In books on quantum physics you often see things like $\left(A|\Psi\rangle\right)^\dagger=\langle\Psi|A^\dagger$. However, $|\Psi\rangle:=\Psi\in\mathcal{H}$ is a vector and $\langle\Psi|:=\Psi^\ast\in\mathcal{H}^\ast$ is its dual counterpart, while $A$ is an operator, i.e. $A\in Aut(\mathcal{H})$. From what I understand, the property 'adjoint' is only defined with respect to operators.

Let $A,B:\mathcal{H}\rightarrow\mathcal{H}$ be operators. Then $B=A^\dagger$, iff $\langle B\Psi,\Phi\rangle=\langle\Psi,A\Phi\rangle$ for all $\Psi,\Phi\in\mathcal{H}$.

Then how is $|\Psi\rangle^\dagger$ defined and justified as notation and why do we have $|\Psi\rangle^\dagger=\langle\Psi|$?

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    $\begingroup$ Have you seen this? $\endgroup$
    – Tendero
    Jan 13, 2017 at 19:34
  • $\begingroup$ No, but it is of great help to me. Thanks a bunch! $\endgroup$ Jan 13, 2017 at 19:35
  • $\begingroup$ Glad to hear that! $\endgroup$
    – Tendero
    Jan 13, 2017 at 19:36
  • $\begingroup$ You can view a bra as a linear functional that takes a vector in a hilbert space and gives you a number. There are bras that take a basis ket and spit out a number one for example...but for all the other basis kets, they spit zero. These make a dual basis in a dual space of these linear functionals. So this dual space is also a vector space. $\endgroup$ Jan 13, 2017 at 21:59
  • $\begingroup$ Rigorously, let $\Psi \in H$ with $H$ a Hilbert space. Then $H^\star= \mathrm{Hom}(H,\mathbb{C})$ and $\Psi^\dagger \in H^\star$. In physics then we often dress the elements of $H$ and $H^\star$ as kets and bras without much gain actually. $\endgroup$ Sep 29, 2019 at 20:13

1 Answer 1


A vector is an arrow in spatial dimensions, it is not related to states in a hilbert space of the quantum mechanics. "Adjoint" is defined only in relation to the inner product of the states, not the operators. What you make us of when writing things like $|\Psi\rangle^\dagger=\langle\Psi|$ is the inner product of your complex state space, which relates complex conjugacy to the symmetries of said inner product !

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    $\begingroup$ I'm afraid, that is actually wrong since a Hilbert space is a vector space in particular, i.e. its elements are vectors. But you are right: adjointness is defined with respect to the inner product of the vector space that the operators are defined on. $\endgroup$ Jan 13, 2017 at 19:50
  • $\begingroup$ But the elements of a vector space are not vectors but states, a vector is just an arrow pointing at things in the real world $\endgroup$ Jan 13, 2017 at 19:51
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    $\begingroup$ No, that's wrong. A vector space is concisely defined in mathematics: link. The analogy to our space in reality makes sense for the vector space $\mathbb{R}^3$ with the standard scalar product and the Euklidean norm. In quantum physics the span of a vector in a Hilbert space, i.e. a one-dimensional subspace of the Hilbert space is called a state. $\endgroup$ Jan 13, 2017 at 19:59
  • $\begingroup$ But how should this be a vector, on what does this point in this Hilbert space ? You can't point to states ! $\endgroup$ Jan 13, 2017 at 20:04
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    $\begingroup$ @StringTheoretican You use a very nonstandard (confused) mathematical terminology. The elements of a vector (or linear) space are called vectors. The states are actually vectors as well, in the continuous dual of the linear space of (bounded) observables. This means that observables are vectors as well. This is the standard mathematical terminology. For you "vector" refers to a translation or displacement vector in an affine space. $\endgroup$
    – yuggib
    Jan 13, 2017 at 20:10

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