Please help im not sure how to do this. Consider a particle in a ID half-infinit

Question

Please help im not sure how to do this.

Consider a particle in a ID half-infinite well. The potential in region III is infinite and is zero in region I. In this case however, the potential is not infinite in region II but has a constant value V_0. Region III Write the Schrodinger equation in region III. What is the only acceptable solution? Region I Write the Schrodinger equation in region I. Show that Psi_I = Asin(kx)+Bcos (kx) is solution of this equation and give the constant k as a function of the total energy E of the particle. Using continuity of the wave function at the boundary between region III and I, determine the constant B. Region II Write the Schrodinger equation in region II. Show that Psi_II Ce^lambdax + De^-lambdax is solution of this equation and give the constant as a function of the total energy E

Explanation / Answer

From the given potentials in three regions we consider the problem similar to the problem of particle in a box. First of all we check the coordinate of the wavefunctions in each region to determine that result is infinite. Finally we need to find a continious solution in each region.

a) In the region III, potential V(x)= infinity and the well is 1-D

Therefore we use infinity in the position of V(x) in the general schrodinger equation

The expression is:

{(-hcut^2/2m)d^2/dx^2 +infinity}(x) =E(x)

If we take the constant infinity term to the right side that means E-infinity=-infinity

therefore the equation becomes

{(-hcut^2/2m)d^2/dx^2}(x) =-infinity(x)

Now taking infinity as dividend we can prove that 1/infinity {(-hcut^2/2m)d^2/dx^2}(x) =0

or (x) =0

and this is the only acceptable solution.

b) In case of region I, the potential V(x) =0 and again we introduce the value of potential in the general Schrodinger equation

{(-hcut^2/2m)d^2/dx^2 +0}(x) =E(x)

Using trigonometric functions we get the general solution of the equation as (assume that E<V0)

(x) =ASinkx +BCoskx where A,B, K are constant

Now to determine k we introduce the solution in the general schrodinger equation and it looks like

{(-hcut^2/2m)d^2/dx^2 +0}[ASinkx +BCoskx] =E[ASinkx +BCoskx]

Taking second derivative on the left and simplifying we get

(-hcut^2k^2/2m)[ASinkx +BCoskx] =E[ASinkx +BCoskx]

that means (-hcut^2k^2/2m)= E

or k= {2mE/hcut^2}^1/2

c) Considering the position and time continuity aas well as the exponential form of the generaal solution given in b)

we can write E= ihcutB

or B= iE/hcut

d) The schrodinger equation at region II will include potential V0 in the expression

{(-hcut^2/2m)d^2/dx^2 +V0}(x) =E(x)

aiming isolation of 2nd derivative on the left part of the solution we can write

d^2/dx^2 (x) =2m(V0-E)/hcut^2(x)

We can see a similarity with the region I that means the most suitable solution will be in exponential form

(x) = Cexp(lamda x) +Dexp(-lamda x)

we consider lamda as real and positive because k and E are already in relation in region I

To get the value lamda we again introduce the solution in the equation

{(-hcut^2/2m)d^2/dx^2 +V0}[Cexp(lamda x) +Dexp(-lamda x)] =E[Cexp(lamda x) +Dexp(-lamda x)]

Again moving constant terms and second derivative we get

(-hcut^2lamda^2/2m)[Cexp(lamda x) +Dexp(-lamda x)]= (E- V0)[Cexp(lamda x) +Dexp(-lamda x)]

or lamda={2m(E- V0)/hcut}^2

e) Now from the given conditions for region II if x tends to infinity the wave function in the region will go off which also remove the part Cexp(lamda x). Therefore only acceptable solution for C=0

f) To check continuity at x=L we consider region I and II as in these two regions the potentials are finite.

So at x=L

(L)region I =(L)regionII

'(L)region I ='(L)regionII

taking ratio of continuity equation we get

'(L)region I/(L)region I ='(L)regionII/(L)regionII

AkCoskL/ASinkL =-Dexp(-lamdaL)/Dexp(-lamdaL)

simplifying we get

kCotkL=-lamda

g) kCotkL=-lamda we multiply L in both side to get

kLCotkL=-lamdaL

introducing k and lamda solution we get

kLCotkL=-[2mV0L^2/hcut^2-(kL)^2]^1/2

introducing z=kL and alpha=[mV0L^2/hcut^2]^1/2

we get zarctan(z) =-(alpha^2- z^2)^1/2

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