One-Dimensional Bound States

	Quantum Mechanics A
	Schrödinger Equation; The most fundamental equation of quantum mechanics; given a Hamiltonian Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{H}} , it describes how a state Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \|\Psi\rangle} evolves in time.
	Physical Basis of Quantum Mechanics
	Basic Concepts and Theory of Motion; UV Catastrophe (Black-Body Radiation); Photoelectric Effect; Stability of Matter; Double Slit Experiment; Stern-Gerlach Experiment; The Principle of Complementarity; The Correspondence Principle; The Philosophy of Quantum Theory
	Schrödinger Equation
	Brief Derivation of Schrödinger Equation; Relation Between the Wave Function and Probability Density; Stationary States; Heisenberg Uncertainty Principle; Some Consequences of the Uncertainty Principle
	Operators, Eigenfunctions, and Symmetry
	Linear Vector Spaces and Operators; Commutation Relations and Simultaneous Eigenvalues; The Schrödinger Equation in Dirac Notation; Transformations of Operators and Symmetry; Time Evolution of Expectation Values and Ehrenfest's Theorem
	Motion in One Dimension
	One-Dimensional Bound States; Oscillation Theorem; The Dirac Delta Function Potential; Scattering States, Transmission and Reflection; Motion in a Periodic Potential; Summary of One-Dimensional Systems
	Discrete Eigenvalues and Bound States
	Harmonic Oscillator Spectrum and Eigenstates; Analytical Method for Solving the Simple Harmonic Oscillator; Coherent States; Charged Particles in an Electromagnetic Field; WKB Approximation
	Time Evolution and the Pictures of Quantum Mechanics
	The Heisenberg Picture: Equations of Motion for Operators; The Interaction Picture; The Virial Theorem
	Angular Momentum
	Commutation Relations; Angular Momentum as a Generator of Rotations in 3D; Spherical Coordinates; Eigenvalue Quantization; Orbital Angular Momentum Eigenfunctions
	Central Forces
	General Formalism; Free Particle in Spherical Coordinates; Spherical Well; Isotropic Harmonic Oscillator; Hydrogen Atom; WKB in Spherical Coordinates
	The Path Integral Formulation of Quantum Mechanics
	Feynman Path Integrals; The Free-Particle Propagator; Propagator for the Harmonic Oscillator
	Continuous Eigenvalues and Collision Theory
	Differential Cross Section and the Green's Function Formulation of Scattering; Central Potential Scattering and Phase Shifts; Coulomb Potential Scattering

When the energy of a particle moving in one dimension is less than the potential at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x\to\pm\infty} , the particle is trapped in a potential well and is said to be in a bound state. However, when the energy is larger than the potential at either infinity, the particle is said to be in a scattering state. The wave function for a bound state must decay at least exponentially at both infinities.

Basic Properties

1. The bound states of a one-dimensional potential are non-degenerate.

To prove this, let us suppose that there are two different wave functions, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_E^{(1)}} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_E^{(2)},} for a given bound state with energy Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E\!} . They must then both satisfy the same Schrödinger equation:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\frac{\hbar^2}{2m}\frac{d^2\psi_E^{(1)}}{dx^2}+V(x)\psi_E^{(1)}=E\psi_E^{(1)}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\frac{\hbar^2}{2m}\frac{d^2\psi_E^{(2)}}{dx^2}+V(x)\psi_E^{(2)}=E\psi_E^{(2)}}

Multiplying the first equation by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_E^{(2)}} and the second by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_E^{(1)}} , then subtracting one equation from the other, we get

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d^2\psi_E^{(1)}}{dx^2}\psi_E^{(2)}-\frac{d^2\psi_E^{(2)}}{dx^2}\psi_E^{(1)}=0,}

or

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \rightarrow \frac{d}{dx}\left (\frac{d\psi_E^{(1)}}{dx}\psi_E^{(2)}-\frac{d\psi_E^{(2)}}{dx}\psi_E^{(1)}\right )=0.}

We recognize the quatity Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle W(x)=\frac{d\psi_E^{(1)}}{dx}\psi_E^{(2)}-\frac{d\psi_E^{(2)}}{dx}\psi_E^{(1)}} as the Wronskian of the two wave functions. We therefore see that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle W(x)=C,} where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle C} is constant. For bound states, the wave functions must vanish at infinity, and therefore Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle W=0,} or

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_E^{(1)}\frac{d\psi_E^{(2)}}{dx}-\frac{d\psi_E^{(1)}}{dx}\psi_E^{(2)}=0.}

Rearranging, we obtain

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{\psi_E^{(1)}}\frac{d\psi_E^{(1)}}{dx}-\frac{1}{\psi_E^{(2)}}\frac{d\psi_E^{(2)}}{dx}=0.}

We can now integrate this equation to obtain

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ln(\psi_E^{(1)})-\ln(\psi_E^{(2)})=A,}

or

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_E^{(1)}(x)=a\psi_E^{(2)}(x).}

We see that the two wave functions must be proportional to each other; after normalization, they will become completely identical. Therefore, we see that the two wave functions correspond to the same state, and thus all bound states in one dimension are non-degenerate.

2. The wave function for a real potential Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V(x)} can be chosen to be real.

If we take the complex conjugate of the Schrödinger equation for the system,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E\psi=-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}+V(x)\psi,}

then we obtain Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E\psi^{\ast}=-\frac{\hbar^2}{2m}\frac{d^2\psi^{\ast}}{dx^2}+V(x)\psi^{\ast}.}

Since both Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi\!} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi^{\ast}} satisfy the same Schrödinger equation, their sum Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi+\psi^{\ast}} must also be a solution. This sum is real, and thus we see that the wave function, as asserted, can be chosen to be real.

3. The Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} th excited state has Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} nodes.

See the Oscillation Theorem.

Parity and the Symmetry of the Wave Functions

One useful theorem about one-dimensional potentials is that, if the potential Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V(x)\!} is even (i.e. Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V(x)=V(-x)\!} ), then the eigenstates can be taken to be even or odd.

To prove this, we first note that, if Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi(x)} satisfies the Schrödinger equation for the potential,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}+V(x)\psi=E\psi,}

then so will Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi(-x)\!} . We may see this by simply taking Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x\to -x\!} in the above equation. Since both Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi(x)\!} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi(-x)\!} are solutions of the Schrödinger equation, any linear combination of the two must also be a solution. In particular, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_e(x)=\psi(x)+\psi(-x)\!} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_o(x)=\psi(x)-\psi(-x)\!} are solutions. We may easily see that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_e(x)\!} is even, while Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_o(x)\!} is odd.

In particular, this implies that all bound states in such a potential must be either even or odd. Since Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi(x)\!} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi(-x)\!} both solve the Schrödinger equation, and because the eigenstates must be non-degenerate, we see that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi(x)\!} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi(-x)\!} must be proportional to one another. We may easily show that the proportionality constant must be Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \pm 1\!} , corresponding to either an even function or an odd function.

The above discussion is a simple illustration of how the symmetry of the Hamiltonian of a system dictates the transformation properties of the eigenstates, as described earlier.

Infinite Square Well

Let's consider the motion of a particle in an infinite and symmetric square well: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V(x)=+\infty} for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |x| \ge L/2,} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V(x)=0\!} otherwise.

A particle subject to this potential is free everywhere except at the two ends Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x = \pm L/2),} where the infinite potential keeps the particle confined to the well. Within the well the Schrödinger equation takes the form,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}=E\psi,}

or equivalently,

{\frac {d^{2}\psi }{dx^{2}}}=-k^{2}\psi ,

where

k={\frac {\sqrt {2mE}}{\hbar }}.

Writing the Schrödinger equation in this form, we see that our solution are simply

$\psi (x)=A\sin(kx)+B\cos(kx).\!$

As we showed earlier, the eigenstates of this potential, all of which are bound states, must be either even or odd. This means that we must either set $A=0$ or $B=0.$ Let us consider each case in turn.

Case 1: $B=0\!$ .

In this case,

\psi (x)=A\sin {kx}.\!

We require that the wave function vanish at the ends, so that

\sin \left({\frac {kL}{2}}\right)=0.\!

The solutions of this equation are

k={\frac {2n\pi }{L}},\!

where $n=1,2,3,\ldots \!$

Case 2: $A=0.\!$

In this case,

\psi (x)=B\cos {kx}.\!

Again requiring that the wave function vanish at the ends, i.e.

\cos \left({\frac {kL}{2}}\right)=0,\!

we find that

k={\frac {(2n+1)\pi }{L}}.\!

Combining these two results, we find that the energy eigenvalues are $E_{n}={\frac {\hbar ^{2}}{2m}}\left({\frac {\pi n}{L}}\right)^{2}\!$ , where $n=1,2,3,\ldots \!$ The wave numbers are quantized as a result of the boundary conditions, and thus the energies are quantized as well. The lowest energy state is the ground state, and it has a non-zero energy, which is due to quantum zero point motion. The ground state is nodeless, the first excited state has one node, the second excited state has two nodes, and so on. The wavefunctions are also orthogonal.

Finite Asymmetric Square Well

We now consider an asymmetric square well potential $V(x),$ defined as

$V(x)={\begin{cases}V_{1},\,x<0\\0,\,0<x<a\\V_{2},\,x>a\end{cases}},$

where both $V_{1}\!$ and $V_{2}\!$ are positive. We may then split our one-dimensional space into three regions; region I is given by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x<0\!} , region II by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0<x<a\!} , and region III by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x>a\!} . The Schrödinger equation may be written in each region as

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\frac{\hbar^2}{2m}\frac{d^2\psi_{I}(x)}{dx^2}+V_1\psi_{I}(x)=E\psi_{I}(x),}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\frac{\hbar^2}{2m}\frac{d^2\psi_{II}(x)}{dx^2}=E\psi_{II}(x),}

and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\frac{\hbar^2}{2m}\frac{d^2\psi_{III}(x)}{dx^2}+V_2\psi_{III}(x)=E\psi_{III}(x).}

We will be considering bound states here, so that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E<\min(V_1,V_2).\!} In this case, the solutions to the above Schrödinger equations are

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_{I}(x)=Ae^{k_{1}x},}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_{II}(x)=Be^{ik_{2}x}+Ce^{-ik_{2}x},}

and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_{III}(x)=De^{-k_{3}x},}

where, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k_{1}=\frac{\sqrt{2m(V_{1}-E)}}{\hbar},} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k_{2}=\frac{\sqrt{2mE}}{\hbar},} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k_{3}=\frac{\sqrt{2m(V_{2}-E)}}{\hbar}.}

The wave function and it's first derivative must both be continuous at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=0\!} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=a.\!} The condition at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=0\!} yields

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A=B+C\!}

and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k_{1}A=ik_{2}(B-C),\!}

while that at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=a\!} gives us

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Be^{ik_{2}a}+Ce^{-ik_{2}a}=De^{-k_{3}a}}

and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Be^{ik_{2}a}-Ce^{-ik_{2}a}=\frac{ik_{3}}{k_{2}}De^{-k_{3}a}.}

If we reduce the above set of equations to equations only involving Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B\!} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle C\!} , we obtain

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (k_1-ik_2)B+(k_1+ik_2)C=0\!}

and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (k_3+ik_2)Be^{ik_2a}+(k_3-ik_2)Ce^{-ik_2a}=0.}

In order for this system of homogeneous equations to possess a non-trivial solution in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B\!} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle C,\!} the following condition must be met:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (k_1-ik_2)(k_3-ik_2)e^{-ik_2a}-(k_1+ik_2)(k_3+ik_2)e^{ik_2a}=0,}

or

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tan{k_{2}a}=\frac{(k_1+k_3)k_2}{k_2^2-k_1k_3}.}

If we now substitute in our expressions for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k_1,\,k_2,\!} and $k_{3},\!$ we get

$\tan \left({\frac {\sqrt {2mE}}{\hbar }}a\right)={\frac {{\sqrt {E}}({\sqrt {V_{1}-E}}+{\sqrt {V_{2}-E}})}{E-{\sqrt {(V_{1}-E)(V_{2}-E)}}}}.$

The solutions to this equation then give us the bound-state energy spectrum of the system.

Problems

(1) An electron is moving freely inside of a one-dimensional box with walls at $x=0\!$ and $x=a.\!$ If the electron is initially in the ground state of the box and we suddenly increase the size of the box by moving the right-hand wall instantaneously from $x=a\!$ to $x=4a,\!$ then calculate the probability of finding the electron in