I understand that if my operator $A$ is a Hermitian, then suppose it has the following eigen vectors $$A|v_1\rangle=\lambda|v_1\rangle:(A).(V_1)=\lambda(V_1)\\ A|v_2\rangle=\beta|v_2\rangle:(A).(V_2)=\beta(V_2)\\ \langle v_2|A|v_1\rangle=\lambda\langle v_2|v_1\rangle:(V_2)^{\dagger}.(A).(V_1)=\lambda(V_2)^{\dagger}.(V_1)\\ \text{since}(A).(V_2)=\beta(V_2)\quad\text{hence}\quad (V2)^{\dagger}.(A)^{\dagger}=\beta(V_2)^{\dagger}$$ Now, if $A$ is a hermitian operator, then $A=A^{\dagger}$, hence $$(V_2)^{\dagger}.(A)^{\dagger}=(V_2)^{\dagger}.(A)=\beta(V_2)^{\dagger}\\ \text{Hence}\quad (V_2)^{\dagger}.(A).(V_1)=\beta(V_2)^{\dagger}(V_1)\\ \text{implies} \quad\langle v_2|A|v_1\rangle =\lambda\langle v_2|v_1\rangle = \beta\langle v_2|v_1\rangle:\lambda\neq\beta\\ \text{Hence}\quad \langle v_2|v_1\rangle=0$$ So, the vectors must be orthogonal to each other and their inner product must be zero, but how do I know in a differential equation if my operator is a hermitian or not and whether I will have quantized solutions whose inner products will be zero? For example the schrodinger wave equation leads to quantized orthogonal solutions. Or rather, how do I extend a matrix based Hermitian operator to a differential equation?

  • 1
    $\begingroup$ @PatrickAbraham, please respect the author's choice of notations. Your edit attempt borders vandalism. $\endgroup$
    – Alex M.
    Commented Jan 3, 2017 at 15:32

1 Answer 1


Disclaimer: I don't know the answer to the question "how do I extend a matrix based Hermitian operator to a differential equation?". I know the answers to the previous questions raised in the post, so, here they are.

Let us establish a few definitions. The notion of hermiticity is tied to the notion of an inner product. In general, for a complex vector space $V$ if $\langle-,-\rangle:V \times V \to V$ is a hermitian inner product then the hermitian adjoint of an operator $A: V \to V$ is defined to be another operator $A^\dagger: V \to V$ that satisfies: $$ \langle u, A(v) \rangle = \langle A^\dagger(u), v \rangle\,. $$ for any choice of vectors $u, v \in V$. An operator $A$ is called hermitian if it is equal to its hermitian adjoint. In your example, given two vectors $V_1$ and $V_2$ (which are column matrices with complex entries), their inner product is defined as $$\langle V_2, V_1 \rangle := V_2^\dagger \cdot V_1\,. \tag{ip1}$$ With respect to this inner product, hermitian adjoint of a matrix operator is the same as the transpose conjugate (denoted conveniently by $\dagger$) of that matrix, since for any two column matrices $V_1$ and $V_2$ and any square matrix $A$ that acts on these column matrices we get: $$ \langle V_2, A \cdot V_1 \rangle = V_2^\dagger \cdot A \cdot V_1 = (A^\dagger \cdot V_2) \cdot V_1 = \langle A^\dagger \cdot V_2, V_1 \rangle\,. $$ Therefore, with respect to the inner product (ip1), a matrix is hermitian if it satisfies $A^\dagger = A$, as you have mentioned.

how do I know in a differential equation if my operator is a hermitian or not

To know whether a differential operator is hermitian or not first you must establish a definition of a hermitian inner product on the complex vector space of functions (on which the differential operator acts).

Example: As our complex vector space of functions, let us take the space of square integrable complex functions on $\mathbb R$ (such functions will be called "$L^2$") and as our differential operator let us take $\hat D:= \frac{\mathrm{d^2}}{\mathrm{d x^2}}$ where $x$ is the coordinate on $\mathbb R$. We also need an inner product: for any two $L^2$ functions $f$ and $g$ we define their inner product as: $$ \langle f, g \rangle := \int_{\mathbb R} \mathrm dx\, f(x)^* g(x)\,. \tag{ip2}$$ Now we can find out what the hermitian adjoint of $\hat D$ is. Recall that $\hat D^\dagger$ is an operator that satisfies $\langle \hat D^\dagger f, g \rangle = \langle f, \hat D g \rangle$ for any two arbitrary $L^2$ functions $f$ and $g$. Using the integration by parts twice we find: $$ \langle f, \hat D g \rangle = \int_{\mathbb R} \mathrm dx f(x)^* \frac{\mathrm{d^2} g(x)}{\mathrm d x^2} = -\int_{\mathbb R} \mathrm dx \frac{\mathrm df(x)^*}{\mathrm d x} \frac{\mathrm{d} g(x)}{\mathrm d x} = \int_{\mathbb R} \mathrm dx \frac{\mathrm{d^2} f(x)^*}{\mathrm d x^2} g(x) = \langle \hat D f, g\rangle\,. $$ Therefore we conclude that with respect to the inner product (ip2), the operator $\hat D$ satisfies $\hat D^\dagger = \hat D$, therefore, it is hermitian.

Edit: Apart from the functions being $L^2$ I am also assuming them to be at least twice differentiable so that I can define the action of $\frac{\mathrm{d}^2}{\mathrm dx^2}$, I am also assuming that the functions have some boundary condition (such as vanishing at infinty) that allows me neglect the boundary terms coming from integration by parts.

Note that the differential operator $\frac{\mathrm d}{\mathrm d x}$ is not hermitian with respect to the inner product (ip2) since integration by parts now gives $\langle f, \frac{\mathrm d}{\mathrm d x} g \rangle = -\langle \frac{\mathrm d}{\mathrm d x} f, g \rangle$. So the operator is in fact anti-hermitian: $\left(\frac{\mathrm d}{\mathrm dx}\right)^\dagger = -\frac{\mathrm d}{\mathrm dx}$.

and whether I will have quantized solutions whose inner products will be zero?

Your proof of orthogonality of eigenvectors of a hermitian operator holds in general. But whether the eigenvalues will be discrete (quantized) or not depends on the space on which the functions are defined. If the space is compact (such as a circle, or a finite interval (a.k.a. the infinite square well in quantum mechanics)) then the eigenvalues will be discrete, if the space is non-compact (such as $\mathbb R$ from the previous example) the eigenvalues will be continuous unless we impose some extra constraints on the functions to restrict to a space with discrete spectrum (eigenvalues) for the operator. You can only write operators as square matrices and vectors as column matrices if the operators have discrete spectrum.

Edit: (Correction, as pointed out by Chappers in the comment) It is possible to have discrete eigenvalues for differential operators on noncompact spaces if the function space is restricted by strong enough boundary conditions (that allows to normalize them for example).

For example the schrodinger wave equation leads to quantized orthogonal solutions.

The differential operator in the (assuming time-independent) Schrödinger equation, for a particle moving in a one dimensional space, is: $$ \hat D_\psi := -\frac{\hbar^2}{2m} \frac{\mathrm d^2}{\mathrm d^2 x} + V(x)\,, $$ and the inner product of wave functions is exactly as we defined in (ip2). Therefore, if the potential $V(x)$ is a real function, then the operator $\hat D_\psi$ is hermitian, since we have already proved that the second derivative is hermitian and you can check that multiplication by a real function is hermitian as well. Hermiticity leads to orthogonal eigenvector. But once again, the eigenvalues are discrete if the space in which the paticle moves is compact, or if we impose normalizability which is usually assumed in physical scenarios.

how do I extend a matrix based Hermitian operator to a differential equation?

This seems too complicated for me, I'm afraid. The problem is to find a space on which there is some operator that acts on (perhaps a restricted set of) functions defined on that space and the eigenvalues of that operator coincides with the eigenvalues of the matrix. I can not say anything useful about how to proceed, sorry!

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    $\begingroup$ Most of your answer is very good, but just to point out a couple of subtleties: 1) You can't define $(d/dx)^2$ on all of $L^2$, since most $L^2$ functions are not continuously differentiable; you have to restrict to, e.g. the Sobolev space $H^2$ to make sense of that. There's also the issue of discarding the boundary terms. 2) Having discrete eigenvalues on non-compact sets is not quite that simple, and does arise, as a consequence of boundary conditions needed for normalisability, e.g. $f(x) \to 0$ as $x \to \pm \infty$; examples of this are harmonic oscillator and the hydrogen atom. $\endgroup$
    – Chappers
    Commented Jan 4, 2017 at 1:51
  • $\begingroup$ @Nafiz Ishtiaque; I would mark your answer correct, but I'll just wait it out a few more days before I do so and check if I get an answer that translates Hermitian Matrix operators in to continuous operators. I have already up voted your answer btw :). Thaks a lot for the descriptive answer here, really. $\endgroup$
    – Ghosal_C
    Commented Jan 4, 2017 at 2:13
  • $\begingroup$ @Chappers: You're absolutely right, thanks for pointing it out! $\endgroup$
    – nio
    Commented Jan 4, 2017 at 2:57
  • $\begingroup$ @ubuntu_noob: Thanks and you're welcome, I am also curious to know what the answer is to the remaining question, which seems really interesting. $\endgroup$
    – nio
    Commented Jan 4, 2017 at 2:59

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