I was attending a Quantum Mechanics lecture when the instructor casually mentioned the following theorem:
$\langle \alpha \rvert A \rvert \alpha \rangle = 0 ~\forall \alpha \implies A=0$, where $A$ is an operator and $\rvert\alpha\rangle$ is an arbitrary ket in the complex Hilbert space.
I have always assumed that the above theorem was 'obvious', but on second thought, it doesn't seem to be easy or trivial to prove. I tried looking at various sources for the theorem, but it seems to be surprisingly difficult to find this theorem or proof anywhere.
I would be very glad if someone would point me towards the proof of the theorem, and provide a small outline of it if possible.
Answer
Pick any orthonormal basis $\lvert \psi_i\rangle$ of our Hilbert space. Then $\langle \psi_i\vert A \vert \psi_i \rangle = 0$ for all $i$ by assumption, and for $\lvert \phi_{ij}(a,b)\rangle := a\lvert \psi_i\rangle + b\lvert \psi_j\rangle$ we find $$ \langle \phi_{ij} \vert A \vert \phi_{ij}\rangle = a^\ast b \langle \psi_i \vert A \vert \psi_j\rangle + ab^\ast \langle \psi_j \vert A \vert \psi_i\rangle = 0,$$ which for $a,b = 1$ implies $$ \langle \psi_i \vert A \vert \psi_j \rangle = - (\langle \psi_j \vert A \vert \psi_i\rangle )^\dagger = - \langle \psi_i \vert A^\dagger \vert \psi_j \rangle $$ which means that $A = -A^\dagger$, i.e. $A$ is anti-Hermitian. Since anti-Hermitian operators are in particular normal, they are diagonalizable by the spectral theorem, and therefore $\langle \alpha \vert A \vert \alpha\rangle = 0$ means that all eigenvalues are 0, i.e. the diagonalization of $A$ is the zero matrix, which also means $A = 0$.
Note that the application of the spectral theorem relies on the space being a complex vector space, and that the assertion would be false over a real vector space - $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ is a counterexample on $\mathbb{R}^2$ (but not on $\mathbb{C}^2$, since its expectations values do not vanish for all vectors there).
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