I saw it in this pdf, where they state that

$P=P^\dagger$ and thus $P$ is hermitian.

I find this notation confusing, because an operator A is Hermitian if

$\langle \Psi | A \Psi \rangle=\langle A \Psi| \Psi \rangle=\left( \langle \Psi|A\Psi\rangle \right)^\dagger$

or more elaborately

$\int H^\dagger \Psi ^\dagger \Psi dV =\int \Psi^\dagger H \Psi dV $

or in another way ( $\left< X\right>$ being expectation value here, to avoid confusion)


but surely this does not imply $A=A^\dagger$?

Look at the momentum-operator $\vec{p}=-i\hbar\nabla $. But $\vec{p}\neq \vec{p}^\dagger$

Is it just sloppy notation and are they talking about expectation values, or are Hermitian operators actually operators that are equal to their Hermitian conjugate? My QM course that I've been taught would state otherwise.

Let me try to illustrate this with another (1D )example:

$\langle \Psi | P\Psi \rangle \\ =\int \Psi ^\dagger(-i\hbar \frac{\partial}{\partial x})\Psi dx\\ =\Psi ^\dagger \Psi |^{\infty}_{-\infty}+i\hbar\int \Psi \frac{\partial \Psi ^\dagger}{\partial x}dx\\ =0+\int \Psi (-i\hbar\frac{\partial \Psi )}{\partial x})^\dagger dx\\ =\int (-i\hbar\frac{\partial}{\partial x})^\dagger \Psi ^\dagger \Psi dx\\ =\langle P \Psi | \Psi \rangle$

Now this shows that the expectation values might be the same, but the operator between the parentheses itself is different in the first and last expression (look at the parentheses + $\dagger$ as a whole). I think the reason why I am confused is because I don't understand how $\dagger$ works on something that is not a matrix (if that is even possible).


6 Answers 6


The other answers are correct, but the OP still seems confused about how to prove that $\hat{p}$ is hermitian, so I will briefly show how to demonstrate that here using the properties of hermitian operators already given by other posters.

As already stated, the defining property of a hermitian operator is that it is equal to its conjugate transpose, i.e. its matrix elements satisfy $A_{mn} = (A_{nm})^{\ast}$. For the momentum operator you need to use the $L^2$ inner product to find the matrix elements. The OP's confusion stems from the assumption that these can be found from an expectation value, but this is not as general as a matrix element. The expectation value ($\hbar = 1$) $$ \langle\psi|\hat{p}|\psi\rangle = \int\mathrm{d}x\, \psi^{\ast}(x)\left(-i\frac{\partial}{\partial x} \psi (x)\right) $$ only gives a diagonal element in a basis in which $\psi(x)$ is a basis vector.

We want a general (possibly off-diagonal) matrix element in the basis of wave functions $\{\phi_m(x)\}$. This is given by

\begin{eqnarray} p_{mn} = \langle \phi_m|\hat{p}|\phi_n\rangle &=& \int\mathrm{d}x\, \phi^{\ast}_m(x)\left(-i \frac{\partial}{\partial x} \phi_n(x)\right) \\ &=& \int\mathrm{d}x\, \left(+i \frac{\partial}{\partial x} \phi_m^{\ast}(x)\right)\phi_n(x) \\ &=& \left[\int\mathrm{d}x\,\phi^{\ast}_n(x) \left(-i\frac{\partial}{\partial x}\phi_m(x)\right) \right]^{\ast} \\ &=& (p_{nm})^{\ast} \end{eqnarray} The second line follows by integration by parts and dropping boundary terms. The third line is a trivial rearrangement of terms. The final equality showing $p_{mn} = (p_{nm})^{\ast}$ follows by comparison of the third and first lines. Hope that's clear.

Incidentally, the expression for $p_{mn}$ above gives you a recipe for constructing a matrix representation of the operator $\hat{p}$ in some basis, however since the momentum $p$ is a continuous variable the matrix would be infinite-dimensional, so I don't recommend actually doing it!


This isn't just sloppy notation. The symbol $A^\dagger$ denotes the adjoint of the matrix $A$. It's defined by giving its matrix elements $\langle \psi |A^\dagger \phi \rangle = ( \langle \phi| A \psi \rangle)^* = \langle A \psi | \phi \rangle$, where $^*$ is complex-conjugation. This is a new operator, constructed by rearranging the bits of $A$: it's the conjugate transpose. A Hermitian operator is one which is equal to its adjoint.

In particular, the momentum operator $P = -i\hbar \partial_q$ actually is self-adjoint, because the derivative operator is skew-symmetric. $P^\dagger = (-i\hbar)^* \partial_q^\dagger = i\hbar (-\partial_q) = -i\hbar \partial_q = P$. (You can prove the derivative operator is skew symmetric by using integration by parts for $L^2$ functions.)

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    $\begingroup$ So is this notation only correct for the matrix representation, or is it correct in/for all cases/representations/observables? My course never really stated that explicitly so it would be reckless of me to just assume that. $\endgroup$
    – Jan M.
    Dec 9, 2012 at 18:35
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    $\begingroup$ which notation? $P=P^\dagger$? That doesn't make reference to any basis. $\endgroup$
    – user1504
    Dec 9, 2012 at 18:40
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    $\begingroup$ Also, for the thing you said about the proving of the skew symmetry, you prove that by integration, which implies you calculate an expectation value, but isn't that already a step to far? I thought that $P^\dagger$ would just mean taking the complex conjugate (and looking at $P$ as a complex number (i.e. $-i\hbar \left(\frac{\partial}{\partial x}+... \right)$ and just doing $a+ib \rightarrow a-ib$ and you're done) $\endgroup$
    – Jan M.
    Dec 9, 2012 at 18:41
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    $\begingroup$ No, it's not a step too far. $P^\dagger$ doesn't mean complex conjugate; it means conjugate and transpose. You can $A^\dagger = {(A^*)}^T$, if you like, but the point in any case is that $P$ is an operator, not a complex number. $\endgroup$
    – user1504
    Dec 9, 2012 at 18:50
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    $\begingroup$ That's a vector of operators. $P = (P_x, P_y, P_z)$. You're supposed to apply the adjoint to each component. So $P^\dagger = (P^\dagger_x,P^\dagger_y,P^\dagger_z)$. $\endgroup$
    – user1504
    Dec 9, 2012 at 18:58

Hermitian in the sense used by physicists doing quantum mechanics is usually meant to be self-adjoint, or, equivalently, with real eigenvalues. That is, for some matrix representation of $A$:

$$ \left( A^{\dagger} \right)_{ij} \equiv \left(A \right)^{\star}_{ji} = \left(A\right)_{ij}$$

where the first equality is the definition of $A^\dagger$ and the second one means that $A$ is self-adjoint/hermitian, which implies real eigenvalues, i.e. $A$ corresponds to an observable.


I think the main issue is that you are being confused by the Dirac notation, which works fine when it does but can be occasionally confusing when you are worried about that kind of thing. I'll try and give an accessible explanation that also addresses the issues raised by Wouter.

Take a Hilbert space $\mathcal{H}$, and a linear operator $\hat{A}:\mathcal{H}\rightarrow\mathcal{H}$. The Hilbert space comes equipped with an inner product $\langle\cdot,\cdot\rangle:\mathcal{H}\times\mathcal{H}\rightarrow\mathbb{C}$, and this inner product associates to each vector $\psi\in\mathcal{H}$ a linear functional $\langle\psi,\cdot\rangle:\mathcal{H}\rightarrow\mathbb{C}$, which acts in the natural way. Thus there is a correspondence between the Hilbert space and the set $\mathcal{H}^\ast$ of all linear functionals from $\mathcal{H}$ into $\mathbb{C}$. This space is known as the dual space of $\mathcal{H}$ and is where (well-behaved$^1$) kets live.

Consider then a fixed vector $\phi\in\mathcal{H}$ and a linear operator $\hat{A}: \mathcal{H} \rightarrow\mathcal{H}$. These two define a new linear functional which depends only on the one associated with $\phi$: $$ \psi\mapsto\langle\phi,\hat{A}\psi\rangle. $$ If you denote by $\phi^\dagger=\langle\phi,\cdot\rangle\in\mathcal{H}^\ast$ the linear functional associated with $\phi$, then the new functional defined above is a function of $\phi^\dagger$: it is denoted $\hat{A}^\dagger\phi^\dagger\in\mathcal{H}^\ast$, and this function $\hat{A}^\dagger$ is called the adjoint of $\hat{A}$.

Now, since the Hilbert space and its adjoint are in a strict correspondence, we can pull this back to define the action of the adjoint on $\mathcal{H}$ itself. This is done in the obvious way: $\hat{A}^\dagger$ acting on an arbitrary vector $\phi\in\mathcal{H}$ gives the unique vector $\hat{A}^\dagger\phi\in\mathcal{H}$ such that for any $\psi\in\mathcal{H}$ you have $$ \langle\hat{A}^\dagger\phi,\psi\rangle=\langle\phi,\hat{A}\psi\rangle. $$ This is really the definition of the adjoint $\hat{A}^\dagger$.

Since the adjoint $\hat{A}^\dagger$ acts on the same space as the original operator $\hat{A}$, they are comparable and we can ask whether they are equal. With the definition just given, the condition for that is $$ \hat{A}\textrm{ is hermitian} \Leftrightarrow \hat{A}=\hat{A}^\dagger\Leftrightarrow \langle\hat{A}\phi,\psi\rangle=\langle\phi,\hat{A}\psi\rangle\textrm{ for all }\phi,\psi\in\mathcal{H}. $$ The argument I gave above works only for the left-to-right sense of this equivalence. To prove the converse you need the theorem @MarkMitchison mentioned (in essence) earlier, which states modulo subtleties that two operators are equal if and only if all their matrix elements are equal. That is:

$$ \hat{A}=\hat{B}\Leftrightarrow \langle\phi,\hat{A}\psi\rangle=\langle\phi,\hat{B}\psi\rangle\textrm{ for all }\phi,\psi\in\mathcal{H}. $$

That's as far as the maths is concerned; now for some physics. Why did I state Dirac notation can be confusing? Well, when you come across a matrix element like $ \langle\phi|\hat{A}|\psi\rangle$, it can be hard to see what exactly is acting on what. In the terms laid out above, $$ \langle\phi|\hat{A}|\psi\rangle =\langle\phi,\hat{A}\psi\rangle =\langle\hat{A}^\dagger\phi,\psi\rangle =\left(\hat{A}^\dagger\phi^\dagger\right)(\psi). $$ The last term means the linear functional $\hat{A}^\dagger\phi^\dagger$ acting on the vector $\psi$. This is what is meant by saying that $\hat{A}^\dagger$ acts on bras.

$^1$I'm going to ignore the fact that functionals may be discontinuous, operators may be unbounded, domains may be restricted, symmetric operators need not be hermitian, and such other difficulties. All of this can be rigorously dealt with using rigged Hilbert spaces. For resources on that see e.g. this question.


First, regarding the question in your title, the Hamiltonian operator $H$ is real (has real eigenvalues) and $H=H^{*}$ is not "sloppy notation" but a consequence of this.

Second, regarding the definition of Hermitian operator your reference is correct. From Volume I of Quantum Mechanics by Cohen-Tannoudji, Diu, and Laloë; Chapter II; section 4.e "Hermitian operators":

An operator A is said to be Hermitian if it is equal to its adjoint, that is if $$A=A^\dagger$$

Adjoint does not mean conjugate.


It is not just sloppy notation, in the sense that there is no real equality of the two objects, rather a symbolic equality.

Short version

An operator $\hat{A}$ acts on a ket $|\psi\rangle$, while its adjoint $\hat{A}^{\dagger}$ acts on a bra $\langle\psi|$.

We say an operator is hermitian if
$\langle\phi|\hat{A}|\psi\rangle = \langle\phi|\hat{A}^{\dagger}|\psi\rangle$
which is not the same as
$\hat{A} = \hat{A}^{\dagger}$
since the left hand side has the operator acting on the ket, while in the right hand side the operator (namely the adjoint of the first operator) acts on the bra.

Extended version

Let's call the Hilbert space of the kets $\mathcal{V}$ and consider a linear operator $\hat{A}$ which acts on $\mathcal{V}$, i.e.
$\hat{A} : \mathcal{V} \to \mathcal{V}.$
In other words, $\hat{A}$ takes a ket $|a\rangle$, element of $\mathcal{V}$, as input and returns, in general, a different ket $|b\rangle = \hat{A}|a\rangle$ as output.

With the Hilbert space $\mathcal{V}$ of the vectors we call kets, we associate a dual space $\mathcal{V}^*$ of functionals $\langle a|$ we call bras, which act on kets to yield complex numbers, i.e.
$\langle a| : \mathcal{V} \to \mathbb{C}$.
In this dual space every ket $|a\rangle \in \mathcal{V}$ has an associated dual element $\langle a| \in \mathcal{V}^*$. The inproduct of two kets $|a\rangle$ and $|b\rangle$ is defined with the help of these associated dual elements as follows:
$(|a\rangle)\cdot(|b\rangle) = \langle a|b\rangle = \sum_j a_j^* b_j$ (if the dimension of $\mathcal{V}$ is finite or denumerably infinite)
where $\langle a| = (|a\rangle)^{\mathrm{dual}}$.

Now we can define an adjoint operator $\hat{A}^{\dagger} : \mathcal{V}^* \to \mathcal{V}^*$ acting on these dual elements, these bras. And we define it such that
$\hat{A}^{\dagger} : \langle a| \mapsto \langle b|$
$\hat{A} : |a\rangle \mapsto |b\rangle = \hat{A}|a\rangle$
which is saying
$\hat{A}^{\dagger}(\langle a|) = \left(\hat{A}|a\rangle\right)^{\mathrm{dual}}.$
We denote the left hand side of the above equation as $\langle a|\hat{A}^{\dagger}$. Now, let's consider the action of this new bra $\langle a|\hat{A}^{\dagger} = \langle b|$ on some ket $|c\rangle\in\mathcal{V}$:
$\langle a|\hat{A}^{\dagger}|c\rangle = \langle b|c \rangle.$
It follows from the definition of the inproduct that
$\langle b|c \rangle = \left(\langle c|b \rangle\right)^*$
$\langle a|\hat{A}^{\dagger}|c \rangle = \left(\langle c|\hat{A}|a\rangle\right)^*.$
So by definition this holds. For all operators $\hat{A}$ on $\mathcal{V}$.

However sometimes it also holds that
$\langle a|\hat{A}^{\dagger}|c \rangle = \langle a|\hat{A}|c\rangle.$ (where $\hat{A}^{\dagger}$ acts on $\langle a|$ and $\hat{A}$ on $|c\rangle$ !)
If this is true, we call $\hat{A}$ hermitian. From this, it follows that
$\langle a|\hat{A}|c\rangle = \left(\langle c|\hat{A}|a\rangle\right)^*$
and therefore also
$\langle a|\hat{A}|a\rangle = \left(\langle a|\hat{A}|a\rangle\right)^*$
which is the well-known result that hermitian operators yield real expectation values (for which reason observable quantities need to correspond to hermitian operators). Now, although we write $\hat{A} = \hat{A}^{\dagger}$ and the matrix elements of both operators are the same, they are not the same object. They still strictly have a different (but very closely related) domain and image. This is why it's important to remember that it is only a symbolic equality and more than just being sloppy by omitting the bra and ket from
$\langle a|\hat{A}^{\dagger}|c \rangle = \langle a|\hat{A}|c\rangle.$

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    $\begingroup$ But this is not true, the adjoint operator can perfectly well be thought of as acting on a ket. The condition you want is $$ \langle\psi|\hat{A}|\phi\rangle = \langle \phi|\hat{A}^{\dagger}|\psi\rangle \Leftrightarrow \hat{A} = \hat{A}^{\dagger}. $$ $\endgroup$ Dec 9, 2012 at 23:59
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    $\begingroup$ Out of characters so new comment. The reason why you shouldn't write $\hat{A} = \hat{A}^{\dagger}$ is precisely because you are identifying two objects who have a different domain (and image). So the two cannot be equal. It's similar to saying some row vector $a$ is equal to some column vector $b$. They might have the same elements, but they are not the same object, not equal to each other. You could argue then that it is just sloppy notation, but I feel you need to at least realize this. $\endgroup$
    – Wouter
    Dec 10, 2012 at 0:33
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    $\begingroup$ @Wouter, your answer is incorrect. An operator and its adjoint are incomparable (because their domains are different) on a general Banach space. A Hilbert space, however is (anti)isomorphic to its topological dual, and this permits a natural action of $\hat{A}^\dagger$ on $\mathcal{H}$. (That is, $\hat{A}^\dagger$ can act on kets.) It is in precisely that sense that we say $\hat{A}=\hat{A}^\dagger$ for a hermitian operator. $\endgroup$ Dec 11, 2012 at 12:45
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    $\begingroup$ -1: The fact that these things can be seen as acting either on the bra or the ket is precisely the reason for the notation $\langle \psi | A | \phi \rangle$, as opposed to $\psi \cdot (A \phi)$ or $(A \psi) \cdot \phi$. $\endgroup$
    – N. Virgo
    Dec 29, 2012 at 8:15
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    $\begingroup$ No, your answer is still incorrect, and the notation is not symbolic. You are again confused by the (slight) ambiguity of Dirac notation; please see my answer. If you (incorrectly in a Hilbert space) insist on considering $A^\dagger$ as an operator exclusively on the dual space, then the equality is not symbolic but exact, in the sense that a hermitian operator obeys $A=i^{-1}A^\dagger i$, where $i:\mathcal{H}\rightarrow\mathcal{H}^\ast$ is the (unique) inner-product equivalence between the Hilbert space and its dual: $i(\psi)=\langle\psi,\cdot\rangle$. (cont.) $\endgroup$ Jan 3, 2013 at 23:51

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