Definition of Adjoint Operator for Quantum Mechanics

Question

While learning about adjoint operators for quantum mechanics, I encountered two definitions.

The first definition is given by Shankar in The Principle of Quantum Mechanics:

Given a ket $$ A\lvert V \rangle = \lvert A V \rangle$$ the correspoding bra is $$\langle AV \rvert=\langle V \rvert A^\dagger$$ which defines the operator $A^\dagger$.

The second definition is:

For a linear operator $A$, its adjoint is defined so that $$\langle u \rvert A^\dagger \lvert v \rangle = {\langle v \rvert A \lvert u \rangle}^* $$ where $^*$ means to take the complex conjugate.

From the first definition, it seems that the adjoint $A^\dagger$ should only act on bras. But from the second definition the adjoint $A^\dagger$ acts on a ket $\lvert v \rangle$. How does the two definition reconcile with each other? Is the first definition of the adjoint somewhat misleading, since it can also act on kets?

Answer 1

The underlying reason is that for a Hilbert space $H$, its dual is again $H$. What this means is that any bounded linear functional $H\to\mathbb C$ is given by the inner product against some fixed vector. So, any linear functional is given by choosing some $v\in H$ and then doing $u\longmapsto \langle v|u\rangle$ (this is known to mathematicians as the Riesz Representation Theorem).

That is, the "kets" are the elements of the space $H$, and the "bras" are the elements of the dual. Now, the adjoint $A^\dagger$ is defined precisely as the operator (on the dual $H^*$, but for us it is again $H$) such that $$\tag1\langle A^\dagger v|u\rangle=\langle v|Au\rangle.$$ When $A$ is selfadjoint, you have $$ \langle v|Au\rangle=\langle Av|u\rangle=\overline{\langle u|Av\rangle}. $$ As a mathematician, it is hard for me to grasp any advantage from talking about bras and kets. It would make a lot more sense to write $v^*u$ instead of $\langle v|u\rangle$. Here, for a vector $v$, the vector $v^*$ is the conjugate transpose, and then $v^*u$ is precisely the matrix product. When there is an operator in the mix, $$ v^*(Au)=v^*Au=(A^*v)^*u, $$ which is the same as $(1)$ but doesn't require any convention other than the mathematical properties of taking adjoints (namely, "conjugate and tranpose").

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If you really care about the math behind, given an operator $T:X\to Y$, one can always define an adjoint $T^*:Y^*\to X^*$ (where $X^*$ is the dual, i.e. the space of bounded linear functionals on $X$) by $$ (T^*g)(x)=g(Tx). $$ Because, as already mentioned, for a Hilbert $H$ the dual is again $H$, we can see the adjoint again as an operator on $H$.

Answer 2

Suppose the underlying (complex Hilbert) vector space is $V$, with inner product $(-, -)_V$. We then have bras and kets. Kets are the easiest to understand: $| v \rangle$ is just the vector $v$. The bra $\langle v |$ is a linear map $V \to \mathbb{C}$, given by $ \langle v | (| w \rangle ) = (v, w)_V$. For aesthetic reasons, we leave out the parentheses and extra vertical line when feeding a ket into a bra, to get $\langle v | w \rangle = (v, w)_V$.

Any linear operator $A: V \to V$ acts naturally on both bras and kets. Since a ket is a vector, its action is the usual one: $A | v \rangle$ means the vector $Av \in V$. Recall that a bra $\langle v |$ is a map $V \to \mathbb{C}$, so the map $\langle v | A$ is the map $V \to \mathbb{C}$ which first applies $A$, then applies $\langle v|$. We get $$ \langle v | A | w \rangle = \langle v | (A (| w \rangle)) = (v, Aw)_V = (\langle v| A)(| w\rangle)$$ i.e. the expression $\langle v | A | w \rangle$ is consistent with both these interpretations (of course it is, it's just composition of operators).

So what is the adjoint of $A$? An interesting thing that happens because the inner product is conjugate-linear in the first argument is that if $|v\rangle + \lambda |w \rangle$ is a ket, its corresponding bra is $\langle v | + \lambda^* \langle w |$, where $\lambda^*$ means the complex conjugate. The adjoint is just the translation of this for operators: fix an operator $A: V \to V$. It can be shown that there exists an operator $A^\dagger: V \to V$ such that if $A | v \rangle$ is a ket, then its corresponding bra is $\langle v | A^\dagger$. This definition of adjoint was the first one you listed.