Dirac Notation for Operators: Ambiguity in Expectation Values?

[dyn]

Hi
If A is a linear operator but not Hermitian then the expectation value of A² is written as < ψ | A²| ψ >. Now if i write A² as AA then i have seen the expectation value written as < ψ | A⁺A| ψ > but if i only apply the operators to the ket , then could i not write it as < ψ | AA | ψ > ? In other words is the notation slightly ambiguous ?
Thanks

 

[topsquark]

Hi
If A is a linear operator but not Hermitian then the expectation value of A² is written as < ψ | A²| ψ >. Now if i write A² as AA then i have seen the expectation value written as < ψ | A⁺A| ψ > but if i only apply the operators to the ket , then could i not write it as < ψ | AA | ψ > ? In other words is the notation slightly ambiguous ?
Thanks

If A is not Hermitian, then ##AA \neq A^{\dagger} A##, so you can't write it that way.

-Dan

 

[dyn]

I might be confusing myself here but if A is not Hermitian and A² = AA and A³ = AAA then how do i write the expectation values of these 2 quantities ?

 

[topsquark]

I might be confusing myself here but if A is not Hermitian and A² = AA and A³ = AAA then how do i write the expectation values of these 2 quantities ?

The same way you did in the OP:
##\langle A^2 \rangle = \langle \psi \mid A^2 \mid \psi \rangle \equiv \langle \psi \mid AA \mid \psi \rangle##

You would have to calculate ##\mid \phi \rangle = A \mid \psi \rangle##, then ##\mid \zeta \rangle = A \mid \phi \rangle##, then finally ##\langle \psi \mid \zeta \rangle##.

That's as far as you can go until you specify what the operator A looks like.

-Dan

 

[DrClaude]

The same way you did in the OP:
##\langle A^2 \rangle = \langle \psi \mid A^2 \mid \psi \rangle \equiv \langle \psi \mid AA \mid \psi \rangle##

You would have to calculate ##\mid \phi \rangle = A \mid \psi \rangle##, then ##\mid \zeta \rangle = A \mid \phi \rangle##, then finally ##\langle \psi \mid \zeta \rangle##.

That's as far as you can go until you specify what the operator A looks like.

-Dan

Alternatively, you can calculate
$$
\begin{align*}
\ket{\phi} &= A \ket{\psi} \\
\ket{\chi} &= A^\dagger \ket{\psi} \\
\braket{\psi | A^2 | \psi} &= \braket{\chi| \phi}
\end{align*}
$$

 

[vanhees71]

I might be confusing myself here but if A is not Hermitian and A² = AA and A³ = AAA then how do i write the expectation values of these 2 quantities ?

For a pure state, represented by a normalized vector ##|\Psi \rangle## expectation value is
$$\langle f(\hat{A}) = \langle \Psi|f(\hat{A}) \Psi \rangle=\langle f(\hat{A})^{\dagger} \Psi|\Psi \rangle,$$
for an arbitrary function ##f(\hat{A})##. It doesn't matter whether the operator is self-adjoint or not for the identity of the two expressions. Of course, such an operator cannot represent an observable to begin with, and you might argue that it doesn't make sense to call this expression an "expectation value" in the first place.