Solution of Schrodinger equation for a potential step

Solution of Schrodinger equation for a potential step


Schrodinger equation has an great importance to solve many of different cases. In this particular article, we are going to discuss Solution of Schrodinger equation for a Potential step. Also about the different cases and important formulas.

What is the potential step?

When the one dimensional potential function V(x) is of the form V(x) =0 for x<0 and V(x) =V₀ ,then it is called Potential step . Here V₀ is called the positive constant. At x=0, the right-hand limit of V(x) is V₀ and left-hand limit of V(x) is zero.

The solution of the Schrodinger equation for a Potential step (case (a) E> v₀)

A particle of mass m is moving in a region where there is a potential step at x=0, that is,

V(x) =0 for x<0

V(x) = V₀ for x>0.           (1)

Where V₀ is a positive constant. The potential (energy) doesn’t depend on time and the total energy of the particle E remains constant during motion. Therefore the complete wave function for the particle is separated into space dependent and time-dependent parts :

ψ(x,t) = ψ(x) exp (-iEt/h)            (2)

Where ψ(x) satisfied the following time-independent Schrodinger equation

-h²/2m (d²ψ(x)/dx²) + V(x) ψ(x) = Eψ(x)         (3)

We solve this equation for E< V₀

Solutions of the Schrodinger equation

(wave function in the region x<0 and x>0)

Solution of Schrodinger equation for a Potential step

Schrodinger equation in region first,

d²ψ/dx² + 2m(E-V)ψ₁/h² = 0

For region first , V=0

d²ψ/dx² + 2m(E)ψ₁/h² = 0

Put 2mEψ₁/h² = k²₁

Hence above equation can be written as

d²ψ₁/dx² + k₁²ψ₁ = 0              (1)

This is second order differential equation and solution of this equation is given by-

ψ₁ = Aexp (ik₁x) + Bexp (-ik₁x)          (2)

Now similarly Schrodinger equation for region second

d²ψ₂/dx² + 2m(E-V₀)ψ₂/h² = 0

∴ for region second V=V₀

Now put 2m(E-V₀)ψ₂/h² =k₂²

Now using this , above equation can be written as

d²ψ₂/dx² +ψ₂k₂² = 0       (3)

This is again a second order differential equation and solution of this equation is given by-

ψ₂ = Cexp(ik₂x) + D exp(-ik₂x)         (4)


Aexp (ik₁x) = represents the incident wave .

Bexp (-ik₁x) = represents reflected wave.

Cexp(ik₂x) = represents transmitted wave

D exp(-ik₂x) = there is no reflected wave in region second .,so amplitude will be zero .i.e. D =0

Hence equation (4) becomes,

ψ₂ = Cexp(ik₂x)     (5)

Now , Boundary conditions are-

  1. At x=0 , ψ₁ = ψ₂          (6)
  2. At x=0 (dψ₁ /dx) = (dψ₂/dx)         (7)

Applying boundary conditions-

Aexp (ik₁x) + Bexp (-ik₁x)  = Cexp(ik₂x)

At x=0 , A+B =C            (8)

And differentiating at x=0

k₁A -k₁B = k₂C               (9)

Now put the value of C from equation (8) into equation (9)

k₁A -k₁B = k₂(A+B)

B = (k₁-k₂ / k₁+ k₂ ) A       (10)

Where B is the amplitude of reflected wave.

Now, put this value in equation (8) we get,

C = (2k₁ / k₁ + k₂ )A           (11)

Using these values, wave function in the two regions are

ψ₁(x) = A exp(ikx) + (k₁-k₂ / k₁+ k₂ ) A exp(-ikx)       (12)

ψ₂ ( x) = (2k₁ / k₁ + k₂ )Aexp (ik₂x)                                    (13)

Probability current densities

A particle is moving from region first through the potential step to region second. In a general case, a stationary state describing this situation contains three parts. In region first the state is composed of the incoming wave with probability current Ni and a reflected current Nr. In region second there is a transmitted wave of probability current Nt.

Then Ni i.e. incident wave probability current is given by-

Ni = |Aψi ψi *|²v

Ni = |A| ² .v

(Note :- ψi ψi* =1 ,if the incident particle is present.)

Now, we know that k₁² = 2mE/h²

K₁² = 2m(1/2mv²)/ h²

v= k₁h/m

Ni = |A|²k₁h/m               (14)

Similarly reflected flux Nr,

|Nr| = |B|².v

|Nr| = |(k₁-k₂ / k₁+ k₂ )|²|A|². (k₁h/m)

|Nr| = |(k₁-k₂ / k₁+ k₂ )|²Ni      (15)

Similarly , transmitted flux Nt,

Nt = |C|² v₁

Put the value of C,from equation (11)

Nt = |(2k₁ / k₁ + k₂)|²|A|² (k₂ h/ 2m)

Nt = {4k₁ / (k₁ + k₂)²} (k₁k₂h /m) |A|²

Nt = [4k₁k₂/ (k₁+k₂)²] Ni               (16)

Now, from equation (14), (15) and (16)

(Ni-Nr) =|A|²k₁h/m – |(k₁-k₂ / k₁+ k₂ )|²|A|². (k₁h/m)

(Ni-Nr) =|A|²k₁h/m [1- |(k₁−k₂)/(k₁+k₂)|²]

(Ni-Nr) =|A|²k₁h/m [ { (k₁+k₂)²-(k₁−k₂)²}/ (k₁+K₂)² ]

(Ni-Nr) =|A|²k₁h/m [ (4k₁k₂)/ (k₁+K₂)² ]

(Ni-Nr) =Ni [ (4k₁k₂)/ (k₁+K₂)² ]

(Ni-Nr) = Nt          (17)


Reflection coefficient R can be define as follows-

R = magnitude of reflected flux/ magnitude of incident flux

R= Nr/Ni

R= [(k₁−k₂/k₁+k₂)²]           (18)

Transmission cofficient can be written as

T = Nt/Ni

T= 4k₁k₂/(k₁+k₂)²

Now, R+T = [(k₁−k₂/k₁+k₂)²] +4k₁k₂/(k₁+k₂)²

R+T = [(k₁²+k₂²-2k₁k₂) + 4k₁k₂]/ (k₁+k₂)²

R+T = (k₁+k₂)²/(k₁+k₂)²

R + T = 1             (19)

Reflection coefficient R and transmission coefficient T for a potential step V0 high versus energy E (in units V0).
Case 2  when E<v₀

Schrodinger equation in region first

d²ψ₁/dx² + 2m(E)ψ₁/h² = 0        (1)

And solution of above equation is given by-

ψ  = Aexp (ik₁x) + B exp (-ik₁x)       (2)

Now , Schrodinger equation in region second

d²ψ₂/dx² – 2m(V₀-E)ψ₂/h² = 0

Put  2m(V₀-E)ψ₂/h² = k₃²

d²ψ₂/dx² – k₃²/h² = 0       (3)

This is an second order differential equation and solution of this equation is given by-

ψ₂ =Cexp k₃x + D exp (-k₃x)         (4)

When x>0, Or x→∞ then ψ = 0 ,to fulfill this condition

I.e., C =0

So, ψ₂ = D exp(-k₃x)                        (5)

Now applying boundary conditions

  1. At x=0, ψ₁=ψ₂
  2. At x=0, (dψ₁/dx) = (dψ₂/dx)

From equation (3) and (5)

Aexp (ik₁x) + Bexp (-ik₁x) = Dexp(-k₃x)

At x=0, A+B =D           (6)

And differentiating we get-

ik₁A -ik₁B = -k₃D          (7)

Solving above equation and find out the value of D,

(multiply equation (a) by k₃ and adding in (b),

k₃A + k₃B =k₃D

Ik₁A+k₃A-k₃B-ik₁B = k₃D−k₃D

A(k₃+ik₁) + B(k₃-ik₁)  = 0

B = [(ik₁ + k₃)/(ik₁−k₃)]A         (8)

And D = [ (2ik₁/ik₁−k₃)A       (9)

Incident flux Ni= |A|²v

And , reflected flux Nr=|B|²v

Nr = v|(ik₁+k₃) / (ik₁−k₃)|²|A|²

Nr= v.1.|A|²

Nr= |A|²v

Nr= Ni                      (10)

And Ni= 0( for region first when x<0)

Reflection cofficient R= Nr/Ni =1

And T=0

Also, R+T =1

For second region

Nt= |D|²v

Where, ψ₂ = D exp (-k₃x)

Probability of finding the particle in this region,

P(x) =|D|²exp(-2k₃x)

Where probability can be expressed by

P(x) = ψψ*

This equation shows that P(x) , decreases exponentially in second region. And Nt≠0 unless k₃ (i.e. v₀ is ∞) is ∞

If Nt ≠0 ,i.e. there is some probability to find particle in second region.



In this article Solution of Schrodinger equation for a Potential step, the concept of a potential step is clear. Also, we have discussed many important terms like incident flux, reflected wave function (flux), Along with reflection coefficient R, transmission coefficient T etc.









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