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Let $L_iL_j-L_jL_i = i\hbar\varepsilon_{ijk}L_k$ where $i,j,k\in\{1,2,3\}$

Let $u$ be any eigenstate of $L_3$. How might one show that $\langle L_1^2 \rangle = \int u^*L_1^2u = \int u^*L_2^2u = \langle L_2^2\rangle$ ?

I can show that $\langle u|L_3L_1L_2-L_1L_2L_3|u\rangle=\int u^*(L_3L_1L_2-L_1L_2L_3)u =0$. And I know that $L_3 $ can be written as ${1\over C}(L_1L_2-L_2L_1)$. Hence I have $\int u^*L_1L_2^2L_1u=\int u^*L_2L_1^2L_2u$. But I don't seem to be getting the required form... Help will be appreciated.

Added: The $L_i$'s are operators that don't necessarily commute.

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$L_i$ are operators on what? $u$ is a function defined over what space? –  Willie Wong Sep 3 '11 at 17:18
    
@Willie W: The $L_i$'s operate on the space of functions. In fact, I must say this reminds me of the angular momentum operator and eigenstates/eigenfunctions (can't distinguish those 2)... –  Tom L Sep 3 '11 at 17:25
    
@Willie W: You are quite right. I have edited the question to give it verbatim. Hopefully it makes more sense now? :) –  Tom L Sep 3 '11 at 17:45
    
Are the $L_i$ assumed to be self-adjoint? –  Willie Wong Sep 3 '11 at 18:11
    
@Willie W: Yup. Sorry :S –  Tom L Sep 3 '11 at 18:12

1 Answer 1

Okay then. In below, we fix $u$ an eigenfunction of $L_3$, and denote by $\langle T\rangle:= \langle u|T|u\rangle$ for convenience for any operator $T$.

Using self-adjointness of $L_3$, we have that $L_3 u = \lambda u$ where $\lambda \in \mathbb{R}$. And furthermore

$$ \langle L_3T\rangle = \langle L_3^* T\rangle = \lambda \cdot \langle T\rangle = \langle TL_3\rangle $$

which we can also write as

$$ \langle [T,L_3] \rangle = 0 $$

for any operator $T$. This implies

$$ \frac{1}{\hbar}\langle L_1^2\rangle = \langle -iL_1L_2L_3 + iL_1L_3L_2 \rangle = \langle -iL_3L_1L_2 +i L_1L_3L_2\rangle = \frac{1}{\hbar}\langle L_2^2\rangle $$

The first and third equalities are via the defining relationship $[L_i,L_j] = i \hbar \epsilon_{ijk} L_k$. The middle equality is the general relationship derived above, applied to the first summand. (And is precisely the identity that you said you could show in the question.)


Remark: it is important to note that the expression $\langle u| [T,A] |u\rangle = 0$ holds whenever $A$ is self adjoint and $u$ is an eigenvector for $A$. This does not imply that $[T,A] = 0$. This is already clear in a finite dimensional vector space where we can represent operators by matrices: consider $A = \begin{pmatrix} 1 & 0 \\ 0 & 2\end{pmatrix}$ and $T = \begin{pmatrix} 1 & 1 \\ 0 & 0\end{pmatrix}$. The commutator $[T,A] = \begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix}$, which is zero on the diagonals (as required), but is not the zero operator.

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Thank you very much!! –  Tom L Sep 3 '11 at 19:00

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