How does one write Adjoint, Self-adjoint and Hermitian operators in Dirac notation?

Question

The following portion is paraphrased from Mathematical Methods for Physics and Engineering by Riley, Hobson, and Bence.

• The adjoint of a linear operator $\hat{A}$, denoted by $A^\dagger$, is an operator that satisfies $$\int_{a}^{b}\psi_1^*(\hat{A}\psi_2)dx =\int_{a}^{b}(\hat{A}^\dagger\psi_1)^*\psi_2 dx+\text{boundary terms}\tag{1}$$ where the boundary terms are evaluated at the end-points of the interval $[a,b]$.

• An operator is said to be self-adjoint if $A^\dagger=A$. Therefore, for self-adjoint operators, $$\int_{a}^{b}\psi_1^*(\hat{A}\psi_2)dx -\int_{a}^{b}(\hat{A}\psi_1)^*\psi_2 dx=\text{boundary terms}.\tag{2}$$

• In addition, if certain boundary conditions are met by the function $\psi_1$ and $\psi_2$ on which the self-adjoint operator acts, or by the operator itself, such that the boundary terms vanish, then the operator said to be hermitian in the interval $a\leq x\leq b$. In that case, $$\int_{a}^{b}\psi_1^*(\hat{A}\psi_2)dx =\int_{a}^{b}(\hat{A}\psi_1)^*\psi_2 dx.\tag{3}$$

My question is that in terms of Dirac's abstract bra and ket notation, how does one write each of these defining equations?

Answer 1

This is a good question because in Dirac notation because, despite being tempted, one should not write $$ \langle A^\dagger n|m\rangle = \langle n|A m\rangle, $$ as you cannot put the operator acting the state label. You should really write: $$ \langle n|A^\dagger|m\rangle = \langle m|A|n\rangle^*. $$

Now Dirac deliberately does not make a distinction between these various possibilities of Hermitian, self-adjoint and so on (i.e. bounded, compact etc.) -- although he surely knew it. He set up his notation so as to capture the essential features of QM and to sweep all the mathematical subleties under the rug. In that way they do not get in the way of seeing the big picture.

I'm a bit shocked, hwever, by the confusion between self adjoint and Hermitian that you describe in Riley, Hobson and Bence. What they call "self-adjoint" is more correctly called "formally self adjoint." If the boundary conditions are such that the boundary terms are zero then they are correct in that this is what physicists call "Hermitian" and mathematicians call "symmetric." Being actaully self adjoint, however, is a much stronger condition than formally self-adjoint or Hermitian. It requires not only that $A$ be formally self-adjoint, but also the domains of definition of $A$ and $A^\dagger$ be the same. For example the momentum opertor $P= -i\partial_x$ acting on wave functions that vanish at the endpoints of the integral is Hermitian, but it is not self adjoint. Self-adjoint opertors have complete sets of eigenfunctions but $P$ with these boundary conditions has no eigenfunctions at all, never mind a complete set. It is only with some sort of periodic boundary conditions that $P$ on a finite interval has the same domain of definition as $P^\dagger$.

I must check my copy of RHB and see exactly what they say.

Answer 2

1. This is a non-standard definition of (self-)adjointness. From context, your definitions are supposed to be for operators $A$ on the space of square-integrable complex-valued functions $L^2(\mathbb{R})$. This is a Hilbert space, and the abstract definition of the adjoint $A^\dagger$ of any operator $A$ on any Hilbert space $H$ with inner product $\langle -,-\rangle$ is $$ \langle Av,w\rangle = \langle v, A^\dagger w\rangle \label{Ad}\tag{Ad}$$ for all $v \in D(A), w\in D(A^\dagger)$, where $D(A)$ is the domain of definition of $A$ and $D(A^\dagger)$ the inferred domain of definition of the adjoint. The operator $A$ may be a bounded operator defined on the entire Hilbert space, or it may be densely defined only on some dense subspace $D(A)\subset H$. This is usually the case for unbounded operators like the position and momentum operators.

   Note the absence of any "boundary terms" in $\eqref{Ad}$. This is because the notion of "boundary terms" only makes sense for the specific case of $L^2(\mathbb{R})$, but not for a generic Hilbert space.

   The text you are using likely wants to side-step the discussion of domains of definition - for vectors outside the domain of definition of $A$ or $A^\dagger$, you usually get such boundary terms when trying to apply the naive definition of operators on $L^2(\mathbb{R})$. However, in some contexts it is really crucial to pay attention to this subtlety, and the operators really are only defined on the subspace of functions where these boundary terms vanish. See this answer of mine for a detailed discussion of a case where not paying attention to this leads to an apparent contradiction between the operators and their commutation relations.

   It appears that this - the case when the boundary terms vanish - is when your text wants to call the operator "Hermitian". This is a decidedly non-standard usage and I would strongly recommend against accepting that usage into your vocabulary. In almost all other usages, "Hermitian" is either synonymous with self-adjoint or is a weaker condition, e.g. some people call an operator Hermitian (or symmetric) when $A=A^\dagger$ but $D(A)\neq D(A^\dagger)$ and self-adjoint when additionally $D(A) = D(A^\dagger)$). Other people do not use the word "Hermitian" for operators on infinite-dimensional spaces at all and reserve its usage for finite-dimensional matrices on $\mathbb{C}^n$, where there are no domain issues and it is always equivalent to self-adjointness.

2. Dirac notation has trouble expressing a lot of the things that go on with adjoints because in $\langle v \vert A\vert w\rangle$, you can't really tell whether $A$ acts to the left on $\lvert w\rangle$ or to the right on $\langle v\vert$, i.e. it is ambiguous whether this is $\langle v, Aw\rangle$ or $\langle Av,w\rangle$. For self-adjoint operators $A=A^\dagger$ this doesn't matter since $\langle Av,w\rangle = \langle v,A^\dagger w\rangle = \langle v,Aw\rangle$, and so Dirac notation is only unambiguous when you use only self-adjoint operators.

   People usually - but not always - assume that $\langle v\vert A\vert w\rangle$ for non-self-adjoint $A$ means $\langle v,Aw\rangle$, and would express the adjoint condition as something like $$ \langle A^\dagger v\vert w\rangle = \langle v\vert A\vert w\rangle$$ or $$ (A\lvert v\rangle)^\dagger \lvert w\rangle = \langle v\vert A\vert w\rangle$$ or $$ (\langle v\vert A^\dagger)\lvert w\rangle = \langle v\vert A\vert w\rangle$$ but all of these are suboptimal and could arguably be interpreted wrongly. In my opinion, it is best just to not use Dirac notation in cases where this ambiguity can happen.