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The momentum operator generates spatial translations. In a one-dimensional position representation in quantum mechanics, it acts asApplying it to a plane wave $e^{ikx}$ gives momentum eigenvalue $\hslash k$, connecting momentum with the de Broglie wavelength. A localized wave packet generally contains several such momenta.

A differential expression alone does not define a self-adjoint operator: its domain and boundary conditions matter. On a periodic interval, the allowed momentum eigenvalues are discrete; on the whole real line they form a continuous spectrum. In a magnetic field, this operator is the canonical momentum. The mechanical momentum entering the kinetic energy is instead $P-qA$ in SI units, where $A$ is the vector potential.

Orbital angular momentum is the quantum counterpart of $\mathbf{r}\times \mathbf{p}$. In the position representation in quantum mechanics,Its components generate rotations of the spatial wavefunction. For a three-dimensional scalar wavefunction, simultaneous eigenstates of $L^2$ and $L_z$ have eigenvalues ${\hslash }^2\ell (\ell +1)$ and $m\hslash$, with integer $\ell \ge 0$ and $m=-\ell ,\dots ,\ell$.

For a central potential energy, the angular factors are spherical harmonics, and orbital angular momentum organizes the energy eigenstates into rotational multiplets. It differs from spin, which is an intrinsic degree of freedom rather than angular dependence of a scalar spatial wavefunction. The angular momentum operators obey the same rotation algebra in both cases.

For a quantum state that is an eigenstate of the parity operator, parity is its eigenvalue $+1$ or $-1$. In a scalar position representation in quantum mechanics, the parity operator acts by $(P\psi )(\mathbf{r})=\psi (-\mathbf{r})$. Since $P^2=I$, these are its only possible eigenvalues: even wavefunctions have parity $+1$ and odd wavefunctions have parity $-1$.

If the Hamiltonian commutes with the parity operator, its energy eigenstates can be chosen to have definite parity. In the quantum harmonic oscillator, parity alternates as $(-1{)}^n$. This helps establish selection rules: the position operator is odd under spatial reflection, so its matrix element between two states of the same parity vanishes.

In the position representation in quantum mechanics, $\hat{\mathbf{p}}=-i\hslash \nabla$. Therefore the orbital angular momentum operator givesAt fixed $r,\theta$, the spherical coordinates obey ${\partial }_{\varphi }x=-y$, ${\partial }_{\varphi }y=x$, and ${\partial }_{\varphi }z=0$. The chain rule consequently gives ${\partial }_{\varphi }=-y{\partial }_x+x{\partial }_y$, henceThe derivative holds the other two spherical coordinates fixed; it is not a derivative along an arbitrary spatial curve.

Factorization. Put $\alpha =\sqrt{m\omega /(2\hslash )}$ and use the canonical commutation relation $[x,p]=i\hslash$:It follows thatThe $-1/2$ arises from quantum operator ordering; treating $x$ and $p$ as commuting would lose the zero-point energy.

commutators. Direct use of $[x,p]=i\hslash$ gives $[a,a^{†}]=1$. The product rule for commutators then yields $[a^{†}a,a]=-a$ and $[a^{†}a,a^{†}]=a^{†}$, hence

Raising the energy. If $H|\varphi ⟩=E|\varphi ⟩$, thenThus the creation operator produces an energy eigenstate with energy $E+\hslash \omega$. It is nonzero: for a normalized physical oscillator eigenstate, $\Vert a^{†}\varphi {\Vert }^2=⟨a{a}^{†}⟩=E/(\hslash \omega )+1/2>0$. Normalize by this norm if a normalized state is wanted.

Lowest wavefunctions and orthogonality. In the position representation in quantum mechanics,The annihilation condition therefore gives ${\psi }_0^{\prime }=-(m\omega /\hslash )x{\psi }_0$. Applying $a^{†}$ to its solution givesTheir inner product is proportional to ${\int }_{-\infty }^{\infty }x{e}^{-m\omega x^2/\hslash }dx=0$, since the integrand is odd. These are the even and odd lowest wavefunctions shown in A2, with energies $\hslash \omega /2$ and $3\hslash \omega /2$.

Take a normalized vector in the required common operator domains. Expanding the squared norm and using self-adjointness givesThus its final term is $\hslash t⟨C⟩$. Set $a=\Vert A\psi {\Vert }^2$, $b=\Vert B\psi {\Vert }^2$, $c=⟨C⟩$. If $b>0$, minimizing the nonnegative quadratic at $t_{\ast }=-\hslash c/(2b)$ givesEquality holds precisely when the minimum norm vanishes, that is $(A-i{t}_{\ast }B)\psi =0$. If $b=0$, however, then $B\psi =0$ and $c=0$, so equality is automatic even when $A\psi \ne 0$. The printed equality criterion needs this zero-vector exception. For example $A=I$, $B=0$, $C=0$ saturates the inequality but admits no such real $t$. The exact general criterion is linear dependence of $A\psi ,B\psi$ together with vanishing real part of their inner product, as in the uncentered second-moment commutator inequality.

Apply the same calculation to $A-⟨A⟩I$, $B-⟨B⟩I$, whose commutator is unchanged. This gives the Robertson uncertainty principleIf $\psi$ is an eigenvector of $A$, its dispersion is zero; finite dispersion of $B$ then forces $\boxed{⟨C⟩=0}$.

In the position representation, $X\psi =x\psi$ and $P\psi =-i\hslash {\psi }^{\prime }$. The product rule gives $[P,X]\psi =-i\hslash \psi$, hence $\boxed{[P,X]=-i\hslash I}$. Since $V(X)$ commutes with $X$,For an energy eigenstate, $⟨[H,X]⟩=E⟨X⟩-E⟨X⟩=0$, so $\boxed{⟨P⟩=0}$. For the oscillator $[H,P]=i\hslash m{\omega }^{2}X$, and the same argument gives $\boxed{⟨X⟩=0}$. These expectation identities require the stated finite moments and commutator domains; formal position eigenvectors are not normalizable exceptions.

Apply the Robertson uncertainty principle to each angular-momentum pair:If $\psi$ is an eigenvector of $L_3$, then $\Delta L_3=0$. The inequalities for $(L_2,L_3)$ and $(L_3,L_1)$ force $\boxed{⟨L_1⟩=⟨L_2⟩=0}$, assuming the relevant moments exist.

The position operator represents a position measurement. In the position representation in quantum mechanics, its one-dimensional action is multiplication, $(X\psi )(x)=x\psi (x)$. The Born rule associates an interval with the integral of $|\psi (x){|}^2$ over that interval, rather than assigning a finite probability to a single point of a continuous distribution.

The expectation value is $⟨X⟩=\int x|\psi (x){|}^{2}dx$ when that integral exists. Its domain in the Hilbert space of square-integrable wavefunctions requires $x\psi (x)$ to be square integrable; a normalized state need not belong to this domain. Together with the momentum operator, the position operator obeys the canonical commutation relation on a suitable common domain, leading to the Heisenberg uncertainty principle.

The spatial translation operator is the unitary operator generated by momentum. For displacement $\mathbf{y}$,where $\hslash$ is the reduced Planck constant. In a position representation in quantum mechanics, $\mathbf{P}=-i\hslash \nabla$, so its action on a wavefunction isA localized wave packet is therefore displaced by $+\mathbf{y}$. A momentum eigenstate of eigenvalue $\mathbf{p}$ instead acquires the phase $e^{-i\mathbf{p}\cdot \mathbf{y}/\hslash }$.

The canonical commutation relation givesThus the expectation value of the position operator in the translated quantum state increases by $\mathbf{y}$.

For an operator-valued scalar field, the compatible translation convention isInfinitesimally this means $[P_j,\varphi (\mathbf{x})]=i\hslash {\partial }_j\varphi (\mathbf{x})$. The field's coordinate argument labels where a local observable is evaluated; $\mathbf{X}$ is the position operator of a particle. These meanings explain their differently ordered conjugation formulas. ^[ref]