# 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? Jan 5 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. Jan 5 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) Jan 5 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.) Jan 5 at 19:37
• Possible duplicates: physics.stackexchange.com/q/743398/2451 and links therein. Jan 5 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 $$\langle V$$. It represents the action of $$\hat{\Omega}$$. 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 $$\langle V \lvert \left(\hat{\Omega}^{\dagger} \rvert U \rangle\right)^{\dagger}$$,.

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.