Scattering States, Transmission and Reflection

	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

We will now discuss scattering states in a one-dimensional potential. Scattering states are states that are not bound. Such states have energies larger than the potential at at least one of 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,} and their energy spectrum forms a continuous band, rather than a discrete set as the bound states do. Unlike the bound case, the wave function does not have to vanish at infinity, though a particle cannot reflect from infinity, often giving a useful boundary condition. At any discontinuous changes in the potentials, the wave function must still be continuous and differentiable as for the bound states.

We are interested in obtaining the wave functions for these scattering states in order to discuss transmission and reflection of waves from one-dimensional potentials, and to find the transmission and reflection coefficients 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 T\!} 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 R,\!} which give the probability that an incident wave will be transmitted and reflected, respectively.

The Step Potential

As a first example, let us consider a one-dimensional potential step. The potential in this case 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 V(x) = \begin{cases} 0, & x < 0, \\ V_0, & x > 0. \end{cases} }

The corresponding Schrödinger equation is

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 \left( - \frac{\hbar^2}{2m} \frac{d^2}{dx^2} + V(x) \right) \psi(x) = E \psi(x). }

Let us first consider states 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>V_0.\!} We will divide the one-dimensional space into two regions - region I, for which 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 region II, for which 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.\!} The Schrödinger equations for the two regions 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 - \frac{\hbar^2}{2m} \frac{d^2 \psi_{I}(x)}{dx^2} = E \psi_{I}(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 \left( - \frac{\hbar^2}{2m} \frac{d^2}{dx^2} + V_0 \right) \psi_{II}(x) = E \psi_{II}(x), }

and thus the corresponding wave functions 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) = A e^{i k_0x} + B e^{-ik_0x} }

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_{II}(x) = C e^{i k x} + D e^{-i k 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_0 = \sqrt{\frac{2mE}{\hbar^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 k = \sqrt{\frac{2m(E-V_0)}{\hbar^2}}.} We may consider the first term 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 \psi_{I}(x),\,Ae^{ik_0x},} to be an incident wave from the left, in which case the second term, 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_0x},} is a reflected wave from the potential barrier and the first term 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 \psi_{II}(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 Ce^{ikx},\!} is a transmitted wave. Similarly, we can think of the second term 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 \psi_{II}(x),\,De^{-ikx},} to be an incident wave from the right, in which case 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 Ce^{ikx}\!} is now the reflected wave 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_0x}\!} is the transmitted wave. These interpretations of the various terms will become more obvious shortly.

The boundary conditions 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\! } require

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(0) = \psi_{II}(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 \left. \frac{d\psi_I(x)}{dx} \right|_{x=0} = \left. \frac{d\psi_{II}(x)}{dx} \right|_{x=0}, }

which give 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 A + B = C + D \!}

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_0\left(A-B\right) = k \left(C-D\right). }

If we assume that the wave is incident from the left, then we can set 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 D=0\!} . In this case, reflection occurs at the potential step, and there is transmission to the right. We then have

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 \left(A-B\right) = \frac{k}{k_0} C. }

From the above equations, 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{B}{A} = \frac{k_0-k}{k_0+k}}

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{C}{A} = \frac{2k}{k_0+k}. }

To determine the probability of reflection and transmission, we must now find the current density,

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 j = \frac{\hbar}{2m i} \left (\psi^* \frac{d\psi}{dx} - \frac{d\psi^*}{dx} \psi \right ),}

on each side of the barrier. Doing so, 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 j = \begin{cases} \displaystyle \frac{\hbar k_0}{m} \left( \left|A\right|^2 - \left|B\right|^2 \right), & x < 0, \\ {} & {} \\ \displaystyle \frac{\hbar k}{m} \left|C\right|^2, & x > 0. \end{cases} }

We thus see more clearly that the terms that we earlier identified as incident, reflected, and transmitted waves are as we labeled them. We define the ratio of the reflected current density to the incident current density as the reflection coefficient, 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 R=\frac{\left |B\right |^2}{\left |A\right |^2},} and the ratio of the transmitted current density to the incident current density as the transmission coefficient, 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 T=\frac{k}{k_0}\frac{\left |C\right |^2}{\left |A\right |^2}.}

The continuity equation for the current density in one dimension implies conservation of current density, and thus 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 R+T=1.\!}

If we determine the reflection and transmission coefficients for the problem at hand, 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 R = \frac{\left|B\right|^2}{\left|A\right|^2} = \frac{\left( k_0 - k \right)^2}{\left(k_0 + k \right)^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 T = \frac{k}{k_0} \frac{\left|C\right|^2}{\left|A\right|^2} = \frac{4k_0k}{\left(k_0 + k \right)^2}. }

We see that these expressions satisfy 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 R + T = 1,\!} as expected.

Now let us consider the case in which 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 < E < V_0.\!} If we again assume that the wave is incident from the left, the wave functions become

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) = A e^{i k_0x} + B e^{-ik_0x} }

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_{II}(x) = C e^{-\kappa 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 \kappa=\frac{\sqrt{2m(V_0-E)}}{\hbar}.} Applying the same boundary conditions as before, 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{B}{A} = \frac{k_0-i\kappa}{k_0+i\kappa}}

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{C}{A} = \frac{2i\kappa}{k_0+i\kappa}. }

From the second wave function, 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},\!} we see that the transmitted wave decreases exponentially over a length scale 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 \frac{1}{\kappa}.} .

If we determine the current density in this case, we find 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 j = \begin{cases} \displaystyle \frac{\hbar k_0}{m} \left( \left|A\right|^2 - \left|B\right|^2 \right), & x < 0, \\ {} & {} \\ \displaystyle 0, & x > 0. \end{cases} }

We see that the wave is completely reflected; i.e., $R=1\!$ and $T=0.\!$

The reflected wave acquires a phase difference relative to the incident wave. To see this, we simply rewrite the wave function $\psi _{I},\!$ as

{\begin{aligned}\psi _{I}(x)&=e^{ik_{0}x}+{\frac {k_{0}^{2}-\kappa ^{2}-2i\kappa k_{0}}{k_{0}^{2}+\kappa ^{2}}}e^{-ik_{0}x}\\&=e^{ik_{0}x}+e^{i\theta }e^{-ik_{0}x},\end{aligned}}

where the phase difference $\theta \!$ of the reflected wave with respect to the incident wave is given by

e^{i\theta }={\frac {k_{0}^{2}-\kappa ^{2}-2i\kappa k_{0}}{k_{0}^{2}+\kappa ^{2}}},

or

\theta =\tan ^{-1}\left({\frac {2\kappa k_{0}}{\kappa ^{2}-k_{0}^{2}}}\right).

The Square Potential Barrier

For a square potential barrier, given by

V(x)={\begin{cases}0,&x<-a,\\V_{0},&-a<x<a,\\0,&x>a,\end{cases}}

we can write the general solution of the Schrödinger equation for $0<E<V_{0}:\!$

\psi (x)={\begin{cases}Ae^{ik_{0}x}+Be^{-ik_{0}x},&x<-a,\\Ce^{-\kappa x}+De^{\kappa x},&-a<x<a,\\Fe^{ik_{0}x}+Ge^{-ik_{0}x},&x>a,\end{cases}}

where $k_{0}={\frac {\sqrt {2mE}}{\hbar }}\!$ and $\kappa ={\frac {\sqrt {2m(V_{0}-E)}}{\hbar }}.\!$

The boundary conditions are the same as before; the boundary conditions at $x=-a,\!$ give us

Ae^{-ik_{0}a}+Be^{ik_{0}a}=Ce^{\kappa a}+De^{-\kappa a},

Ae^{-ik_{0}a}-Be^{ik_{0}a}={\frac {i\kappa }{k_{0}}}\left(Ce^{\kappa a}-De^{-\kappa a}\right).

Similarly, the boundary conditions at $x=a\!$ give us

Ce^{-\kappa a}+De^{\kappa a}=Fe^{ik_{0}a}+Ge^{-ik_{0}a},

Ce^{-\kappa a}-De^{\kappa a}=-{\frac {ik_{0}}{\kappa }}\left(Fe^{ik_{0}a}-Ge^{-ik_{0}a}\right).

For the convenience, let us express the coefficients of these linear homogeneous relations in terms of matrices:

{\begin{bmatrix}A\\B\end{bmatrix}}={\frac {1}{2}}{\begin{bmatrix}\left(1+{\frac {i\kappa }{k_{0}}}\right)e^{\kappa a+ik_{0}a}&&\left(1-{\frac {i\kappa }{k_{0}}}\right)e^{\kappa a-ik_{0}a}\\\left(1-{\frac {i\kappa }{k_{0}}}\right)e^{-\kappa a+ik_{0}a}&&\left(1+{\frac {i\kappa }{k_{0}}}\right)e^{-\kappa a-ik_{0}a}\end{bmatrix}}{\begin{bmatrix}C\\D\end{bmatrix}}

{\begin{bmatrix}C\\D\end{bmatrix}}={\frac {1}{2}}{\begin{bmatrix}\left(1-{\frac {ik_{0}}{\kappa }}\right)e^{\kappa a+ik_{0}a}&&\left(1+{\frac {ik_{0}}{\kappa }}\right)e^{-\kappa a+ik_{0}a}\\\left(1+{\frac {ik_{0}}{\kappa }}\right)e^{\kappa a-ik_{0}a}&&\left(1-{\frac {ik_{0}}{\kappa }}\right)e^{-\kappa a-ik_{0}a}\end{bmatrix}}{\begin{bmatrix}F\\G\end{bmatrix}}

If we combine these two equations, we have

{\begin{bmatrix}A\\B\end{bmatrix}}={\begin{bmatrix}\left(\cosh {2\kappa a}+{\frac {i\varepsilon }{2}}\sinh {2\kappa a}\right)e^{2ik_{0}a}&&-{\frac {i\eta }{2}}\sinh {2\kappa a}\\{\frac {i\eta }{2}}\sinh {2\kappa a}&&\left(\cosh {2\kappa a}-{\frac {i\varepsilon }{2}}\sinh {2\kappa a}\right)e^{-2ik_{0}a}\end{bmatrix}}{\begin{bmatrix}F\\G\end{bmatrix}}

where $\varepsilon ={\frac {\kappa }{k_{0}}}-{\frac {k_{0}}{\kappa }}\!$ and $\eta ={\frac {\kappa }{k_{0}}}+{\frac {k_{0}}{\kappa }}.\!$

Note that $\eta ^{2}-\varepsilon ^{2}=4.\!$

Let us now assume that there is only an incident wave coming from the left; i.e., $G=0.\!$ By similar arguments as before, we may identify the transmission coefficient as

$T={\frac {|F|^{2}}{|A|^{2}}}={\frac {4}{4\cosh ^{2}{2\kappa a}+\varepsilon ^{2}\sinh ^{2}{2\kappa a}}}$

and the reflection coefficient as

$R={\frac {|B|^{2}}{|A|^{2}}}={\frac {\eta ^{2}\sinh ^{2}{2\kappa a}}{4\cosh ^{2}{2\kappa a}+\varepsilon ^{2}\sinh ^{2}{2\kappa a}}}.$

Finite Asymmetric Square Well

We now consider an asymmetric square well potential, given by

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

In this case, the wave functions are given by

\psi (x)={\begin{cases}A_{1}e^{ik_{1}x}+A_{2}e^{-ik_{2}x},&x<-a,\\B_{1}e^{ik_{2}x}+B_{2}e^{-ik_{2}x},&-a<x<a,\\C_{1}e^{ik_{3}x}+C_{2}e^{-ik_{3}x},&x>a,\end{cases}}

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(E-V_{1})}}{\hbar},\,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(E-V_{2})}}{\hbar}.}

Applying the boundary conditions 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,\!} 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 A_{1}e^{-ik_{1}a}+A_{2}e^{ik_{1}a}=B_{1}e^{-ik_{2}a}+B_{2}e^{ik_{2}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 ik_{1}A_{1}e^{-ik_{1}a}-ik_{1}A_{2}e^{ik_{1}a}=ik_{2}B_{1}e^{-ik_{2}a}-ik_{2}B_{2}e^{ik_{2}a},}

while those 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\!} give 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 B_{1}e^{ik_{2}a}+B_{2}e^{-ik_{2}a}=C_{1}e^{ik_{3}a}+C_{2}e^{ik_{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 ik_{2}B_{1}e^{ik_{2}a}-ik_{2}B_{2}e^{-ik_{2}a}=ik_{3}C_{1}e^{ik_{3}a}+C_{2}e^{-ik_{3}a}.}

We may express these in matrix form 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 \begin{bmatrix} A_{1} \\ A_{2} \end{bmatrix} =\tfrac{1}{2} \begin{bmatrix} \left (1+\frac{k_2}{k_1}\right )e^{i(k_1-k_2)a} && \left (1-\frac{k_2}{k_1}\right )e^{i(k_1+k_2)a} \\ \left (1-\frac{k_2}{k_1}\right )e^{-i(k_1+k_2)a} && \left (1+\frac{k_2}{k_1}\right )e^{-i(k_1-k_2)a} \end{bmatrix} \begin{bmatrix} B_{1} \\ B_{2} \end{bmatrix} }

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 \begin{bmatrix} B_{1} \\ B_{2} \end{bmatrix} =\tfrac{1}{2} \begin{bmatrix} \left (1+\frac{k_3}{k_2}\right )e^{i(k_2-k_3)a} && \left (1-\frac{k_3}{k_2}\right )e^{i(k_2+k_3)a} \\ \left (1-\frac{k_3}{k_2}\right )e^{-i(k_2+k_3)a} && \left (1+\frac{k_3}{k_2}\right )e^{-i(k_2-k_3)a} \end{bmatrix} \begin{bmatrix} C_{1} \\ C_{2} \end{bmatrix}. }

If we combine these, 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 \begin{bmatrix} A_{1} \\ A_{2} \end{bmatrix} =\tfrac{1}{2} \begin{bmatrix} \left (1+\frac{k_3}{k_1}\right )e^{i(k_1-k_3)a} && \left (1-\frac{k_3}{k_1}\right )e^{i(k_1+k_3)a} \\ \left (1-\frac{k_3}{k_1}\right )e^{-i(k_1+k_3)a} && \left (1+\frac{k_3}{k_1}\right )e^{-i(k_1-k_3)a} \end{bmatrix} \begin{bmatrix} C_{1} \\ C_{2} \end{bmatrix}. }

Let us once again assume that there is only an incident wave from the left, 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 C_{2}=0.\!} Using the same arguments as before, we may identify the transmission coefficient 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 T=\frac{k_3|C_1|^2}{k_1|A_1|^2}=\frac{4k_3}{k_1}\frac{k_1^2}{(k_1+k_3)^2}=\frac{4k_1k_3}{(k_1+k_3)^2}}

and the reflection coefficient 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 R=\frac{|A_2|^2}{|A_1|^2}=\frac{(k_1-k_3)^2}{(k_1+k_3)^2}.}

The Dirac Delta Function Potential

We now consider scattering from a Dirac delta function 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)=V_0\delta(x).\!} 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\neq 0,\!} the Schrödinger equation is just that for a free particle,

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(x)}{dx^2}=-\frac{2mE}{\hbar^2}\psi(x) = -k^{2}\psi,}

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

The general solution 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<0\!} is

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_{1}(x)=Ae^{ikx}+Be^{-ikx}, \!}

while that 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>0 \!} is

\psi _{2}(x)=Fe^{ikx}+Ge^{-ikx}.\!

As in the other cases, the wave function must be continuous at $x=0,\!$ so

F+G=A+B.\!

The derivative of the wave function, however, is not continuous, as noted when we studied the bound states. The discontinuity of the derivative is given by

\left[{\frac {d\psi _{2}(0)}{dx}}-{\frac {d\psi _{1}(0)}{dx}}\right]-{\frac {2mV_{0}}{\hbar ^{2}}}\psi (0)=0,

which yields the relation,

ik(F-G-A+B)=-{\frac {2mV_{0}}{\hbar ^{2}}}(A+B),

or

F-G=A(1+2i\beta )-B(1-2i\beta ),\!

where

\beta ={\frac {mV_{0}}{\hbar ^{2}k}}.\!

We once again assume that the incoming particles are coming from the left, so that $G=0.\!$ Consideration of the current densities on each side of the potential tells us that the reflection coefficient $R={\frac {|B|^{2}}{|A|^{2}}}$ and that the transmission coefficient is $T={\frac {|F|^{2}}{|A|^{2}}}.$ We thus wish to solve for $B\!$ and $F\!$ in terms of $A;\!$ doing so, we obtain