# Help Verifying Schrodinger's equation in different potentials

Homework Statement:
I am asked to find the probability current density. The set up is one dimensional along the z-axis and it goes as follows: for z>0, V=0 which has the solution of the time independent SE. For 0<=z<d, V= Constant, which has the solution of the time dependent SE. In addition to V being nonzero, V> E, the energy of the electron. For z>d, then V=0 which is once again the time independent SE. I need verification on the equations I will be using. In addition I amy
Relevant Equations:
Schrodinger's Wave equation
There are 3 regions, to which I split the function as follows. I can derive the solutions myself. However I need to verify whether I am using them properly.

There are two principles/ideas that I am not sure if I am misinterpreting.
1) Anytime a wave is incident on a discontinuity(such as when a light ray enters a water, there is a transmitted wave and a reflected wave.
2) Tunneling is the phenomena when an electron can appear on the other side of a potential barrier and still have the same energy as before.

In the first region where z>0 and V=0, I have ##exp(-i \frac{E}{ \hbar }t) \left[ Aexp(ik_1z) + Bexp(-ik_1z) \right]##
where ##k_1^2= \frac{2mE}{\hbar ^2}## . E is the energy of the electron. The ##+i_k1z## term represents a wave traveling in the +z direction and the ##-k_1z## term represents the wave reflected at z=0.

In region 2, 0<=z<d and V=V. Here I have ##exp(-i \frac{E}{ \hbar }t) \left[ Cexp(ik_2^{'}z) + Dexp(ik_2^{'}z) \right]## where ##(k_2^{'})^2 = \frac{2mE}{ \hbar ^2}(E-V)## . Let ##k_2=\frac{2m}{ \hbar^2}(V−E)## and thus ##(k_2^{'})^2 = -k_2^2##. Which leads to ##k_2^{'}=ik_2##.
So in effect this is the actual wave equation in Region 2, ##exp(-i \frac{E}{ \hbar }t) \left[ Cexp(-k_2z) + Dexp(k_2z) \right]##.
So here I am saying that there is a wave moving in the +z direction, and because of the discontinuity at z=d, there is a reflected wave.

Finally in region 3 I have ##exp(-i \frac{E}{ \hbar }t)Fexp(ik_3z)##.
Since there are no more discontinuities in front of region 3, there is no reflected wave. In addition I am saying that because the electron has the same energy as before crossing the potential barrier, then k3=k_i and thus I can simply find that. So I end up with ##exp(-i \frac{E}{ \hbar }t)Fexp(ik_1z)##.

I am unsure about how to find the probability current of the transmitted current. My issue here is that I end up with the constant C. I am unsure about how to normalize the wavefunctions here.

Last edited:

## Answers and Replies

kuruman
Science Advisor
Homework Helper
Gold Member
Please use LaTeX to write your equations. They are very hard to read. If you don't know how to use LaTeX, click on the LaTeX Guide link lower left, above "Attach files".

Please use LaTeX to write your equations. They are very hard to read. If you don't know how to use LaTeX, click on the LaTeX Guide link lower left, above "Attach files".
Sorry about that. I have made those edits.

kuruman
Science Advisor
Homework Helper
Gold Member
Thank you for the edits.

Do yourself a favor and drop the time dependence, i.e. solve the TISE. Then you will have
$$\psi(z) = \begin{cases} Ae^{ik_1z}+Be^{-ik_1z} & z \leq 0 \\ Ce^{k_2z}+De^{-k_2z} & 0 \leq z \leq d \\ Fe^{ik_1z} & z \geq d \end{cases}$$
Note that the term ##Ae^{ik_1z}## is the incident wave on the barrier. At the first boundary ##z=0##, the fraction of the wave that is reflected is what counts not the value of ##B## by itself. So set ##A=1##, apply the 4 boundary conditions and then use the expressions for the probability current.

So set ##A=1##, apply the 4 boundary conditions and then use the expressions for the probability current.
Thank you for the response. I have been trying to figure this out but I can't seem to get the answer the book gets. Here is my work for the Probability Current
In region 3,
##J_3 \propto \psi_3^{*} \frac{d \psi}{dz}-\psi_3 \frac{d \psi_3^{*}}{dz}##
## = Fexp(-ik_1z)(ik_1)Fexp(ik_1z)-Fexp(ik_1z)(-ik_1)Fexp(-ik_1z) = ik_1F^2--ik_1F^2 = 2ik_1F^2##

In region 1,
##J_1 \propto \psi_1^{*} \frac{d \psi_1}{dz}-\psi_1 \frac{d \psi_1^{*}}{dz}##
## = Aexp(-ik_1z)(ik_1)Aexp(ik_1z)-Aexp(ik_1z)(-ik_1)Aexp(-ik_1z) = ik_1A^2--ik_1A^2 = 2ik_1A^2##
Note here I use only the transmitted wave and not the Transmitted+Reflected Wave Function. Doing so would lead to,
##J_1^{'} = [Aexp(-ik_1z)+Bexp(ik_1z)][ik_1Aexp(ik_1z)+-ik_1Bexp(-ik_1z)] - [Aexp(ik_1z)+Bexp(-ik_1z)][-ik_1Aexp(-ik_1z)+ik_1Bexp(ik_1z)] = ik_1A^2exp(ik_1z-ik_1z)-ik_1ABexp(-2ik_1z)+ik_1ABexp(2ik_1z)-ik_1B^2exp(ik_1z-ik_1z)-[-ik_1A^2exp(ik_1z-ik_1z)+ik_1ABexp(2ik_1z)-ik_1ABexp(-2ik_1z)+ik_1B^2exp(ik_1z-ik_1z)] = ik_1A^2--ik_1A^2-ik_1ABexp(-2ik_1)+ik_1ABexp(-2ik_1)+ik_1ABexp(2ik_1z)-ik_1ABexp(2ik_1z)-ik_1B^2+-ik_1B^2 = 2ik_1(A^2-B^2)##

For continuity I have,
##\psi_1(0,t) = \psi_2(0,t)##
##Aexp(ik_1*0)+Bexp(-ik_1(0)) = Cexp(k_z*0)+Dexp(-k_2*0) = A+B = C+D##
##\frac{d \psi_1(0,t)}{dz} = \frac{d \psi_2(0,t)}{dz}##
##ik_1Aexp(ik_1*0)+-ik_1Bexp(-ik_1(0)) = k_2Cexp(k_z*0)+-k_2Dexp(-k_2*0) = ik_1A-ik_1B = k_2C+-k_2B##
##\psi_3(d,t) = \psi_2(d,t)##
##Fexp(ik_1d) = Cexp(k_2d)+Dexp(-k_2d)##
##\frac{d \psi_3(d,t)}{dz} = \frac{d \psi_2(d,t)}{dz}##
##ik_1Fexp(ik_1d) = k_2Cexp(k_2d)+-k_2Dexp(-k_2d) ##

I just can't see how the book arrives at their answer of
##\frac{J_3}{J_1} = [cosh^2(k_2d)+\frac{1}{4}(\frac{k_2}{k_1} - \frac{k_1}{k_2})^2sinh^2(k_2d)]^-1##
Looking at the ##\frac{J_3}{J_1}##, it seems like both regions should have continuity satisfied via exponential's but only Region 3 has these continuities. Setting ##A=1##, I still have to deal with B.