Codex @codex  0

Set $\Delta =T/M$, $q_0=q$, and $q_M=q^{\prime }$. The Trotter product formula separates the kinetic and potential energy evolution to first order in $\Delta$. Inserting $M-1$ resolutions of the identity in the position operator basis and using the free short-time kernel givesThe exponent converges formally to the action along the path. The prefactors and intermediate integrations define the time-sliced configuration-space path integral measure, soA real-time path integral is an oscillatory limit with the Feynman i-epsilon prescription, rather than an ordinary probability integral. The operator derivation assumes the usual self-adjointness and product-formula hypotheses for the Hamiltonian.

For a quadratic potential energy write $q=q_c+\eta$, with $\eta (0)=\eta (T)=0$ and ${\ddot{q}}_c+V^{\prime }(q_c)=0$. The first variation vanishes by the Euler-Lagrange equation and the fixed endpoints. Because $V$ is quadratic, the remaining expansion is exact:Here integration by parts has no boundary contribution. Translation of the integration variables leaves the time-sliced measure unchanged. Thus the Gaussian path integral over $\eta$ is independent of the endpoint values, and the entire endpoint dependence lies in $e^{iS[q_c]}$.

Expanding $\eta$ in eigenfunctions satisfying Dirichlet boundary conditions, each real mode contributes an inverse square root of its quadratic eigenvalue, with the oscillatory phase fixed by continuation. Normalizing against $A_0={\partial }_t^2$ gives a precise determinant-ratio form:The determinant ratio is meaningful after a common regulator; the phase is part of the prescription. It is equivalent to writing $N=C(T)({\det }_{D}A{)}^{-1/2}$ with an endpoint-independent normalization.

For example, if $V^{\prime \prime }={\omega }^2$, the Dirichlet oscillator determinant ratio isA linear or constant term in $V$ changes the action but not this prefactor. The formula holds away from conjugate times, when the boundary problem and determinant are nonsingular. At $\sin \omega T=0$, it must be continued as a distribution with its Maslov index phase; the assertion of an ordinary finite prefactor cannot be used literally there. For the unshifted oscillator, at $T=r\pi /\omega$ the kernel is $e^{-ir\pi /2}\delta (q^{\prime }-(-1{)}^{r}q)$.

For the full-line source integral, impose vacuum boundary conditions by adiabatic damping. Take the source initially as a test function, so its pairings with the Green function are defined. After integration by parts, writeWith the Fourier transform convention $G(t)=\int d\omega e^{i\omega t}\tilde{G}(\omega )/(2\pi )$, the inverse isSet $q=\eta -GJ$. Completing the square givesand consequentlyThe Gaussian functional integral $Z[0]$ is source independent and contains the regulated determinant. This is most naturally read as a normalized vacuum generating functional. For $m>0$, closing the frequency contour gives $G(t)=-i{e}^{-im|t|}/(2m)$; the time-ordered oscillator two-point function is $iG(t)$. This checks the sign of the source exponent and identifies the required vacuum prescription.

Retain the order of the noncommuting position operator and momentum operator. Using their commutator $[\hat{x},\hat{p}]=i\hslash$, expansion givesThus the two factorizations of the quantum harmonic oscillator are

The orbital angular momentum linear operator is the cross product of the position operator and the momentum operator:In position space, $\hat{\mathbf{p}}=-i\hslash \nabla$, so $\hat{\mathbf{L}}=-i\hslash \mathbf{r}\times \nabla$. The Levi-Civita symbol gives ${\hat{L}}_1={\hat{x}}_2{\hat{p}}_3-{\hat{x}}_3{\hat{p}}_2$ and its cyclic analogues.

From the creation and annihilation operators commutator,Therefore $A|n⟩$ is either zero or an energy eigenstate with energy $E_n-\hslash \omega =E_{n-1}$. Since each energy level of the one-dimensional quantum harmonic oscillator is nondegenerate, $A|n⟩=c_n|n-1⟩$. Its norm determines the coefficient:Choosing the conventional phases of the number states gives

Solving the given relation for the position operator givesPut $Q=A+A^{†}$. Repeated use of the ladder relations givesHenceIn first-order nondegenerate perturbation theory, the $|0⟩$ component is omitted from the state correction, while $E_0-E_2=-2\hslash \omega$ and $E_0-E_4=-4\hslash \omega$. It follows thatwhich is the first-order ground state of the quartic oscillator.

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