# Shankar's definition of adjoint [duplicate]

I think Shankar's definition of adjoint operator (in his QM book) differs from many other sources. On page 26, he made the definition

$$\langle \Omega V|=\langle V|\Omega^\dagger \quad .$$

Now $$\langle V|$$ belongs to the dual space (mathematicians use the notation $$H'$$) of $$H$$, i.e. the space of all complex linear maps $$H\to {\mathbb C}$$. In math text books, the action of $$\Omega^\dagger$$ on $$\langle V|$$ is, $$\langle V|$$ compose with $$\Omega$$, i.e. $$\langle V|\Omega^\dagger$$ should be the linear map which maps $$|W\rangle$$ to $$\langle V|\Omega W\rangle$$. Definition in some physics books, e.g. Griffiths, Peter Woit, seems to be equivalent to this. However, if you use Shankar's definition, $$\langle V|\Omega^\dagger$$ would have been the linear map which maps $$|W\rangle$$ to $$\langle \Omega V|W\rangle$$. If $$\Omega$$ is not self-adjoint, this is different from the math (and Peter Woit) definition.

Am I understanding this matter correctly?

• Which math books? Can you provide a reference for this? Commented Jan 5, 2023 at 19:24
• Most functional analysis books, e.g. John Conway "A course in functional analysis" p31, or Peter Woit's text book page 47. Commented Jan 5, 2023 at 19:27
• Out of curiosity, in the math text books, is the inner product also anti-linear in the first entry? Despite that, check e.g. eq. 4.1 of this. I don't have the other references (and it would be great if you could include edition/page/eq. number) Commented Jan 5, 2023 at 19:29
• I have no access to the math text books you mentioned, but usually mathematicians define the scalar product in a complex Hilbert space being linear in the left entry and anti-linear in the right, whereas in physics the opposite convention is chosen. Might this be reason for the confusion? (Shankar's definition is correct with the physicist's convention.) Commented Jan 5, 2023 at 19:37
• Possible duplicates: physics.stackexchange.com/q/743398/2451 and links therein. Commented Jan 5, 2023 at 19:55

Generally speaking, mathematical notations can fail. This is one place where Dirac notation fails, and it fails partially by being ambiguous. Here is how I understand how this works. Note first that I don't like pulling operators into kets. I think of the symbols inside a ket as labels for the ket, and you can't act with operators on labels. Thus, here's my way of being careful with matrix elements of non-Hermitian operators.

First, always assume that operators act to the right. That is, $$\langle \psi \lvert \hat{\Omega} \rvert \varphi \rangle \to \langle v_\psi,\Omega v_\varphi\rangle\,,$$ where the second expression is the typical$$^1$$ mathematical notation for the inner product on a Hilbert space:

• $$v_{\psi}$$ is an alternative notation for $$\lvert \psi \rangle$$,
• $$\Omega v_\varphi$$ is the operation of the operator $$\Omega$$ on the vector $$v_{\varphi}$$, represented in Dirac notation as $$\hat{\Omega}\lvert \phi \rangle$$, and
• $$\langle v_\psi,\Omega v_\varphi\rangle$$ is the inner product of $$\Omega v_\varphi$$ with $$v_{\psi}$$.

Thus, I always interpret $$\langle \psi \lvert \hat{\Omega} \rvert \varphi \rangle$$ as $$\langle \psi \lvert \left(\hat{\Omega} \rvert \varphi \rangle\right)$$, even if there is an adjoint, i.e., it's also true that $$\langle \psi \lvert \hat{\Omega}^{\dagger} \rvert \varphi \rangle = \langle \psi \lvert \left(\hat{\Omega}^{\dagger} \rvert \varphi \rangle\right)\,.$$

Now, what about if we wanted to take the inner product in the other direction? Let's trace this through: $$\langle \Omega v_\psi, v_\varphi\rangle = \langle v_\psi, \Omega^{\dagger}v_\varphi\rangle "=" \langle \psi \lvert \left(\hat{\Omega}^{\dagger} \rvert \varphi \rangle\right) = \langle \psi \lvert \hat{\Omega}^{\dagger} \rvert \varphi \rangle\,.$$ This here is what is really meant by the notation $$\langle \Omega \psi \lvert = \langle \psi \lvert\hat{\Omega}^{\dagger}$$: it's that the $$\hat{\Omega}^{\dagger}$$ actually acts to the right on whatever kets are going to be put in that spot. You do not think of this as $$\Omega^{\dagger}$$ acting to the left. This is the answer to your following complaint:

In math text books, the action of $$\Omega^\dagger$$ on $$\langle V|$$ is, $$\langle V|$$ compose with $$\Omega$$, i.e. $$\langle V|\Omega^\dagger$$ should be the linear map which maps $$|W\rangle$$ to $$\langle V|\Omega W\rangle$$.

Well, $$\langle V \lvert\hat{\Omega}^{\dagger}$$ does not represent the action of $$\hat{\Omega}^{\dagger}$$ on $$\langle V\rvert$$. It represents the action of $${\hat{\Omega}}$$ on $$\lvert V\rangle$$, than conjugated. That is, again, $$\langle V \lvert\hat{\Omega}^{\dagger} \rvert U \rangle "=" \langle V, \Omega^{\dagger} U\rangle = \langle \Omega V, U\rangle\,,$$ with a slight abuse of (my) notation where I am identifying $$V$$ (math notation) with $$\lvert V \rangle$$ (Dirac notation).

We could take this one step further and try to represent this more explicitly in the Dirac notation as $$\left(\hat{\Omega}\lvert V \rangle\right)^{\dagger}\lvert U \rangle = \langle V \lvert \hat{\Omega}^{\dagger} \rvert U \rangle\,,$$ where the second expression is understood to be equivalent to $$\langle V \lvert \left(\hat{\Omega}^{\dagger} \rvert U \rangle\right)$$.

Note that this is consistent with the mathematical notation as well: \begin{align} \langle \psi \lvert \hat{\Omega} \rvert \varphi \rangle "=" \langle v_\psi,\Omega v_\varphi\rangle = \langle \Omega^{\dagger}v_\psi, v_\varphi\rangle = \langle v_\varphi, \Omega^{\dagger} v_\psi\rangle^* "=" \left(\langle \varphi \lvert \left(\hat{\Omega}^{\dagger} \rvert \psi \rangle\right)\right)^* = \langle \varphi \lvert \hat{\Omega}^{\dagger} \rvert \psi \rangle^*\,. \end{align}

$$^1$$ Up to the order of the vectors. I've often seen this where the inner product is linear in the first argument rather than the second and anti-linear in the second rather than the frst, but I'll do it the other way that matches the Dirac notation better.