Inner product of a vector with an operator

[EquationOfMotion]

So say our inner product is defined as $\int_a^b f^*(x)g(x) dx$, which is pretty standard. For some operator $\hat A$, do we then have $\langle \hat A ψ | \hat A ψ \rangle = \langle ψ | \hat A ^* \hat A | ψ \rangle = \int_a^b ψ^*(x) \hat A ^* \hat A ψ(x) dx$? This seems counter-intuitive. Say our operator is $\frac{d}{dx}$ and $ψ(x)=x$. Then, evidently $\langle \hat A ψ | \hat A ψ \rangle = b-a$ and $\langle ψ | \hat A ^* \hat A | ψ \rangle = 0$, which is obviously incorrect. What am I missing?

 

[fresh_42]

So say our inner product is defined as $\int_a^b f^*(x)g(x) dx$, which is pretty standard. For some operator $\hat A$, do we then have $\langle \hat A ψ | \hat A ψ \rangle = \langle ψ | \hat A ^* \hat A | ψ \rangle = \int_a^b ψ^*(x) \hat A ^* \hat A ψ(x) dx$? This seems counter-intuitive. Say our operator is $\frac{d}{dx}$ and $ψ(x)=x$. Then, evidently $\langle \hat A ψ | \hat A ψ \rangle = b-a$ and $\langle ψ | \hat A ^* \hat A | ψ \rangle = 0$, which is obviously incorrect. What am I missing?

You must consider the correct brackets. $\langle A\psi , A\psi \rangle = \int_a^b (A(\psi))^*(x)(A(\psi))(x)dx$ and $\frac{d}{dx}(\psi)=\psi$ if $\psi = \operatorname{id}$.

 

[PeroK]

So say our inner product is defined as $\int_a^b f^*(x)g(x) dx$, which is pretty standard. For some operator $\hat A$, do we then have $\langle \hat A ψ | \hat A ψ \rangle = \langle ψ | \hat A ^* \hat A | ψ \rangle = \int_a^b ψ^*(x) \hat A ^* \hat A ψ(x) dx$? This seems counter-intuitive. Say our operator is $\frac{d}{dx}$ and $ψ(x)=x$. Then, evidently $\langle \hat A ψ | \hat A ψ \rangle = b-a$ and $\langle ψ | \hat A ^* \hat A | ψ \rangle = 0$, which is obviously incorrect. What am I missing?

Careful! The differential operator is not Hermitian on the set of all functions. Try with a valid wave-function on an interval.

 

[EquationOfMotion]

Careful! The differential operator is not Hermitian on the set of all functions. Try with a valid wave-function on an interval.

Doesn't moving around the operators in an inner product work regardless of whether or not it's Hermitian?

 

[PeroK]

Doesn't moving around the operators in an inner product work regardless of whether or not it's Hermitian?

Yes, but you have to work out what the Hermitian conjugate of your operator is. That's assuming it has one.

 

[EquationOfMotion]

Yes, but you have to work out what the Hermitian conjugate of your operator is. That's assuming it has one.

Ooh, so it does always work as long as you know what the Hermitian conjugate is, but you of course *need* the correct conjugate. Thanks!

 

[PeroK]

Ooh, so it does always work as long as you know what the Hermitian conjugate is, but you of course *need* the correct conjugate. Thanks!

The proof of The Hermicity of $d/dx$ involves integration by parts and taking the term evaluated on the boundary to be zero, where the range is either $R$ or a finite interval. The function $x$ does not vanish on the boundary, so the operator is only Hermitian on a suitable subset of all functions.

Technically, therefore, all valid wavefunctions must be zero on the boundary.