2 removed discussion of a different operator edited Feb 2 '16 at 15:42 ACuriousMind♦ 76.1k1818 gold badges139139 silver badges356356 bronze badges This is what happens if one cares not for the differerence between self-adjointness and Hermiticity, or, more general, if one ignoressubtlety that quantum mechanical operators are typically only defined on subspaces of the full Hilbert space. Let's set $$a=1$$ for convenience. The operator $$p =-\mathrm{i}\hbar\partial_x$$ acting on wavefunctions with periodic boundary conditions defined on $$D(p) = \{\psi\in L^2([0,1])\mid \psi(0)=\psi(1)\land \psi'\in L^2([0,1])\}$$ is Hermitian, but not self-adjoint, that is, on the domain of definition of $$p$$, we have $$p=p^\dagger$$, butand $$p^\dagger$$ admits a largerthe same domain of definition. The Hermiticityself-adjointness of $$p$$ follows from the periodic boundary conditions killing the surface terms that appear in the $$L^2$$ inner product $$\langle \phi,p\psi\rangle - \langle p^\dagger \phi,\psi\rangle = \int\overline{\phi(x)}\mathrm{i}\hbar\partial_x\psi(x) - \overline{\mathrm{i}\hbar\partial_x\phi(x)}\psi(x) = 0$$ for every $$\psi\in D(p)$$ and every $$\phi\in D(p^\dagger) = \{\phi\in L^2([0,1])\mid \phi'\in L^2([0,1])\}$$. Since $$D(p)\subset D(p^\dagger)$$ but $$D(p)\neq D(p^\dagger)$$, the operator p is Hermitian, but not self-adjoint, and hence not the correct observable for momentum. In particular, the functions $$\psi_{p_0}(x) = \exp(\mathrm{i}p_0x),p_0\in\mathbb{C}$$ belong to $$D(p^\dagger)$$$$\phi\in D(p^\dagger) = D(p)$$, but not to $$D(p)$$, so they are not eigenvectors of $$p$$ and thus not solutions to the Schrödinger equation. $$p$$ has no eigenvalues, but those of $$p^\dagger$$ are all complex numbers $$p_0\in\mathbb{C}$$. The operator $$p$$ can be made-self-adjoint by enlarging its domain of definition. Recall that a phase is unphysical - this is the motivation to relax the boundary condition to have $$D(p) := \{\psi\in L^2([0,1])\mid \exists\alpha\in\mathbb{R}:\psi(0) = \mathrm{e}^{\mathrm{i}\alpha}\psi(1)\land \psi'\in L^2([0,1])\}$$ and for this new operator, we have $$p=p^\dagger$$ and$$\phi$$ with $$D(p^\dagger) = D(p)$$$$\phi(0)\neq\phi(1)$$. Its spectrum is the real line, the eigenfunctions are $$\psi_{p_0},p_o\in\mathbb{R}$$. Now, for the question of the commutator: the multplication operator $$x$$ is defined on the entire Hilbert space, since for $$\psi\in L^2([0,1])$$ $$x\psi$$ is also square-integrable. For the product of two operators $$A,B$$, we have the rule $$D(AB) = \{\psi\in D(B)\mid B\psi\in D(A)\}$$ and $$D(A+B) = D(A)\cap D(B)$$ so we obtain \begin{align} D(px) & = \{\psi\in L^2([0,1])\mid x\psi\in D(p)\} \\ D(xp) & = D(p) \end{align} and $$x\psi\in D(p)$$ means $$0\cdot \psi(0) = \mathrm{e}^{\mathrm{i}\alpha}\psi(1)$$$$0\cdot \psi(0) = 1\cdot\psi(1)$$, that is, $$\psi(1) = 0$$. Hence we have $$D(px) = \{\psi\in L^2([0,1])\mid \psi'\in L^2([0,1]) \land \psi(1) = 0\}$$ and finally $$D([x,p]) = D(xp)\cap D(px) = \{\psi\in L^2([0,1])\mid \psi'\in L^2([0,1])\land \psi(0)=\psi(1) = 0\}$$ meaning the plane waves $$\psi_{p_0}$$ do not belong to the domain of definition of the commutator $$[x,p]$$ and you cannot apply the naive uncertainty principle to them. However, for self-adjoint operators $$A,B$$, you may rewrite the uncertainty principle as $$\sigma_\psi(A)\sigma_\psi(B)\geq \frac{1}{2} \lvert \langle \psi,\mathrm{i}[A,B]\rangle\psi\rvert = \frac{1}{2}\lvert\mathrm{i}\left(\langle A\psi,B\psi\rangle - \langle B\psi,A\psi\rangle\right)\rvert$$ where the r.h.s. and l.h.s. are now both defined on $$D(A)\cap D(B)$$. Applying this version to the plane waves yields no contradiction. This is what happens if one cares not for the differerence between self-adjointness and Hermiticity, or, more general, if one ignores that quantum mechanical operators are typically only defined on subspaces of the full Hilbert space. Let's set $$a=1$$ for convenience. The operator $$p =-\mathrm{i}\hbar\partial_x$$ acting on wavefunctions with periodic boundary conditions defined on $$D(p) = \{\psi\in L^2([0,1])\mid \psi(0)=\psi(1)\land \psi'\in L^2([0,1])\}$$ is Hermitian, but not self-adjoint, that is, on the domain of definition of $$p$$, we have $$p=p^\dagger$$, but $$p^\dagger$$ admits a larger domain of definition. The Hermiticity of $$p$$ follows from the periodic boundary conditions killing the surface terms that appear in the $$L^2$$ inner product $$\langle \phi,p\psi\rangle - \langle p^\dagger \phi,\psi\rangle = \int\overline{\phi(x)}\mathrm{i}\hbar\partial_x\psi(x) - \overline{\mathrm{i}\hbar\partial_x\phi(x)}\psi(x) = 0$$ for every $$\psi\in D(p)$$ and every $$\phi\in D(p^\dagger) = \{\phi\in L^2([0,1])\mid \phi'\in L^2([0,1])\}$$. Since $$D(p)\subset D(p^\dagger)$$ but $$D(p)\neq D(p^\dagger)$$, the operator p is Hermitian, but not self-adjoint, and hence not the correct observable for momentum. In particular, the functions $$\psi_{p_0}(x) = \exp(\mathrm{i}p_0x),p_0\in\mathbb{C}$$ belong to $$D(p^\dagger)$$, but not to $$D(p)$$, so they are not eigenvectors of $$p$$ and thus not solutions to the Schrödinger equation. $$p$$ has no eigenvalues, but those of $$p^\dagger$$ are all complex numbers $$p_0\in\mathbb{C}$$. The operator $$p$$ can be made-self-adjoint by enlarging its domain of definition. Recall that a phase is unphysical - this is the motivation to relax the boundary condition to have $$D(p) := \{\psi\in L^2([0,1])\mid \exists\alpha\in\mathbb{R}:\psi(0) = \mathrm{e}^{\mathrm{i}\alpha}\psi(1)\land \psi'\in L^2([0,1])\}$$ and for this new operator, we have $$p=p^\dagger$$ and $$D(p^\dagger) = D(p)$$. Its spectrum is the real line, the eigenfunctions are $$\psi_{p_0},p_o\in\mathbb{R}$$. Now, for the question of the commutator: the multplication operator $$x$$ is defined on the entire Hilbert space, since for $$\psi\in L^2([0,1])$$ $$x\psi$$ is also square-integrable. For the product of two operators $$A,B$$, we have the rule $$D(AB) = \{\psi\in D(B)\mid B\psi\in D(A)\}$$ and $$D(A+B) = D(A)\cap D(B)$$ so we obtain \begin{align} D(px) & = \{\psi\in L^2([0,1])\mid x\psi\in D(p)\} \\ D(xp) & = D(p) \end{align} and $$x\psi\in D(p)$$ means $$0\cdot \psi(0) = \mathrm{e}^{\mathrm{i}\alpha}\psi(1)$$, that is, $$\psi(1) = 0$$. Hence we have $$D(px) = \{\psi\in L^2([0,1])\mid \psi'\in L^2([0,1]) \land \psi(1) = 0\}$$ and finally $$D([x,p]) = D(xp)\cap D(px) = \{\psi\in L^2([0,1])\mid \psi'\in L^2([0,1])\land \psi(0)=\psi(1) = 0\}$$ meaning the plane waves $$\psi_{p_0}$$ do not belong to the domain of definition of the commutator $$[x,p]$$ and you cannot apply the naive uncertainty principle to them. However, for self-adjoint operators $$A,B$$, you may rewrite the uncertainty principle as $$\sigma_\psi(A)\sigma_\psi(B)\geq \frac{1}{2} \lvert \langle \psi,\mathrm{i}[A,B]\rangle\psi\rvert = \frac{1}{2}\lvert\mathrm{i}\left(\langle A\psi,B\psi\rangle - \langle B\psi,A\psi\rangle\right)\rvert$$ where the r.h.s. and l.h.s. are now both defined on $$D(A)\cap D(B)$$. Applying this version to the plane waves yields no contradiction. This is what happens if one cares not for the subtlety that quantum mechanical operators are typically only defined on subspaces of the full Hilbert space. Let's set $$a=1$$ for convenience. The operator $$p =-\mathrm{i}\hbar\partial_x$$ acting on wavefunctions with periodic boundary conditions defined on $$D(p) = \{\psi\in L^2([0,1])\mid \psi(0)=\psi(1)\land \psi'\in L^2([0,1])\}$$ is self-adjoint, that is, on the domain of definition of $$p$$, we have $$p=p^\dagger$$, and $$p^\dagger$$ admits the same domain of definition. The self-adjointness of $$p$$ follows from the periodic boundary conditions killing the surface terms that appear in the $$L^2$$ inner product $$\langle \phi,p\psi\rangle - \langle p^\dagger \phi,\psi\rangle = \int\overline{\phi(x)}\mathrm{i}\hbar\partial_x\psi(x) - \overline{\mathrm{i}\hbar\partial_x\phi(x)}\psi(x) = 0$$ for every $$\psi\in D(p)$$ and every $$\phi\in D(p^\dagger) = D(p)$$, but not for $$\phi$$ with $$\phi(0)\neq\phi(1)$$. Now, for the question of the commutator: the multplication operator $$x$$ is defined on the entire Hilbert space, since for $$\psi\in L^2([0,1])$$ $$x\psi$$ is also square-integrable. For the product of two operators $$A,B$$, we have the rule $$D(AB) = \{\psi\in D(B)\mid B\psi\in D(A)\}$$ and $$D(A+B) = D(A)\cap D(B)$$ so we obtain \begin{align} D(px) & = \{\psi\in L^2([0,1])\mid x\psi\in D(p)\} \\ D(xp) & = D(p) \end{align} and $$x\psi\in D(p)$$ means $$0\cdot \psi(0) = 1\cdot\psi(1)$$, that is, $$\psi(1) = 0$$. Hence we have $$D(px) = \{\psi\in L^2([0,1])\mid \psi'\in L^2([0,1]) \land \psi(1) = 0\}$$ and finally $$D([x,p]) = D(xp)\cap D(px) = \{\psi\in L^2([0,1])\mid \psi'\in L^2([0,1])\land \psi(0)=\psi(1) = 0\}$$ meaning the plane waves $$\psi_{p_0}$$ do not belong to the domain of definition of the commutator $$[x,p]$$ and you cannot apply the naive uncertainty principle to them. However, for self-adjoint operators $$A,B$$, you may rewrite the uncertainty principle as $$\sigma_\psi(A)\sigma_\psi(B)\geq \frac{1}{2} \lvert \langle \psi,\mathrm{i}[A,B]\rangle\psi\rvert = \frac{1}{2}\lvert\mathrm{i}\left(\langle A\psi,B\psi\rangle - \langle B\psi,A\psi\rangle\right)\rvert$$ where the r.h.s. and l.h.s. are now both defined on $$D(A)\cap D(B)$$. Applying this version to the plane waves yields no contradiction. 1 answered Feb 2 '16 at 14:57 ACuriousMind♦ 76.1k1818 gold badges139139 silver badges356356 bronze badges This is what happens if one cares not for the differerence between self-adjointness and Hermiticity, or, more general, if one ignores that quantum mechanical operators are typically only defined on subspaces of the full Hilbert space. Let's set $$a=1$$ for convenience. The operator $$p =-\mathrm{i}\hbar\partial_x$$ acting on wavefunctions with periodic boundary conditions defined on $$D(p) = \{\psi\in L^2([0,1])\mid \psi(0)=\psi(1)\land \psi'\in L^2([0,1])\}$$ is Hermitian, but not self-adjoint, that is, on the domain of definition of $$p$$, we have $$p=p^\dagger$$, but $$p^\dagger$$ admits a larger domain of definition. The Hermiticity of $$p$$ follows from the periodic boundary conditions killing the surface terms that appear in the $$L^2$$ inner product $$\langle \phi,p\psi\rangle - \langle p^\dagger \phi,\psi\rangle = \int\overline{\phi(x)}\mathrm{i}\hbar\partial_x\psi(x) - \overline{\mathrm{i}\hbar\partial_x\phi(x)}\psi(x) = 0$$ for every $$\psi\in D(p)$$ and every $$\phi\in D(p^\dagger) = \{\phi\in L^2([0,1])\mid \phi'\in L^2([0,1])\}$$. Since $$D(p)\subset D(p^\dagger)$$ but $$D(p)\neq D(p^\dagger)$$, the operator p is Hermitian, but not self-adjoint, and hence not the correct observable for momentum. In particular, the functions $$\psi_{p_0}(x) = \exp(\mathrm{i}p_0x),p_0\in\mathbb{C}$$ belong to $$D(p^\dagger)$$, but not to $$D(p)$$, so they are not eigenvectors of $$p$$ and thus not solutions to the Schrödinger equation. $$p$$ has no eigenvalues, but those of $$p^\dagger$$ are all complex numbers $$p_0\in\mathbb{C}$$. The operator $$p$$ can be made-self-adjoint by enlarging its domain of definition. Recall that a phase is unphysical - this is the motivation to relax the boundary condition to have $$D(p) := \{\psi\in L^2([0,1])\mid \exists\alpha\in\mathbb{R}:\psi(0) = \mathrm{e}^{\mathrm{i}\alpha}\psi(1)\land \psi'\in L^2([0,1])\}$$ and for this new operator, we have $$p=p^\dagger$$ and $$D(p^\dagger) = D(p)$$. Its spectrum is the real line, the eigenfunctions are $$\psi_{p_0},p_o\in\mathbb{R}$$. Now, for the question of the commutator: the multplication operator $$x$$ is defined on the entire Hilbert space, since for $$\psi\in L^2([0,1])$$ $$x\psi$$ is also square-integrable. For the product of two operators $$A,B$$, we have the rule $$D(AB) = \{\psi\in D(B)\mid B\psi\in D(A)\}$$ and $$D(A+B) = D(A)\cap D(B)$$ so we obtain \begin{align} D(px) & = \{\psi\in L^2([0,1])\mid x\psi\in D(p)\} \\ D(xp) & = D(p) \end{align} and $$x\psi\in D(p)$$ means $$0\cdot \psi(0) = \mathrm{e}^{\mathrm{i}\alpha}\psi(1)$$, that is, $$\psi(1) = 0$$. Hence we have $$D(px) = \{\psi\in L^2([0,1])\mid \psi'\in L^2([0,1]) \land \psi(1) = 0\}$$ and finally $$D([x,p]) = D(xp)\cap D(px) = \{\psi\in L^2([0,1])\mid \psi'\in L^2([0,1])\land \psi(0)=\psi(1) = 0\}$$ meaning the plane waves $$\psi_{p_0}$$ do not belong to the domain of definition of the commutator $$[x,p]$$ and you cannot apply the naive uncertainty principle to them. However, for self-adjoint operators $$A,B$$, you may rewrite the uncertainty principle as $$\sigma_\psi(A)\sigma_\psi(B)\geq \frac{1}{2} \lvert \langle \psi,\mathrm{i}[A,B]\rangle\psi\rvert = \frac{1}{2}\lvert\mathrm{i}\left(\langle A\psi,B\psi\rangle - \langle B\psi,A\psi\rangle\right)\rvert$$ where the r.h.s. and l.h.s. are now both defined on $$D(A)\cap D(B)$$. Applying this version to the plane waves yields no contradiction.