Past exam of the quantum theory course of the University of Oxford / 2015 / 1

The time-independent Schrödinger equation is the energy eigenvalue equation for the Hamiltonian operator:Here $\hslash$ is the reduced Planck constant. A corresponding stationary wavefunction evolves as $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$, which follows by substitution in the Time-dependent Schrödinger equation. Its position probability density is time independent because the time factor has unit modulus. Physical energy eigenstates must also satisfy the system's boundary conditions; for a bound state, wavefunction normalization requires ${\int }_{-\infty }^{\infty }|\psi (x){|}^{2}dx=1$.

Define the reflected function $\varphi (x)=\psi (-x)$. The chain rule gives ${\varphi }^{\prime }(x)=-{\psi }^{\prime }(-x)$ and ${\varphi }^{\prime \prime }(x)={\psi }^{\prime \prime }(-x)$. Evaluating the time-independent Schrödinger equation for $\psi$ at $-x$ and using the even function identity $V(-x)=V(x)$ givesThus reflection preserves the energy eigenvalue. The same substitution $y=-x$ in the norm givesThis is the action of the parity operator $R\psi (x)=\psi (-x)$, which is a unitary operator. For a reflection-symmetric physical problem it preserves the Hamiltonian operator's operator domain as well as its differential expression.

By linearity of the time-independent Schrödinger equation, the even and odd partseach satisfy the same energy eigenvalue equation whenever they are nonzero. They obey $R{\psi }_{+}={\psi }_{+}$ and $R{\psi }_{-}=-{\psi }_{-}$, so they have definite parity. At least one is nonzero, and $\psi ={\psi }_{+}+{\psi }_{-}$. Each nonzero part can be normalized separately.

Therefore a basis of stationary solutions can be chosen to have definite parity, and solving the even and odd sectors loses no energy eigenvalues. This does not assert that every possible quantum superposition has definite parity; a superposition of the two sectors need not. For one-dimensional bound states in the usual regular setting, nondegeneracy of one-dimensional bound states further implies that each individual energy eigenstate has definite parity.

In either exterior region of the finite square well, the time-independent Schrödinger equation reduces to ${\psi }^{\prime \prime }=k^2\psi$, since $E<0$. Its general solution is a linear combination of $e^{kx}$ and $e^{-kx}$. On the left, $e^{-kx}$ grows as $x\to -\infty$ and cannot be a square-integrable function, so its coefficient must be zero. On the right, the excluded growing term is $e^{kx}$. Hence the surviving tails are $A{e}^{kx}$ for $x\le -a$ and $D{e}^{-kx}$ for $x\ge a$.

Inside the well, $E+V_0>0$ and the equation becomes ${\psi }^{\prime \prime }=-{\kappa }^2\psi$, whose general solution is $B\cos \kappa x+C\sin \kappa x$. ThusThe coefficients are initially independent; wavefunction matching at a finite potential step imposes the remaining relations. Here the paper's $k$ is the exterior decay constant and $\kappa$ is the interior oscillation constant.

For an even function, the interior sine term must vanish, because sine is odd and cosine is even. Comparing reflected exterior tails at any $x>a$ gives $D{e}^{-kx}=A{e}^{-kx}$. Hence the parity condition alone isIt does not yet impose the matching condition on $B$. Continuity at $x=a$, when imposed in the next part, additionally gives $A=D=B{e}^{ka}\cos (\kappa a)$.

At a finite jump in potential energy, the wavefunction and its first derivative must be continuous. A jump in the derivative would create a Dirac delta function in ${\psi }^{\prime \prime }$, for which the finite potential has no balancing term. For the even solution, matching at $x=a$ givesA nonzero solution cannot have $B=0$. It also cannot have $\cos \beta =0$: the first condition would give $D=0$, while the second would force $B\kappa \sin \beta =0$, a contradiction. Division is therefore valid and yieldsParity makes the conditions at $x=-a$ identical to those already imposed. Finally the definitions of the decay and oscillation constants giveThe circle relation fixes the well strength; the tangent relation selects its even bound states.

Write $\rho =a\sqrt{2m{V}_0}/\hslash >0$. On the first tangent branch, $0<\beta <\pi /2$, the matching relations are equivalent toThe function $g(\beta )=\beta /\cos \beta$ is continuous and strictly increasing there:The intermediate value theorem therefore gives a unique first-branch root for every $\rho >0$. At this root $\beta <\rho$ and $\alpha =\beta \tan \beta >0$, so the energy eigenvalueobeys $-V_0<E<0$. This proves the even bound state in every attractive square well result.

To verify all the requested properties, take the resulting even wavefunction with exterior tails written in terms of their boundary values:The matching equation makes both the values and first derivatives agree at the endpoints. The exterior tails are square-integrable because $k>0$, and the interior interval is finite. More explicitly,so choosing $B$ as the positive inverse square root of the bracket gives wavefunction normalization. Thus an even, continuously differentiable, normalizable stationary solution always exists, however small the positive depth.