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Made corrections based on comment.
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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: $\left(\frac{\mathrm d}{\mathrm dx}\right)^\dagger = -\frac{\mathrm d}{\mathrm dx}$.

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 underlyingoperators 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 compactrestricted by strong enough boundary conditions (that allows to normalize them for example).

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.

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. The only thing I can say is that if this can be done then the space will turn out to be compact since the matrix has discrete eigenvalues. Beyond that, I can not say anything useful about how to proceed, sorry!

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: $\left(\frac{\mathrm d}{\mathrm dx}\right)^\dagger = -\frac{\mathrm d}{\mathrm dx}$.

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. You can only write operators as square matrices and vectors as column matrices if the underlying space is compact.

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, which is usually assumed in physical scenarios.

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. The only thing I can say is that if this can be done then the space will turn out to be compact since the matrix has discrete eigenvalues. Beyond that, I can not say anything useful, sorry!

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: $\left(\frac{\mathrm d}{\mathrm dx}\right)^\dagger = -\frac{\mathrm d}{\mathrm dx}$.

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).

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.

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!

clarified that I didn't answer all the questions.
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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.

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 functionoperator coincides with the eigenvalues of the matrix. The only thing I can say is that if this can be done then the space will turn out to be compact since the matrix has discrete eigenvalues. Beyond that, I can not say anything useful, sorry!

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.

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 functions defined on that space and the eigenvalues of that function coincides with the eigenvalues of the matrix. The only thing I can say is that if this can be done then the space will turn out to be compact since the matrix has discrete eigenvalues. Beyond that, I can not say anything useful, sorry!

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.

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. The only thing I can say is that if this can be done then the space will turn out to be compact since the matrix has discrete eigenvalues. Beyond that, I can not say anything useful, sorry!

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nio
nio
  • 792
  • 4
  • 10

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.

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. You can only write operators as square matrices and vectors as column matrices if the underlying space is compact.

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, 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 functions defined on that space and the eigenvalues of that function coincides with the eigenvalues of the matrix. The only thing I can say is that if this can be done then the space will turn out to be compact since the matrix has discrete eigenvalues. Beyond that, I can not say anything useful, sorry!