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Expand the state in the energy eigenstates of $H_0$ asSubstitution into the Time-dependent Schrodinger equation and projection onto $⟨k|$ gives the exact interaction picture equationsIn first-order time-dependent perturbation theory, put $c_j(t)={\delta }_{j0}$ on the right-hand side and integrate from $0$ to $t$:

For a uniform electric field, the electric-dipole interaction is $\Delta (t)=e{\mathcal{E}}_{0}z{e}^{-t/\tau }$ up to an irrelevant overall sign. The supplied hydrogen atom wavefunctions giveFor $t\gg \tau$, the time integral isThe long-time transition fraction is consequentlyBoth $|100⟩$ and $|200⟩$ have even parity, whereas $z$ is odd, so the electric-dipole selection rule gives $⟨200|z|100⟩=0$. Thus the requested fraction in $|200⟩$ is zero.

The Time-dependent Schrodinger equation for the one-dimensional Quantum harmonic oscillator isA stationary state has $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$, so cancellation of the common time factor gives

SetThe equation becomesWith $\psi =f(y)e^{-y^2/2}$,and hence $f$ obeys the Hermite differential equation

If $f$ is monic of degree $N$, the coefficient of $y^N$ in this equation is $ϵ-1-2N$. It must vanish, soWriting $f={\sum }_{k=0}^N{a}_k{y}^k$ gives the recurrenceThe coefficient of $y^{N-1}$ first gives $a_{N-1}=0$, and the recurrence preserves parity. If $N$ is even, every odd coefficient vanishes, so $f$ is an even function. The Gaussian factor is also even; therefore the stationary state $\psi$ has even parity.

In the Parity basis of a one-dimensional S-matrix, parity acts as $\mathcal{P}=\mathrm{diag}(1,-1)$. Parity invariance requiresand thereforeIndependently, unitarity givesCombining parity and unitarity leaves two phases,

For the symmetric double-delta potential, consider the odd channel on $x\ge 0$. Choose the asymptotic conventionand write $psi(x)=C\sin (kx)$ for $0<x<a$; the sine enforces odd parity at the origin. Continuity and the delta potential derivative jump at $x=a$ giveEliminating $C$ yields the Odd-parity S-matrix of a symmetric double-delta potentialUsing $2i\sin (ka)=e^{ika}-e^{-ika}$ rewrites this as the requested expressionFor real $k$, the two factors in the ratio are complex conjugates, directly confirming $|S_{--}|=1$.

A bound state occurs at a pole $k=i\kappa$, $kappa>0$. Setting the denominator to zero givesorFor $U_0<0$, write $g=-U_0>0$. The right-hand side $g(1-e^{-2\kappa a})$ is increasing and concave, starts at zero with slope $2ga$, and approaches $g$, whereas the left-hand side is the straight line $2\kappa$. A positive intersection therefore exists exactly whenThis is the odd bound state of a symmetric double-delta potential; equality is its zero-energy threshold.

For $U_0>0$, analytically continue the same pole equation into the lower half of the complex $k$-plane. PutIt becomesWhen $G\gg 1$, the lowest odd mode lies near the hard-wall value $z=\pi$. Writing $z=\pi +\eta$ and expanding givessoIn particular, the odd resonance of a strongly repulsive symmetric double-delta potential hasThe negative imaginary part identifies a decaying scattering resonance. Ifthen, to leading nonzero order,Its time-dependent amplitude is proportional toso its survival probability decays as $e^{-\Gamma t/\hslash }$ and its lifetime isPhysically, this is the lowest odd standing wave trapped for a long time between two strong repulsive barriers and leaking out by quantum tunnelling.

Using the zero test $s$ from part (i), define by primitive recursionSince $s(0)=1$ and $s(1)=0$, the values alternate:Thus $r$ is the required parity function and, like $s$, is primitive recursive without minimization.

Since $X=L(a+a^{†})$, the perturbation isEvery diagonal matrix element vanishes by parity, so all first-order shifts are zero. For the ground state,The second-order shift is thereforeThusFor every nonzero real $\lambda$, the cubic potential is unbounded below on one side, so there are no genuine bound-state eigenvalues continuously connected to the harmonic levels. The formal perturbation series does not converge; it is at best asymptotic to resonance energies.

The parity of a wavefunction is its eigenvalue under spatial reflection: $\psi (-x)=+\psi (x)$ for even parity and $\psi (-x)=-\psi (x)$ for odd parity. If $V(-x)=V(x)$, the Hamiltonian commutes with reflection, so its energy eigenspaces admit a parity eigenbasis. In particular, every nondegenerate bound-state energy eigenfunction has definite parity; within a degenerate scattering eigenspace one must choose the even and odd linear combinations.

To first order in $F$, time-dependent perturbation theory replaces the state on the right-hand side by its unperturbed value $e^{-i{E}_{0}t/\hslash }|0⟩$. The bound state is even, while the position operator $x$ is odd. The parity selection rule therefore givesso ${\dot{b}}_k=0$ and, with no initial continuum component,

Projection onto the odd state givesWriting ${\Omega }_k=(E_k-E_0-\hslash \omega )/\hslash$ and integrating from zero to $t$ yields

Write $m=|m|e^{i\theta }$. Parity reverses the spatial coordinates and must exchange the two chiral Weyl spinors. One convenient choice isThe identities ${\sigma }^0={\bar{\sigma }}^0$ and ${\sigma }^i=-{\bar{\sigma }}^i$ exchange the two kinetic terms after $\mathbf{x}\mapsto -\mathbf{x}$. The mass bilinear $m{\bar{\psi }}_R{\psi }_L$ maps to its complex conjugate $m^{\ast }{\bar{\psi }}_L{\psi }_R$, so the two mass terms are exchanged and the action is invariant. More generally, independent unit phases $a,b$ may multiply the two transformations provided $b^{\ast }a=m^{\ast }/m$.

Apply parity to the mixed anticommutator. The transformed left-hand side isbecause $|{\eta }_P|=1$. The Pauli-matrix identity represented by this product leaves the temporal matrix unchanged and reverses the three spatial matrices, so it equalsBut a parity transformation acts on momentum as $P^0\mapsto P^0$ and $P^i\mapsto -P^i$. This is exactly the parity transform ofso $\{Q_{\alpha },{\bar{Q}}_{\dot{\beta }}\}=2{\sigma }_{\alpha \dot{\beta }}^{\mu }{P}_{\mu }$ is consistent with the stated transformation law. The arbitrary intrinsic phase cancels.

With matrix-valued fields, charge conjugation acts asUnder parity, $A_0$ is even and $A_i$ is odd:Consequently $F_{0i}$ is odd and $F_{ij}$ is even. Under time-reversal symmetry, which is antiunitary, $A_0$ is even and $A_i$ is odd after $t\mapsto -t$; hence $F_{0i}$ is even and $F_{ij}$ is odd. Equivalently, the non-Abelian electric and magnetic fields transform as

A meson with zero spin angular momentum and negative parity. The pion and kaon are familiar examples. Between two pseudoscalar mesons, parity conservation in the strong interaction forces the axial current matrix element to vanish, while the vector current has pseudoscalar-to-pseudoscalar form factors. The weak decay as a whole need not conserve parity.

The symmetric double-delta potentialseparates into even and odd parity channels. At either interaction, the wavefunction is continuous and its derivative has the jump