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?

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    $\begingroup$ Which math books? Can you provide a reference for this? $\endgroup$
    – J. Murray
    Jan 5 at 19:24
  • $\begingroup$ Most functional analysis books, e.g. John Conway "A course in functional analysis" p31, or Peter Woit's text book page 47. $\endgroup$
    – Yuval
    Jan 5 at 19:27
  • $\begingroup$ 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) $\endgroup$ Jan 5 at 19:29
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    $\begingroup$ 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.) $\endgroup$
    – Hyperon
    Jan 5 at 19:37
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/743398/2451 and links therein. $\endgroup$
    – Qmechanic
    Jan 5 at 19:55

1 Answer 1


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.


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