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It's my first time working with the bra-ket notation, and I'm still not too familiar with it. Schwarz inequality states that $|<v_1|v_2>|^2$$\leq$ $<v_1|v_1> <v_2|v_2>$.

I need to deduce that $|<\psi|AB|\psi>|^2$$\leq$ $<\psi|A^2|\psi> <\psi|B^2|\psi>$,where $A$ and $B$ are self-adjoint operators.

which I can't seem to figure out. The "bra" operator is defined by its action on the "ket" vector by $<v| : |w> \rightarrow <v|w>$. So, doesn't the notation $|<\psi|AB|\psi>|$ mean that the operator $<\psi|$ is getting applied to the vector $AB|\psi>$? How do I deduce that inequality from there?

user3461126
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2 Answers2

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Note that $\langle\psi| AB |\psi\rangle$ is also the inner product of $B|\psi\rangle$ and $A^\dagger|\psi\rangle$, which corresponds more nicely to what you want to prove.

(In $\LaTeX$ you can use "\langle" and "\rangle" for better angle brackets.)

mjqxxxx
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  • As you've added to the question that $A$ is self-adjoint, you don't need the $\dagger$ anymore. – mjqxxxx May 31 '18 at 16:08
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As proposed... more explicitly:

Pose $\begin{array}{c}|v1\rangle= A|\psi\rangle\\|v2\rangle= B|\psi\rangle\end{array} $ so the dual are $\begin{array}{c}\langle v1|= \langle\psi|A^{\dagger}\\\langle v2|= \langle\psi|B^{\dagger}\end{array}$

Then Schartz inequality becomes:

$|\langle\psi|A^{\dagger} B|\psi\rangle|^2 \le \langle\psi|A^{\dagger} A|\psi\rangle + \langle\psi|B^{\dagger} B|\psi\rangle$

But A and B are self-adjoint so $A^{\dagger}=A$ and $B^{\dagger}=B$, finaly

$|\langle\psi|A B|\psi\rangle|^2 \le \langle\psi|A^2|\psi\rangle + \langle\psi|B^2|\psi\rangle$

Loic Huguel
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