For one-dimensional potentials that are piecewise constant, solutions to the Schrödinger equation

$[p2 2m +V(x)] ψ(x) =−ℏ2 2m d2ψ dx2 +Vψ(x) =Eψ(x)$

can be expressed simply in terms of exponentials. In regions where the energy exceeds the potential, the component solution are traveling waves,

$ψ(x) ≈eikx, e−ikx k2 =2m ℏ2 (E-V)$

and in regions where the potential exceeds the energy the wave function tunnels:

$ψ(x) ≈eκx, e−κx κ2 =2m ℏ2 (V-E)$

This presentation will evaluate the solutions to the Schrödinger equation for a variety of configurations of potentials, along with visualizations of the resulting wave functions.

Consider first a potential step:

$V(x) ={ 0 ,x<0 V0 ,x>0$

For $E>{V}_{0}$ and an incoming wave from the left, the wave functions are of the form

$ψ(x) ={ eikx +c1e −ikx ,x<0 c2e ik0x ,x>0$

where

$k2=2mE ℏ2 k02 =2m ℏ2 (E-V0)$

The wave function on the left includes a term representing reflection at the step. The wave function on the right represents transmission past it to the right. The wave functions and their derivatives must be matched at the discontinuity of the step:

$1+c1 =c2 k-k c1 =k0 c2 → c1 =k-k0 k+k0 c2=2k k+k0$

For simplicity here and in the sequel, set $\hslash =m=1$ . With the real part in blue and imaginary in red, the behavior of these wave functions with respect to energy is

Since the wave number is lower over the step, the wave length of the function is greater: this can be seen more clearly by lowering the energy value. As the energy increases the difference of wave numbers before and over the step decreases, and the wave functions on either side of the step become more equal. The step thus has less and less effect on the incoming wave.

This latter effect is made mathematically more precise by defining reflection and transmission probabilities relative to the incoming unit wave:

$R=|c1 |2 =(k-k0 )2 (k+k0 )2 T=1-R =4kk0 (k+k0 )2$

These expressions arise formally from reflected and transmitted fluxes, which are derived from an expression for probability current. The simple evaluation of the transmission coefficient means that the incoming wave must be reflected or transmitted with total probability of one for statistical certainty. With transmission in blue and reflection in magenta, these coefficients look like

and transmission becomes the more likely outcome by far for high energies.

When $E={V}_{0}$ then ${k}_{0}=0$ and the transmitted wave is constant. This does not present an issue because for this special case one also has

$R≡1 T≡0 , E=V0$

and the transmitted wave provides no transmission. The apparent contradiction has to do with the fact that these wave functions are nonnormalizable, and hence overall constants are not physically meaningful.

For $E<{V}_{0}$ and again an incoming wave from the left, the wave functions are now of the form

$ψ(x) ={ eikx +c1e −ikx ,x<0 c2e −κ0x ,x>0$

where

$k2=2mE ℏ2 κ02 =2m ℏ2 (V0-E)$

Matching wave functions and their derivatives at the step now gives

$1+c1 =c2 k-k c1 =iκ0 c2 → c1 =k-iκ0 k+iκ0 c2=2k k+ik0$

and the behavior of the wave functions is now

Even though there is now penetration into the step, there is no transmission because

$R=|c1 |2 ≡1 T=1-R ≡0$

The special case $E={V}_{0}$ has already been discussed. The other special case $E=0$ gives $\mathrm{k=0}$ and both wave functions are identically zero, which just means there is no incoming wave.

Next consider a potential barrier of finite width and height:

$V(x) ={ 0 , x<−a V0 , −aa$

For $E>{V}_{0}$ and an incoming wave from the left, the wave functions are of the form

$ψ(x) ={ eikx +c1e −ikx ,x<−a c2e ik0x +c3e −ik0x , −aa$

where again

$k2=2mE ℏ2 k02 =2m ℏ2 (E-V0)$

The wave functions and their derivatives must be matched at both discontinuities of the barrier:

Combining pairwise to eliminate ${c}_{1}$ and ${c}_{4}$ gives

so that then

$c1=2 (k2 -k02) e−2ika sinh2ik0a (k-k0 )2 e2i k0a -(k+k0 )2 e−2i k0a c4=−4k k0e −2ika (k-k0 )2 e2i k0a -(k+k0 )2 e−2i k0a$

The behavior of these wave functions with respect to energy is

The common denominator of the coefficients can be rewritten a bit more simply as

$(k-k0 )2 e2i k0a -(k+k0 )2 e−2i k0a =−4kk0 cos2k0a +2i(k2 +k02) sin2k0a$

whose absolute square simplifies nicely. The reflection and transmission coefficients are then

$R=|c1 |2 =(k2 -k02 )2 sin2 2k0a 4k2 k02 +(k2 -k02 )2 sin2 2k0a T=1-R =4k2 k02 4k2 k02 +(k2 -k02 )2 sin2 2k0a$

With transmission in blue and reflection in magenta, these coefficients look like

The interesting structure here is because reflection is identically zero for $2{k}_{0}a=0$ , where n is an integer. This enhancement of transmission is due to a resonance for such values.

When $E={V}_{0}$ then again ${k}_{0}=0$ , but the reflection and transmission coefficients remain finite:

$R=k2 a2 k2a2 +1 T=1k2 a2+1 , E=V0$

Oddly enough, although ${c}_{1}$ and ${c}_{4}$ remain finite as $E\to {V}_{0}$ , ${c}_{2}$ and ${c}_{3}$ do not...

For $E<{V}_{0}$ and again an incoming wave from the left, the wave functions are of the form

$ψ(x) ={ eikx +c1e −ikx ,x<−a c2e κ0x +c3e −κ0x , −aa$

where

$k2=2mE ℏ2 κ02 =2m ℏ2 (V0-E)$

The wave function inside the barrier represents tunneling. Matching wave functions and derivatives at both discontinuities now gives

Again combining pairwise to eliminate ${c}_{1}$ and ${c}_{4}$ gives

so that then

$c1=2 (k2 +κ02) e−2ika sinh2κ0a (k+iκ0 )2e 2κ0a -(k-iκ0 )2 e−2 κ0a c4=4ik κ0e −2ika (k+iκ0 )2e 2κ0a -(k-iκ0 )2 e−2 κ0a$

These coefficients are identical to the previous evaluation under the substitution ${k}_{0}\to -i{\kappa }_{0}$ . This explicit evaluation acts as a double check.

The behavior of these wave functions with respect to energy is

The simplification of the common denominator of the coefficients is now

$(k+iκ0 )2e 2κ0a -(k-iκ0 )2 e−2 κ0a =4ikκ0 cosh2κ0a +2(k2 -κ02) sinh2κ0a$

and the reflection and transmission coefficients are

$R=|c1 |2 =(k2 +κ02 )2 sinh2 2κ0a 4k2 κ02 +(k2 +κ02 )2 sinh2 2κ0a T=1-R =4k2 κ02 4k2 κ02 +(k2 +κ02 )2 sinh2 2k0a$

which are again identical to the previous evaluation under the substitution ${k}_{0}\to -i{\kappa }_{0}$ . With transmission in blue and reflection in magenta, these coefficients look like

When $E={V}_{0}$ these coefficients remain finite as before, so one could plot the two regimes on the same graph continuously. When $E=0$ there is no transmission and the finite barrier resembles the infinite step.

And again ${c}_{1}$ and ${c}_{4}$ remain finite as $E\to {V}_{0}$ , while ${c}_{2}$ and ${c}_{3}$ do not...

Next consider a potential well of finite width and depth:

$V(x) ={ 0 , x<−a −V0 , −aa$

For $E>0$ and an incoming wave from the left, the wave functions are of the form

$ψ(x) ={ eikx +c1e −ikx ,x<−a c2e ik0x +c3e −ik0x , −aa$

where now

$k2=2mE ℏ2 k02 =2m ℏ2 (E+V0)$

The only change from the potential barrier for $E>{V}_{0}$ is the sign change in the wave number inside the well. That means all of the formulae for that case can be immediately applied with the modified wave number.

The behavior of these wave functions with respect to energy is

Again with transmission in blue and reflection in magenta, these coefficients look like

Reflection is again identically zero for $2{k}_{0}a=0$ , where n is an integer, but the transmission resonance is less pronounced.

For both the potential step and barrier, one could attempt to evaluate wave functions for $E<0$ , which would necessarily be composed of appropriate real exponentials in each region and no incoming wave. The explicit calculation leads to coefficients identically equal to zero, so no immediate solutions are available for those potentials in that case. That is not true of the potential well as long as $-{V}_{0} . The wave functions are then of the form

$ψ(x) ={ c1e κx ,x<−a c2e ik0x +c3e −ik0x , −aa$

where now

$κ2=2m (−E) ℏ2 k02 =2m ℏ2 (E+V0)$

Matching wave functions and derivatives at both discontinuities of the well gives

Attempting to solve for coefficients as before leads to identically zero values. One can however divide the equations pairwise for the expressions

$κ=ik0 c2e −ik0a -c3e ik0a c2e −ik0a +c3e ik0a =ik0 −c2e ik0a +c3e −ik0a c2e ik0a +c3e −ik0a$

Setting ${c}_{3}={c}_{2}$ followed by ${c}_{3}=-{c}_{2}$ leads to the two equations

$κ=k0 tank0a κ=−k0 cotk0a$

which have real positive solutions and act as quantization conditions on the energy of the state. The corresponding equations for real exponentials contain negatives of hyperbolic functions and do not have real positive solutions, so that bound states are not possible in such cases. The negative sign here on the cotangent is not an issue because it inverts the function to resemble the tangent.

More visually, set $z={k}_{0}a$ and plot the trigonometric functions against the combination

$κk0 =1z 2ma2 V0ℏ2 -z2$

For ${c}_{3}={c}_{2}$ the states have even parity, and the eigenvalues are the intersections of the curves in the vicinity of odd multiples of $\frac{\pi }{2}$ , where the tangent has asymptotes:

For ${c}_{3}=-{c}_{2}$ the states have odd parity, and the eigenvalues are the intersections of the curves in the vicinity of multiples of $\pi$ , where the cotangent has asymptotes:

In both cases the number of eigenvalues is determined by how many zeros of the appropriate trigonometric function are less than the quantity

$zmax =2ma2 V0ℏ2$

which is where the falling curve crosses the horizontal axis. For the tangent the zeros are at multiples of $\pi$ and for cotangent at multiples of $\frac{\pi }{2}$ .

These bound states differ from previous evaluations in that they are now normalizable, unlike plane waves. When ${c}_{3}={c}_{2}$ then ${c}_{4}={c}_{1}$ , and when ${c}_{3}=-{c}_{2}$ then ${c}_{4}=-{c}_{1}$ . That leaves one constant undetermined, which can be assigned arbitrarily as the normalization of these states.

With these comments in mind, the behavior of the even parity states with respect to potential depth is

and the behavior of the odd parity states with respect to potential depth is

In both cases the wave functions are shifted downward by their energies for illustrative purposes.

As the well gets deeper there will be more bound states possible. This could be shown by adjusting the parameters of the interactive graphics and adding additional cases for efficient numerical evaluation of roots.

As the depth of the well becomes infinite, the left-hand sides of the eigenvalue conditions also become infinite. For solutions to continue to exist, the right-hand sides must be evaluated at the asymptotes of the trigonometric functions. The eigenvalues then occur at odd multiples of $\frac{\pi }{2}$ for the tangent and multiples of $\pi$ for the cotangent. Wave functions for this infinite well do not have contributions outside the well, since for finite z one has $\kappa \to \infty$ and the real exponentials vanish outside the well.

Possible future additions include double potential wells with and without a barrier between them. The method is straightforward, merely requiring significantly more algebra...