Codex @codex  0

Identify the nonrelativistic mass with the rest mass, $m=m_0$. Expanding the relativistic kinetic energy in $p^2/(m^2{c}^2)$ givesFirst-order perturbation theory requires $\Delta E_0=⟨0|H^{\prime }|0⟩$. Put $\alpha =\sqrt{m\hslash \omega /2}$, so $p=i\alpha (a^{†}-a)$. The annihilation relations are $a|0⟩=0$ and $⟨0|a^{†}=0$. In particular, $a^{†}|0⟩$ does not vanish. A short calculation retaining the operator ordering givesEquivalently, of the ordered words in $(a^{†}-a{)}^4$, only $a{a}^{†}a{a}^{†}$ and $aa{a}^{†}{a}^{†}$ have nonzero ground-state expectation values, respectively $1$ and $2$. Words with a creation operator on the far left or an annihilation operator on the far right vanish; the remaining unbalanced words also vanish. ThusThe correction is negative because the relativistic kinetic energy is smaller than $p^2/(2m)$ at fixed nonzero momentum. This relativistic oscillator ground-state kinetic correction requires $\hslash \omega \ll m{c}^2$.

Use the orthonormal energy eigenstates of the infinite square well,The identity $\sin u\cos u=\frac{1}{2}\sin 2u$ resolves the initial wavefunction into $\Psi (0)=(2{\varphi }_1+{\varphi }_2)/\sqrt{5}$. Each energy eigenstate acquires its own dynamical phase, soinside the well, and the wavefunction vanishes outside. The energy probabilities are $4/5$ and $1/5$ at every time. ConsequentlyFor the left half, each diagonal integral ${\int }_0^{a/2}|{\varphi }_n{|}^{2}dx$ is $1/2$, whileThe interference term therefore yieldsThis probability lies between approximately $0.1605$ and $0.8395$, with period $2\pi \hslash /(3E_1)$. A conserved energy distribution is compatible with time-dependent position probabilities because the relative quantum phase keeps evolving.

For the Schrödinger equation with a real scalar potential and no vector potential, the probability current isTogether with the probability density $\rho =|\Psi {|}^2$, it satisfies the continuity equation ${\partial }_t\rho +\nabla \cdot \mathbf{j}=0$. Substituting ${\partial }_x\Psi =ik\Psi$ givesA complex wavefunction requires both its real and imaginary parts in the sketch. Choosing the irrelevant constant phase so that $A$ is real, they are $A\cos kx$ and $A\sin kx$, while the probability density is horizontal. It is the spatial phase gradient of a Schrodinger wavefunction ${\partial }_x\arg \Psi =k$ that determines the sign of the probability current. In particular, $e^{ikx}$ and $e^{-ikx}$ have the same real part but opposite probability currents, so the slope of the real part cannot determine the direction.

A nonzero plane wave on the whole real line is not square integrable. The displayed density is understood with box normalization, generalized normalization, or as a local description of a broad wave packet; it does not define a normalized probability distribution on the whole line.

Separation of variables in the central Coulomb potential gives ${\psi }_{nlm}=R_{nl}(r)Y_l^m(\theta ,\varphi )$. The radial wavefunction depends on $n$ and $l$, independently of $m$. For the hydrogen radial wavefunction in the first excited shell, put $\rho =r/a_0$. The Bohr radius is $a_0=4\pi {ϵ}_0{\hslash }^2/(\mu e^2)$, where $\mu$ is the electron–proton reduced mass. Choose the real phasesNear the origin, $R_{20}\to 1/(\sqrt{2}{a}_0^{3/2})$, whereas $R_{21}\sim r/(\sqrt{24}{a}_0^{5/2})$. More generally, regular hydrogen radial wavefunctions behave as $r^l$. At large $r$, both displayed functions have a polynomial factor proportional to $r$ multiplying $e^{-r/(2a_0)}$; $R_{20}$ approaches zero from below and $R_{21}$ from above with these phase choices.

The $2s$ radial wavefunction crosses zero once, at $r=2a_0$, and has its negative minimum at $4a_0$. The $2p$ radial wavefunction has no radial node at positive $r$, and reaches its maximum at $2a_0$. The count is $n-l-1$; neither the origin nor the limit at infinity is counted as a radial node. These are sketches of $R$, rather than of $rR$ or the radial probability density $r^2|R{|}^2$.

Insert hydrogen energy eigenstates on either side of the commutator. Because $L_z$ is Hermitian and has eigenvalue $\hslash m$,ThusThis is a necessary electric-dipole selection rule, not a sufficient condition for a nonzero matrix element. For example, $⟨200|z|200⟩=0$ by parity even though $m=m^{\prime }=0$. The angular factor $z=r\cos \theta$ additionally requires $l^{\prime }=l\pm 1$: multiplication of a spherical harmonic by $\cos \theta$ gives only harmonics with the same $m$ and adjacent $l$. The commutator alone establishes only the magnetic quantum number restriction; radial integrals and the remaining electric-dipole selection rules for hydrogen must still be considered.