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In the context of mathematical quantum mechanics, a well known no-go theorem known as Hellinger-Töplitz tells us that an unbounded, symmetric operator cannot be defined everywhere on the Hilbert space $\mathcal H$. Thus, for many operators of interest in quantum mechanics, we must resort to restricting the domain on which they may act. Along with the Hamiltonian, the momentum operator $$ p:=-i\frac{\rm d}{\mathrm{d}x}$$ is the most prominent example of such an operator. Now, one of the most important questions is clearly: What is the correct domain of definition for the momentum operator?

In order to answer this question, it is of crucial importance to consider the question of self-adjointness: Only self-adjoint operators are admissible as observables of the theory, and there are certain vital mathematical results, such as the spectral theorem and Stone's theorem, which make it clear that any 'good' momentum operator must certainly be self-adjoint: $ p^\star= p$. So, the original question is refined slightly: What is the correct domain $\mathcal D( p)\subset \mathcal H$ that allows us to construct a self-adjoint momentum operator?

Recently, one of my lecturers treated a concrete incarnation of this (general and possibly vague) question: Consider a Hilbert space $\mathcal H=L^2(0,1)$ and two 'candidate momenta', $ p_0$ and $ p_\alpha$, with domains

$$\mathcal D(p_0)=\{\psi\in \mathcal H \,|\, \psi\in \text{AC}(0,1),\psi'\in\mathcal H, \psi(0)=0=\psi(1)\} $$ $$ \mathcal D(p_\alpha)=\{\psi\in \mathcal H \,|\, \psi\in \text{AC}(0,1),\psi'\in\mathcal H, \psi(0)=\alpha\psi(1)\},\hspace{.5cm}|\alpha|=1 $$ Are these operators self-adjoint? As it turns out, it can be shown that $p_0$ is symmetric but not self-adjoint. However, $p_\alpha$ is self-adjoint. It seems that the domain of $p_0$ is 'too small'.

Now, my question is: What does the conclusion that $p_0\neq p_0^\star$ imply for the canonical freshman QM example of the infinite potential well? As far as I'm aware, it is conventional to take exactly the boundary conditions that we assume to define the domain of $p_0$ when solving the Schrödinger equation in this case. Can we conclude that the notion of momentum cannot be rigorously defined in this elementary example? Or, at the very least, do we have to admit some $\psi$ that obey non-physical boundary conditions?

For those interested, this is a related question, also touching upon domain issues of momentum operators.

In the context of mathematical quantum mechanics, a well known no-go theorem known as Hellinger-Töplitz tells us that an unbounded, symmetric operator cannot be defined everywhere on the Hilbert space $\mathcal H$. Thus, for many operators of interest in quantum mechanics, we must resort to restricting the domain on which they may act. Along with the Hamiltonian, the momentum operator $$ p:=-i\frac{\rm d}{\mathrm{d}x}$$ is the most prominent example of such an operator. Now, one of the most important questions is clearly: What is the correct domain of definition for the momentum operator?

In order to answer this question, it is of crucial importance to consider the question of self-adjointness: Only self-adjoint operators are admissible as observables of the theory, and there are certain vital mathematical results, such as the spectral theorem and Stone's theorem, which make it clear that any 'good' momentum operator must certainly be self-adjoint: $ p^\star= p$. So, the original question is refined slightly: What is the correct domain $\mathcal D( p)\subset \mathcal H$ that allows us to construct a self-adjoint momentum operator?

Recently, one of my lecturers treated a concrete incarnation of this (general and possibly vague) question: Consider a Hilbert space $\mathcal H=L^2(0,1)$ and two 'candidate momenta', $ p_0$ and $ p_\alpha$, with domains

$$\mathcal D(p_0)=\{\psi\in \mathcal H \,|\, \psi\in \text{AC}(0,1),\psi'\in\mathcal H, \psi(0)=0=\psi(1)\} $$ $$ \mathcal D(p_\alpha)=\{\psi\in \mathcal H \,|\, \psi\in \text{AC}(0,1),\psi'\in\mathcal H, \psi(0)=\alpha\psi(1)\},\hspace{.5cm}|\alpha|=1 $$ Are these operators self-adjoint? As it turns out, it can be shown that $p_0$ is symmetric but not self-adjoint. However, $p_\alpha$ is self-adjoint. It seems that the domain of $p_0$ is 'too small'.

Now, my question is: What does the conclusion that $p_0\neq p_0^\star$ imply for the canonical freshman QM example of the infinite potential well? As far as I'm aware, it is conventional to take exactly the boundary conditions that we assume to define the domain of $p_0$ when solving the Schrödinger equation in this case. Can we conclude that the notion of momentum cannot be rigorously defined in this elementary example? Or, at the very least, do we have to admit some $\psi$ that obey non-physical boundary conditions?

For those interested, this is a related question, also touching upon domain issues of momentum operators.

In the context of mathematical quantum mechanics, a well known no-go theorem known as Hellinger-Töplitz tells us that an unbounded, symmetric operator cannot be defined everywhere on the Hilbert space $\mathcal H$. Thus, for many operators of interest in quantum mechanics, we must resort to restricting the domain on which they may act. Along with the Hamiltonian, the momentum operator $$ p:=-i\frac{\rm d}{\mathrm{d}x}$$ is the most prominent example of such an operator. Now, one of the most important questions is clearly: What is the correct domain of definition for the momentum operator?

In order to answer this question, it is of crucial importance to consider the question of self-adjointness: Only self-adjoint operators are admissible as observables of the theory, and there are certain vital mathematical results, such as the spectral theorem and Stone's theorem, which make it clear that any 'good' momentum operator must certainly be self-adjoint: $ p^\star= p$. So, the original question is refined slightly: What is the correct domain $\mathcal D( p)\subset \mathcal H$ that allows us to construct a self-adjoint momentum operator?

Recently, one of my lecturers treated a concrete incarnation of this (general and possibly vague) question: Consider a Hilbert space $\mathcal H=L^2(0,1)$ and two 'candidate momenta', $ p_0$ and $ p_\alpha$, with domains

$$\mathcal D(p_0)=\{\psi\in \mathcal H \,|\, \psi\in \text{AC}(0,1),\psi'\in\mathcal H, \psi(0)=0=\psi(1)\} $$ $$ \mathcal D(p_\alpha)=\{\psi\in \mathcal H \,|\, \psi\in \text{AC}(0,1),\psi'\in\mathcal H, \psi(0)=\alpha\psi(1)\},\hspace{.5cm}|\alpha|=1 $$ Are these operators self-adjoint? As it turns out, it can be shown that $p_0$ is symmetric but not self-adjoint. However, $p_\alpha$ is self-adjoint. It seems that the domain of $p_0$ is 'too small'.

Now, my question is: What does the conclusion that $p_0\neq p_0^\star$ imply for the canonical freshman QM example of the infinite potential well? As far as I'm aware, it is conventional to take exactly the boundary conditions that we assume to define the domain of $p_0$ when solving the Schrödinger equation in this case. Can we conclude that the notion of momentum cannot be rigorously defined in this elementary example?

For those interested, this is a related question, also touching upon domain issues of momentum operators.

In the context of mathematical quantum mechanics, a well known no-go theorem known as Hellinger-Töplitz tells us that an unbounded, symmetric operator cannot be defined everywhere on the Hilbert space $\mathcal H$. Thus, for many operators of interest in quantum mechanics, we must resort to restricting the domain on which they may act. Along with the Hamiltonian, the momentum operator $$ p:=-i\frac{\rm d}{\mathrm{d}x}$$ is the most prominent example of such an operator. Now, one of the most important questions is clearly: What is the correct domain of definition for the momentum operator?

In order to answer this question, it is of crucial importance to consider the question of self-adjointness: Only self-adjoint operators are admissible as observables of the theory, and there are certain vital mathematical results, such as the spectral theorem and Stone's theorem, which make it clear that any 'good' momentum operator must certainly be self-adjoint: $ p^\star= p$. So, the original question is refined slightly: What is the correct domain $\mathcal D( p)\subset \mathcal H$ that allows us to construct a self-adjoint momentum operator?

Recently, one of my lecturers treated a concrete incarnation of this (general and possibly vague) question: Consider a Hilbert space $\mathcal H=L^2(0,1)$ and two 'candidate momenta', $ p_0$ and $ p_\alpha$, with domains

$$\mathcal D(p_0)=\{\psi\in \mathcal H \,|\, \psi\in \text{AC}(0,1),\psi'\in\mathcal H, \psi(0)=0=\psi(1)\} $$ $$ \mathcal D(p_\alpha)=\{\psi\in \mathcal H \,|\, \psi\in \text{AC}(0,1),\psi'\in\mathcal H, \psi(0)=\alpha\psi(1)\},\hspace{.5cm}|\alpha|=1 $$ Are these operators self-adjoint? As it turns out, it can be shown that $p_0$ is symmetric but not self-adjoint. However, $p_\alpha$ is self-adjoint. It seems that the domain of $p_0$ is 'too small'.

Now, my question is: What does the conclusion that $p_0\neq p_0^\star$ imply for the canonical freshman QM example of the infinite potential well? As far as I'm aware, it is conventional to take exactly the boundary conditions that we assume to define the domain of $p_0$ when solving the Schrödinger equation in this case. Can we conclude that the notion of momentum cannot be rigorously defined in this elementary example? Or, at the very least, do we have to admit some $\psi$ that obey non-physical boundary conditions?

For those interested, this is a related question, also touching upon domain issues of momentum operators.