1
$\begingroup$

My understanding is that for an inner product in state-space, since we want the value to always be a real number we say that $$\langle\psi|\phi\rangle= {\langle\phi|\psi\rangle}^* $$

where * denotes complex conjugate.

But when we have an adjoint operator acting on $|\phi\rangle$ we say:

$$\langle\psi|A\phi\rangle = \langle A\phi|\psi\rangle^*$$

then we define : $$\langle A^\dagger\phi|\psi\rangle=\langle A\phi|\psi\rangle^*$$

my questions are:

  • Where do we denote that $A^\dagger = (A^T)^*$ ? it seems that we just say that $ A^\dagger = A^*$

    • is the A being inside a bra instead of a ket implicitly stating the transpose?
  • when we say $\langle A\phi|\psi\rangle^*$ , what exactly does the * get applied to?, because in our dagger definition it seems that the complex conjugate is just absorbed into the operator, why is it not applied to the inner product of the bra - kets any more?

$\endgroup$
1
  • $\begingroup$ Hi, you are right in your first comment. The notation in which an operator is acting on vectors belonging in the dual space is actually used to denote the dagger operator, i.e. $|A\psi\rangle^{\dagger}=A^{\dagger}\langle\psi|$... $\endgroup$
    – schris38
    Jul 3 at 6:42

3 Answers 3

5
$\begingroup$

IMHO it is better to avoid the use Dirac notation when first encountering the notion of adjoint operators. In the following, we'll assume that we're working with a finite-dimensional complex Hilbert space $H$. This will save us from discussing many technical details in infinite dimensions. To start, let $(\cdot,\cdot):H\times H \longrightarrow \mathbb C$ denote an inner product on $H$ which is anti-linear in the first and linear in the second argument. For $u,v \in H$ it is part of the definition of the inner product that $$(u,v) = (v,u)^* \quad .\tag{1} $$

Now let $A: H\longrightarrow H$ denote an (everywhere defined) operator on $H$. For any $v\in H$, we have that $Av \in H$ is simply a vector again. Using equation $(1)$ we find $$ (u,Av) = (Av,u)^* \quad .\tag{2}$$ The adjoint of $A$ is an operator denoted by $A^\dagger: H\longrightarrow H$ which obeys the following equation: $$ (u,Av) = (A^\dagger u,v) \quad .\tag{3}$$ Note that we've not complex conjugated or transposed anything in the definition$^1$ of the adjoint. In particular, the complex conjugation used in $(1)$ and $(2)$ is only applied to $(u,v) \in \mathbb C$ - nothing is 'absorbed' into an operator. Finally, your last equation can be derived as follows: $$(Av,u)^* \overset{(1)}{=} (u,Av) \overset{(3)}{=}(A^\dagger u,v) \tag{4} $$ and thus coincides with $(3)$. Again: The complex conjugation was applied on $\mathbb C$-numbers only and we nowhere transposed anything.

The upshot of this is that it is important to distinguish between an operator and an associated matrix. However, if you have chosen an ordered orthonormal basis $\{e_j\}_{j=1,\ldots,\dim H}$ and define the matrix elements of $A$ and $A^\dagger$ in this basis as $(A)_{ij}:= (e_i,Ae_j) $ and $(A^\dagger)_{ij}:=(e_i,A^\dagger e_j) $, respectively, then it is easy to show that $(A)_{ji}^* = (A^\dagger)_{ij}$ by just using the above properties $(1)$ - $(3)$. Then you can say that the matrix representing $A^\dagger$ in the said basis is the complex conjugated and transposed matrix of $A$.


$^1$ The definition is as follows: For $u,v\in H$ and an operator $A$, consider the linear functional $(u,A\cdot) :H\longrightarrow \mathbb C$ defined by $$ (u,A \cdot) (v) :=(u,Av) \tag{5} \quad .$$ We can now apply the Riesz representation theorem which guarantees the existence of a $z_u \in H$ such that for all $v \in H$ it holds that $$(u,A \cdot) (v) = (z_u,v) \tag{6} \quad.$$ This in turn allows us to define an operator, denoted by $A^\dagger$, as follows: $$ A^\dagger u:=z_u\tag{7} \quad ,$$ for all $ u \in H$. By definition, we then arrive at equation $(3)$. It is a standard exercise to verify that $A^\dagger$ defined through $(7)$ is indeed a well-defined linear map on $H$ with $\left(A^\dagger \right)^\dagger = A$ by using the properties of the inner product.

As mentioned in the beginning, things get much more complicated in infinite dimensions.

$\endgroup$
13
  • $\begingroup$ I agree with you about transposition, but why does complex conjugation of an operator $A$ only make sense if we work in a particular basis? Couldn't we define $A^*$ such that $(A v)^* = A^* v^*$ for all $v$, or something like that? $\endgroup$ Jul 3 at 15:05
  • $\begingroup$ Dear @MichaelSeifert What should $A^*$ or $v^*$ be/ denote? Even the complex conjugate of a vector does not make sense at all. $\endgroup$ Jul 3 at 15:13
  • 1
    $\begingroup$ Ah, I missed the second part of your comment due to editing. So you're saying that there is not a natural notion of complex conjugation on $H$ itself? That may be the part that I'm missing, though it seems counterintuitive to me. $\endgroup$ Jul 3 at 15:19
  • 1
    $\begingroup$ @MichaelSeifert Yes, I think so: Suppose that $v=\sum\limits_j v_j e_j$ and $v^*:=\sum\limits_j v_j^* e_j$ for some orthonormal basis. Then create a new basis $e_j^\prime : = i e_j$, which is an orthonormal basis again. Then $v= \sum\limits _j v_j^\prime e_j^\prime = \sum\limits_j v_j e_j$ but $v^* =\sum\limits_j (v_j^\prime)^*\, e_j^\prime \neq \sum\limits_j v_j^* \, e_j$. Rather, we obtain $v^* = - v^*$ just by expressing it in two different orthonormal bases. $\endgroup$ Jul 3 at 15:33
  • 1
    $\begingroup$ @JasonFunderberker: That's a very helpful link; thank you! $\endgroup$ Jul 3 at 18:38
0
$\begingroup$

you are right in your first comment. The notation in which an operator is acting on vectors belonging in the dual space is actually used to denote the dagger operator, i.e. $|A\psi\rangle^{\dagger}=\langle\psi|A^{\dagger}$.

As far as the second point is concerned, the complex congugate notation can be thought of as the complex conjugate of the complex number, equal to the result of the inner product between a ket and a bra. You can insert an operator whose eigenvalues you know between a ket and a bra and take the complex conjugate od the result to see how the adjoint operator acts on the dual space.

If there are any questions, please comment

$\endgroup$
0
$\begingroup$

If you think of $\vert \psi\rangle$ as a row vector, then a bra $\langle \phi\vert$ is not only conjugate but also transpose. For instance, if $$ \vert\psi\rangle=\left(\begin{array}{c}a\\ b\end{array}\right) $$ then $$ \langle \psi\vert=(a^*,b^*) $$ so that $\langle \phi\vert\psi\rangle$ is a number, i.e. the inner product is an array multiplication of an array of 1 row and two columns with an array of 2 rows and one column.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.