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?

  • $\begingroup$ We have $\langle u\mid v\rangle =\langle v\mid u\rangle^*$ for all vectors $u, v$. $\endgroup$ – Berci Oct 27 '18 at 15:48
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    $\begingroup$ This is one reason Mathematicians do not use the bra-ket notation: you cannot tell what the operator in the middle acts on. If the operator is selfadjoint, then everything is okay, but it is useful to be able to study symmetric operators that aren't selfadjoint, in order to determine the deficiency spaces, which determine the boundary conditions for the problem. $\endgroup$ – DisintegratingByParts Oct 27 '18 at 16:14
  • $\begingroup$ I have seen notation $\langle x | A \} | y\rangle$ and $\langle x | \{ A | y \rangle$. That fixes a lot of problems, but it sure is awkward. $\endgroup$ – DisintegratingByParts Oct 27 '18 at 16:42
  • $\begingroup$ @DisintegratingByParts But $\left(\langle x | A\right) | y \rangle$ and $\langle x | \left(A | y \rangle\right)$ are exactly the same, aren't they? Isn't this the whole point of why this notation is used in the first place? $\endgroup$ – Joppy Oct 28 '18 at 10:45
  • $\begingroup$ @Joppy : The notation $\langle x| A| y\rangle$ is not adequate. That's the point of my remark. And the substitute notation that you and I mentioned is needlessly complex and awkward. That's why I don't like bra-ket notation, and have no interest in using it. The usual inner product defined by von Neumann was perfectly adequate and was around well before Dirac devised his notation. $\endgroup$ – DisintegratingByParts Oct 28 '18 at 15:49

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").

If you really care about the math behind, given an operator $T:X\to Y$, one can always defined 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 $H$.

  • $\begingroup$ The last $\langle Av | u \rangle = \langle u | Av \rangle$ does not hold - there should be a complex conjugation in there. $\endgroup$ – Joppy Oct 28 '18 at 11:55
  • $\begingroup$ Nice catch, thanks. $\endgroup$ – Martin Argerami Oct 28 '18 at 14:10

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.


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