# Is hermiticity a basis-dependent concept?

Is the equation:

$$\langle Ay | x \rangle = \langle y | A x \rangle \text{ for all } x \in \text{Domain of } A.$$

independent of basis?

• There is absolutely no reference at all to a basis in that equation, so yes, it's basis independent. – DanielSank Jan 23 '17 at 21:01
• I think that the subtlety here is that on one's first pass through these things, the definition of the inner product is given in terms of a standard basis. That's not really the definition of an inner product. The real story is that, given a basis, there is a unique inner product for which those basis vectors form an orthornormal set. On the other hand, an inner product on a vector space can be defined in a basis-free way, and for a given vector space with inner product there are many ways to choose a "standard" orthonormal basis. – Tobias Hagge Jan 24 '17 at 2:51
• @TobiasHagge: true, I'd forgotten that; still, one would expect that in the normal course of events that if one is introduced to the formalism indicated above, then one would have been introduced to the basis-free definition of an inner product. – Mozibur Ullah Jan 24 '17 at 4:06
• It is dependent on the inner product (but not on the basis). – Fabian Jan 24 '17 at 6:35

The relation $$\langle Ay | x \rangle = \langle y | A x \rangle \text{ for all } x \in \text{Domain of } A\tag1$$ makes no reference to any basis at all, so it is indeed basis-independent.

In fact, this definition, which seems pretty strange when you first meet it, arises precisely out of a desire to make things basis-independent. The particular observation that sparks the definition is this:

Let $V$ be a complex vector space with inner product $⟨·,·⟩$, and let $\beta=\{v_1,\ldots,v_n\}$ be an orthonormal basis for $V$ and $A:V\to V$ a linear operator with matrix representation $A_{ij}$ over $\beta.$ Then, if this matrix representation is hermitian, i.e. if $$A_{ji}^*=A_{ij}\tag2$$ when $A$ is represented on any single orthonormal basis, then $(2)$ holds for all such orthonormal bases.

(Similarly, for a real vector space simply remove the complex conjugate.)

Now this is a weird property: it makes an explicit mention of a basis, and yet it is basis independent. Surely there must be some invariant way to define this property without any reference to a basis at all? Well, yes: it's the original statement in $(1)$.

To see how we build the invariant statement out of the matrix-based porperty, it's important to keep in mind what the matrix elements are: they are the coefficients over $\beta$ of the action of $A$ on that basis, i.e. they let us write $$Av_j = \sum_i A_{ij}v_i.$$ Moreover, in an inner product space, the coefficients of a vector on any orthonormal basis are easily found to be the inner products of the vector with the basis: if $v=\sum_j c_j v_j$, then taking the inner product of $v$ with $v_i$ gives you $$\langle v_i,v\rangle = \sum_j c_j \langle v_i,v_j\rangle = \sum_j c_j \delta_{ij} = c_i,$$ which then means that you can always write $$v=\sum_i \langle v_i,v \rangle v_i.$$ (Note that if $V$ is a complex inner product space I'm taking $⟨·,·⟩$ to be linear in the second component and conjugate-linear in the first one.)

If we then apply this to the action of $A$ on the basis, we arrive at $$Av_j = \sum_j A_{ij}v_i = \sum_i \langle v_i, Av_j\rangle v_i, \quad\text{i.e.}\quad A_{ij} = \langle v_i, Av_j\rangle,$$ since the matrix coefficients are unique. We have, then, a direct relation between matrix element and inner products, and this looks particularly striking when we use this language to rephrase our property $(2)$ above: the matrix for $A$ over $\beta$ is hermitian if and only if $$A_{ji}^* = \langle v_j, Av_i\rangle^* = \langle v_i, Av_j\rangle = A_{ij},$$ and if we use the conjugate symmetry $\langle u,v\rangle^* = \langle v,u\rangle$ of the inner product, this reduces to $$\langle Av_i, v_j\rangle = \langle v_i, Av_j\rangle. \tag 3$$ Now, here is where the magic happens: this expression is exactly the same as the invariant property $(1)$ that we wanted, only it is specialized for $x,y$ set to members of the given basis. This means, for one, that $(1)$ implies $(2)$, so that's one half of the equivalence done.

In addition to this, there's a second bit of magic we need to use: the equation in $(3)$ is completely (bi)linear in both of the basis vectors involved, and this immediately means that it extends to any two vectors in the space. This is a bit of a heuristic statement, but it is easy to implement: if $x=\sum_j x_j v_j$ and $y=\sum_i y_i v_i$, then we have \begin{align} \langle A y, x\rangle & = \left\langle A \sum_i y_i v_i, \sum_j x_j v_j\right\rangle && \\ & = \sum_i \sum_j y_i^* x_j \langle A v_i, v_j\rangle &&\text{by linearity} \\ & = \sum_i \sum_j y_i^* x_j \langle v_i, Av_j\rangle &&\text{by }(3) \\ & = \left\langle \sum_i y_i v_i, A\sum_j x_j v_j\right\rangle &&\text{by linearity} \\ & = \langle y, A x\rangle,&& \end{align} and this shows that you can directly build the invariant statement $(1)$ out of its restricted-to-a-basis version, $(3)$, which is itself a direct rephrasing of the matrix hermiticity condition $(2)$.

Pretty cool, right?

The definition that you have cited is indeed basis-independent as it only makes reference to the inner product $\langle\cdot,\cdot\rangle$ and the domain of $A$, neither of which is basis-dependent.

Note that "symmetric" in your above sense and "self-adjoint" in the broader sense are connected by the Hellinger-Toeplitz theorem which says that if the domain is the full Hilbert space, then the operator is self-adjoint: and this in turn means that what physicists mean by "self-adjoint" or "Hermitian" is in fact your notion of "symmetric;" operators like the Hamiltonian usually are not defined over the whole Hilbert space since they're symmetric but not bounded.

This has been argued to lead to a sort of mathematical incompleteness of quantum mechanics (Warning: PDF, warning: philosophy), however it doesn't impact most day-to-day applications in physics and isn't even present in most references about it. Here's a set of lecture notes which mentions it.

Symmetric operators are usually employed when working on real vector space, whereas Hermitian operators are usually employed when working on complex vector spaces.

In finite dimension, the associated matrix is symmetric in the first case ($a_{ij}=a_{ji}$ for all $i,\,j$), whereas it is equal to its complex conjugate transposed matrix in the second case ($a_{ij}=\overline{a_{ji}}$ for all $i,\,j$).

In both cases, the property (of being symmetric / Hermitian) is independent of the choice of basis but dependent on the choice of scalar product for the first case, Hermitian product for the second case.

There is a definite sense in which self-adjointness is indeed a basis-dependent concept. It really depends on what you start with. If you have a particular inner product in mind, then the truth of your equation is entirely independent of what basis you choose to represent $x, y$, and $A$ with respect to. However, if you do not have a particular inner product in mind, then the choice of inner product itself plays a role.

The idea is that choosing an inner product makes some bases special: namely, those which are orthonormal with respect to it. It is not so restricting as choosing a single basis, but it still means that we can make a choice that determines whether or not the matrix (or rather the linear transformation) is self-adjoint.

More precisely, we can find a matrix $A$ and two inner products $\left< \cdot, \cdot \right>_{1}$ and $\left< \cdot, \cdot \right>_{2}$ such that $A$ is self-adjoint with respect to the first but not the second. For instance, take

$$A = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$$

and consider the inner product $\left< \cdot, \cdot \right>_{1}$ such that

$$\beta_{1} = \left\{ \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \end{pmatrix} \right\}$$

is orthonormal (i.e. the standard dot product), and the inner product $\left< \cdot, \cdot \right>_{2}$ such that

$$\beta_{2} = \left\{ \begin{pmatrix} 1 \\ 2 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \end{pmatrix} \right\}$$

is orthonormal. Then $A$ is self-adjoint with respect to the first (since it is hermitian in that basis) but not with respect to the second. To see this, note that with respect to the second basis, $A$ takes the following form. $$[A]_{\beta_{2}} = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix}^{-1} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} = \begin{pmatrix} 3 & 1 \\ -3 & -1 \end{pmatrix}$$

The main point is that the notion of being basis-independent is a vague concept. When only talking about vector spaces (i.e. without inner products), we can speak precisely about what it means to be basis-independent since we can distinguish between things that require us to pick a specific basis and those that do not. However, when we add the structure of an inner product, we cannot really talk about something being basis-independent in the same absolute terms since the choice of inner product is basis-dependent. However, we can talk in a weaker sense about something being basis-independent assuming we already have an inner product. In this way, self-adjointness is basis-independent.