Dirac expression derivation

Question

In Quantum Mechanics, 2nd Edition by Davies & Betts on page 78 it states that there is a symmetry implied by the following Hermitian operator equation:

$${\displaystyle \int \phi^{*}(A \psi)d \,\tau} = {\displaystyle \int \psi(A \phi)^{*}d \,\tau} \tag{1}$$

in the way the Hermitian operator $A$ acts. It may either be regarded as acting forwards on $\psi$ or, complex conjugated backwards on $\phi$. This means:

$$\langle\phi A|\psi\rangle=\langle\phi|A\psi\rangle\tag{2}$$

so one usually writes both equivalent expressions in symmetric form:

$$\langle\phi| A|\psi\rangle\tag{3}$$

Does anyone know of a proof of this derivation (it looks odd as (1) swaps the order of the $\phi$ and $\psi$ functions, and has the operator $A$ applied to its right, but (2) does not).

Answer 1

I will neglect all rigorous description and just provide the usual physics arguments. Let $A$ be a quantum operator and $\psi,\phi$ arbitrary wave functions. Then using the inner product of the underlying Hilbert space $H$ we can write

$$\langle \psi|A\phi\rangle = \int \psi^\star(x)A\phi(x)dx.$$

By definition the adjoint of $A$ is defined by $\langle \psi|A\phi \rangle = \langle A^\dagger \psi|\phi\rangle$. If $A$ is self-adjoint (symmetric) then $\langle \psi|A\phi \rangle = \langle A \psi |\phi \rangle$ for all $\psi,\phi \in H$. So equation (2) seems bad to me. However, I can also show for you that an extra bar does in matter in Dirac notation, i.e. : $\langle \psi|A\phi \rangle = \langle \psi|A|\phi \rangle.$ This means that so far I worked in $x$-representation (and $A$ should be $A_x$ in the previous part to be precise) and now I will move to abstract notation. The idea is as follows using the resolution of the identity:

$$\langle \psi|A|\phi \rangle = \int \langle \psi|x\rangle \langle x|A|\phi \rangle dx = \int \psi^\star(x) A_x \phi(x)dx \equiv \langle \psi|A\phi \rangle,$$

where $A_x$ is the $x$-representation of $A$.

Answer 2

The notation here seems to have gone seriously awry. As taught to me at Cambridge by the students of Dirac, kets are vectors and bras are vectors in the dual space. I have never seen anyone write an operator inside a ket or a bra. That would not make sense as the symbol inside the ket is a label not a vector. E.g. one writes $|p\rangle$ for a momentum ket or $|x\rangle$ for a position ket.

Whether Hermitian or not the operator can always be seen as acting both ways, as one has $$ (\langle g| A )|f\rangle = \langle g|( A |f\rangle) =\langle g| A |f\rangle.$$ so that the round brackets can always be omitted. The Hermitian conjugate is defined such that $A^\dagger|g\rangle$ is the ket corresponding to $\langle g|A$, so there isn't really anything to derive here, beyond the fact that the integral defines an inner product.

If you were using algebraic notation you would write, for a Hermitian operator $$ (A\psi,\phi) = (\psi,A\phi) $$ and again there would be nothing to prove, but it does not pay to conflate the algebraic notation with ket notation, which has very real advantages when correctly used.