# 1 Question for Bounty

In the context of an infinite-dimensional vector space, below I present a supposed "proof" – designated by $$(*)$$ – that

$$\Omega_{ij} = \overline{\Omega}_{ji}$$

implies that $$\Omega$$ is Hermetian. I know this proof is false, but I am persistently confused as to what specific step of it is wrong. Since this is my second bounty for this question I should be clearer than the last one: I am not interested in different perspectives/proofs as to why the converse of $$(*)$$ is true. I am only interested in what specific step of my supposed "proof" (*) is wrong, and why.

# 2 Hermetian definition

Let us use the definition that $$\Omega$$ is Hermitian iff

$$\langle v | \Omega | w \rangle = \overline{\langle w | \Omega | v \rangle}$$

$$\langle v | \Omega | w \rangle = \overline{\langle w | \Omega^{\dagger} | v \rangle}$$

for arbitrary vectors $$|v\rangle$$ and $$|w\rangle$$.

# 3 The supposed "proof" (*)

Suppose we assume by hypothesis that $$K_{x x'} = \overline{K_{x'x}}$$, or in the notation of the textbook that

$$\left\langle x'\left|K^\dagger\right| x\right\rangle = \left\langle x'|K| x\right\rangle.$$

Then why can't we just say:

\begin{aligned} &\langle g|K| f\rangle \\ =& \int_a^b \int_a^b\langle g \mid x\rangle\left\langle x|K| x^{\prime}\right\rangle\left\langle x' \mid f\right\rangle \, d x \, d x' \\ =&\int_a^b \int_a^b \overline{\left\langle x| g\right\rangle}\overline{\left\langle x' \left|K^\dagger\right| x\right\rangle} \overline{\left\langle f \mid x^{\prime}\right\rangle} \, d x \, d x' \quad(\text {conjugate symmetry}) \\ =&\int_a^b \int_a^b \overline{\left\langle f \mid x^{\prime}\right\rangle} \overline{\left\langle x' \left|K^\dagger\right| x\right\rangle} \overline{\left\langle x| g\right\rangle} \, d x \, d x' \quad(\text {commutativity}) \\ =& \int_a^b \int_a^b \overline{\left\langle f \mid x^{\prime}\right\rangle\left\langle x'|K| x\right\rangle\langle x \mid g\rangle} \, d x \, d x' \quad\left(\text{hypothesis}\right) \\ =&\left(\int_{a}^{b} \int_a^b \left\langle f \mid x^{\prime}\right\rangle\left\langle x^{\prime}|K| x\right\rangle\langle x \mid g\rangle \, d x \, d x' \right)^{*} \quad\left(\text{pulling * out}\right) \\ =&\langle f|K| g\rangle^{*} \end{aligned}

What specific step of this proof is wrong?

# 4 The book's counter-example to (*)

The following argument is from Principles of Quantum Mechanics, and shows why $$(*)$$ must be wrong (though it doesn't tell us which step of $$(*)$$ is wrong and why): # 5 My attempted refutation of (*)

My original understanding of why (*) was mistaken:

…the supposed "counter-proof" is wrong in its very first step:

\begin{aligned} &\langle g|K| f\rangle \\ =& \int \int\langle g \mid x\rangle\left\langle x|K| x^{\prime}\right\rangle\left\langle x^{\prime} \mid f\right\rangle d x d x^{\prime} \\ \end{aligned}

Notice here I was inserting twice the so called "resolution of identity" into this integral:

$$\int \mid x \rangle \langle x \mid dx$$

…but this presupposes we can construct an orthonormal basis $$\{x\}$$ in our infinite-dimensional vector space to generate such a "Resolution of Identity". In the finite-dimensional case, we can use the Spectral theorem to always generate such a basis using eigenvectors. Unfortunately, we cannot necessarily do this in the infinite dimensional case (and in general: infinite-dimensional vector spaces don't always have to have orthonormal bases). Thus the "counter-proof" fails in the first step.

But now I am not so sure I am right, since when the author asks under what conditions the following equality holds

$$\begin{array}{l} \int_{a}^{b} \int_{a}^{b}\langle g \mid x\rangle\left\langle x|K| x^{\prime}\right\rangle\left\langle x^{\prime} \mid f\right\rangle d x d x^{\prime} \\ \quad \stackrel{?}{=}\left(\int_{a}^{b} \int_{a}^{b}\langle f \mid x\rangle\left\langle x|K| x^{\prime}\right\rangle\left\langle x^{\prime} \mid g\right\rangle d x d x^{\prime}\right) \end{array}$$

he is assuming (in the left-hand side of this equation) that we can already plug in a resolution of identity (against, presumably, the orthonormal basis of eigenvectors generated by the Spectral Theorem). But under this assumption, why can't we just take the left-hand side of this equation

$$\int_{a}^{b} \int_{a}^{b}\langle g \mid x\rangle\left\langle x|K| x^{\prime}\right\rangle\left\langle x^{\prime} \mid f\right\rangle d x d x^{\prime}$$

and proceed forward using the steps from $$(*)$$ to obtain that the equality holds in all cases?

• @Cosmas Zachos: Can you expand upon your comment? I'm not sure I follow. Jul 14, 2021 at 5:50
• ${\mathbb I}= \int_{-\infty}^{\infty}dx~|x\rangle \langle x|= (\int_{a }^{b}+\int_{-\infty}^{a}+\int_{b}^{\infty} )dx~|x\rangle \langle x|$, so what you inserted in your proof is only a piece of the resolution of the identity. Inserting all three pieces gives you a consistent result. Jul 14, 2021 at 14:31
• @CosmasZachos Doesn't the text limit us to the interval $[a, b]$: "Let $|f\rangle$ and $|g\rangle$ be two kets in the function space, whose images in the X basis are two functions $f(x)$ and $g(x)$ in the interval $a-b$"? So if if we inserted the full resolution, wouldn't the $\int_{-\infty}^a$ and $\int_b^\infty$ just evaluate to zero since the $f$ and $g$ functions evaluate to zero on those portions of the domain? Jul 14, 2021 at 14:56
• The text correctly does that, but the boundary terms, unless f and g vanished there, do not. The missing pieces involve step functions which you might have as well inserted in the original argument to properly monitor the integrations by parts. Jul 14, 2021 at 15:06
• Is my "counter-proof" at least valid in the case we are only interested in a bounded interval? What do you mean by the "step functions" in this context? Jul 14, 2021 at 20:58

I suspect the problem is in integration order. I will write $$\int dx ...\int dx'$$ implying that I first take integral with respect to $$x'$$ and then take the integral with respect to $$x$$. Let's repeat your steps: \begin{aligned} \langle g|K|f\rangle &= \int\limits_a^bdx \langle g|x\rangle \langle x|K|f\rangle =\int\limits_a^bdx \langle g|x\rangle \int\limits_a^bdx'\langle x|K|x'\rangle\langle x'|f\rangle =\\ &= \int\limits_a^bdx \overline{\langle x|g\rangle} \int\limits_a^bdx'\overline{\langle x'|K^\dagger|x\rangle}\overline{\langle f|x'\rangle} = \int\limits_a^bdx \overline{\langle x|g\rangle} \int\limits_a^bdx'\overline{\langle x'|K|x\rangle}\overline{\langle f|x'\rangle} \end{aligned}

Now, if we change the order of integration, we arrive at your result. However, rigorously speaking, we do not have a right to do that. We didn't even specify the functional space yet, but let's assume we work with 'very good' functions which are infinitely differentiable and so everything converges. This works for $$f(x)$$ and $$g(x)$$, however, doesn't have to work for the $$K_{xx'}$$. It doesn't work for unbound operators, for instance. By the way, linear operator $$A$$ is said to be bound if for any function $$f$$ from its domain the following holds: $$\langle Af|Af\rangle \leq C \langle f|f\rangle$$ Where $$|Af\rangle$$ is $$A|f\rangle$$ in your definitions and $$C>0$$. You can check that $$-i\partial_x$$ is not bound in the aforementioned functional space. We may also notice that plugging in $$\overline{\langle x'|K|x\rangle} = K^*_{x'x} = i\delta'(x'-x)$$ gives us the following result: $$\int\limits_a^bdx \overline{\langle x|g\rangle}\int\limits_a^bdx'(i\delta'(x'-x))f(x').$$ According to the book, you are reading, the delta's derivative is $$\delta'(x-y) = \delta(x-y)\partial_y$$ but only when under the integral with respect to $$y$$. This means, that we can use this definition only after we make a transformation: $$\delta'(x'-x) = -\delta'(x-x') = -\delta(x-x')\partial_{x'}$$ So now are back at the starting point: $$\int\limits_a^bdx \overline{\langle x|g\rangle}\int\limits_a^bdx'(i\delta'(x'-x))f(x') = \int\limits_a^bdx \overline{\langle x|g\rangle}\int\limits_a^bdx'(-i\delta'(x-x'))f(x') = \int\limits_a^bdx \overline{\langle x|g\rangle} \left(-if'(x)\right) = \langle g|K|f\rangle.$$

Note that specifying boundary conditions doesn't rid you of this problem, implying that you need some other conditions to change the integration order and make the operator Hermitian.

• Why can't we change the order of integration w.r.t. unbound operators? Is this theorem named? I've tried Googling around for this for an explanation but haven't seen one. Aug 29, 2021 at 0:20

The integrand $$\langle g|x \rangle \langle x | K | x' \rangle \langle x' | f \rangle$$ is not $$L^1$$ so you are not allowed to apply Fubini's theorem. Actually, it's not even a function; it's a distribution: $$\overline{g(x)}\,\delta'(x-x')\,f(x)$$. The calculations should be treated using the laws of distributions.

• Do we even need to appeal to Fubini's theorem here? Since $K|f\rangle$ needn't be integrable, then isn't that alone an issue enough to refute the supposed proof? Sep 1, 2021 at 19:44
• @user1770201. That's of course even worse, but that implies that $\langle x|K|f\rangle$ is not $L^1.$ Sep 1, 2021 at 19:58

You start with the definition of $$\,\Omega\,$$ being Hermitian if $$\langle\Omega v\mid w \rangle \:=\: \langle v\mid\Omega w\rangle \quad\text{for all vectors }v, w\in H\tag{1}$$ where $$H$$ is a complex Hilbert space. This definition is valid independent of $$H$$ being finite-dimensional or infinite-dimensional; and you may take it as an answer to the title question.

Then, when going from section "The finite case" to the next section "The infinite case", two leaps at one stroke are taken: not only from finitely many dimensions to the infinite setting, but also branching from bounded operators to unbounded ones (they only exist in the infinite-dimensional setting), since the differentiation operator $$K = -i\frac{\mathrm{d}}{\mathrm{d}x}$$ is considered. Which is a prominent and important unbounded operator.

Keep these two properties separate. A good border stone is then the Hellinger-Toeplitz theorem:
If $$A:H\to H$$ is linear and satisfies $$\langle Av\mid w \rangle \:=\: \langle v\mid A w\rangle \quad\text{for all }v, w\in H$$ then $$A$$ is continuous, or equivalently, $$A$$ is bounded.
The proof of Hellinger-Toeplitz directly follows from the Closed graph theorem.

One inevitable consequence is that an unbounded operator $$\,\Omega\,$$ cannot satisfy $$(1)$$. Instead a domain of definition $$\operatorname{dom}(\Omega )\subsetneq H$$ needs to be chosen (usually a dense subspace). For example, the surface term vanishing condition $$(1.10.29)$$ is a restriction and corresponds to a choice of an appropriate $$\operatorname{dom}\left(-i\frac{\mathrm{d}}{\mathrm{d}x}\right)\subset H$$.

For unbounded operators you have to distinguish between Hermitian and self-adjoint operators, and only for the latter you have the spectral theorem available -- including eigenvectors, needed to write down the so-called resolution of unity which you exploit / "this supposed proof which I know must be wrong".

'Hermitian' is often called 'symmetric' in the unbounded context, cf https://en.wikipedia.org/wiki/Self-adjoint_operator . The naming is not uniform or standardised.

EDIT: I am now not so sure this is correct. See my original post for a new bounty looking for clarification!

My post asked why my supposed "counter-proof" (which mirrors the reasoning of the finite-dimensional case) was wrong, not for more perspectives on why the book's theorem is true.

The answer is that the supposed "counter-proof" is wrong in its very first step:

\begin{aligned} &\langle g|K| f\rangle \\ =& \int \int\langle g \mid x\rangle\left\langle x|K| x^{\prime}\right\rangle\left\langle x^{\prime} \mid f\right\rangle d x d x^{\prime} \\ \end{aligned}

Notice here I was inserting twice the so called "resolution of identity" into this integral:

$$\int \mid x \rangle \langle x \mid dx$$

...but this presupposes we can construct an orthonormal basis $$\{x\}$$ in our infinite-dimensional vector space to generate such a "Resolution of Identity". In the finite-dimensional case, we can use the Spectral theorem to always generate such a basis using eigenvectors. Unfortunately, we cannot necessarily do this in the infinite dimensional case (and in general: infinite-dimensional vector spaces don't always have to have orthonormal bases). Thus the "counter-proof" fails in the first step.