{L_i ,x^j\} \} +B_j,\tag{8} $$

where $$ B_j ~:=~ L_i[ L_i, x^j] +[ x^j, L_i ]L_i ~\stackrel{(4)}{=}~\mathrm{i}\varepsilon_{ijk}[L_i,x^k] ~\stackrel{(4)}{=}~-\varepsilon_{ijk} \varepsilon_{ik\ell}~x^{\ell}~=~-2x^j. \tag{9}$$

IV) Comparing the LHS and the RHS, we get OP's sought-for identity

$$ [L^2,[L^2,x^j]]~=~2\{L^2, x^j\} . \tag{10}$$

$\endgroup$

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.