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In my textbook, it says that when $L$ is a matrix that represents real($\mathbb{R}$) physical quantity, $L^2$ represents non-negative real physical quantity. What would be the proof of this?

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  • $\begingroup$ Probably there is an assumption that the matrix is diagonalizable and the eigenvalues are the values the physical quantity. Then the square of the matrix will have non-negative eigenvalues. $\endgroup$ – MBN Sep 27 '12 at 11:51
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    $\begingroup$ I'm thinking perhaps this would be better placed on Mathematics - I'll migrate it if other people think the same. $\endgroup$ – David Z Sep 27 '12 at 17:16
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    $\begingroup$ Do mathematicians know about physical observable values corresponding to eigenvalues of operators in quantum mechanics? $\endgroup$ – Kenshin Sep 27 '12 at 23:35
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There exists a set of eigenstates, $\{ |\psi_\lambda\rangle \}$, such that $L|\psi_\lambda\rangle = \lambda|\psi_\lambda\rangle$ where $\lambda$ is a real eigenvalue and $|\psi_\lambda\rangle$ is an eigenstate of $L$. The $\lambda$ represents the value of a physical observable associated with the eigenstate $|\psi_\lambda\rangle$.

There is currently no stipulation as to the sign of $\lambda$. It can be positive or negative, but we are given that it is real. Now we reapply the $L$ operator to our eigenvector equation to arrive at:

$LL|\psi_\lambda\rangle = \lambda L|\psi_\lambda\rangle = \lambda^2|\psi_\lambda\rangle$

Therefore,

$L^2|\psi_\lambda\rangle = \lambda^2|\psi_\lambda\rangle$

$\lambda^2$ must be positive, since $\lambda$ is real. Therefore the eigenvalues, which represent the values of the physical observables associated with the eigenstates of $L^2$, must be positive and real.

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  • $\begingroup$ Your conclusion is not quite correct. If $\lambda$ is real, $\lambda^2$ must be nonnegative, but for $\lambda=0$, it is not positive. $\endgroup$ – Arnold Neumaier Sep 30 '12 at 20:09
  • $\begingroup$ Agreed if $\lambda$ is 0 then $\lambda^2$ will be 0, and thus neither positive or negative. $\endgroup$ – Kenshin Oct 1 '12 at 1:40
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If $\psi$ is a normalized eigenvector of $L^2$ and $\lambda$ the corresponding eigenvalue then

$\lambda=\psi^*\lambda\psi=\psi^*L^2\psi=(L\psi)^*(L\psi)$.

Thus $\lambda$ is manifestly real and nonnegative.

This even holds if $L$ is a vector of noncommuting real (i.e., Hermitian) quantities, such as angular momentum. Then we get in the last step $\sum_i (L_i\psi)^*(L_i\psi)$, which is again manifestly real and nonnegative.

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If $L$ is an operator of a real physical quantity, it means it has its own eigenvectors $|\psi_l\rangle$ and eigenvalues $l$: $$L|\psi_l\rangle=l|\psi_l\rangle$$. If $l$ is real, then $l^2$ (corresponding to the eigenvalue of $L^2$) is positive.

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