To show,

(a) Ψ1(x,t)=Aeikxiωtis the solution of both Klein-Gordon and the Schrodinger equations.

(b) Ψ2(x,t)=Aeikxcosωtis the solution of both Klein-Gordon but not the Schrodinger equations.

(c) TheΨ2 is a combination of positive and negative energy solutions of the Klein-Gordon equation.

(d) To compare the time dependence ofΨ2 for Ψ1and Ψ2.

Short Answer

Expert verified

(a) The solution of both Klein-Gordon and the Schrodinger equations isΨ1(x,t)=Aeikxiωt.

(b) The solution of both Klein-Gordon but not the Schrodinger equations Ψ2(x,t)=Aeikxcosωt.

(c) The first part 12Aeikx+iωt is the negative energy solution, and the second part 12Aeikxiωt is the positive energy.

(d) The function|Ψ1|2 doesn't have the time dependence but |Ψ2|2 has the time dependence.

Step by step solution

01

Given data

The given functions are Ψ1 andΨ2 .

02

Concept of the Wave equation

The Klein-Gordon equation is given as,

c222x2Ψ(x,t)+m2c4Ψ(x,t)=22t2Ψ(x,t)

.

The Schrödinger Equation is given as,

.

22m2x2Ψ(x,t)=itΨ(x,t)

03

Take second order partial derivative of the Candidate wave function Ψ1(x,t)=Aeikx−iωt

(a)

Second order partial derivative of the wave function with respect to x as:

2dx2Ψ1(x,t)=k2Ψ1(x,t) ……(1)

Second order partial derivative of the wave function with respecttas:

2t2Ψ1(x,t)=ω2Ψ1(x,t) …….(2)

After substitution equations (1) and (2) in the Klein-Gordon equation, obtain:

c2k22Ψ1(x,t)+m2c4Ψ1(x,t)=ω22Ψ1(x,t)

Substitute p=k andE=ω in the above equation and simplify as:

c2p2+m2c4=E2

From special relativity, the wave function is proved to be a solution.

04

Take first order partial derivative of Candidate function Ψ1(x,t)=Aeikx−iωt

Taking first order partial derivative of the Candidate wave function with respect to x as:

tΨ1(x,t)=t(Aeikxiωt)=iωΨ1(x,t)

……(3)

After substitution equations (1) and (3) in the Schrödinger equation, obtain:

2k22mΨ1(x,t)=p22mΨ1(x,t)

On the right, we have,

.ωΨ1(x,t)=EΨ1(x,t)

For nonrelativistic particles,

.E=p22m

The wave function is proved to be a solution.

Ψ1(x,t)=Aeikxiωtis the solution of both Klein-Gordon and the Schrodinger equations.

05

Take second order partial derivative of the Candidate wave functionΨ2(x,t)=Aeikxcosωt  

(b)

The Candidate wave function is given as,

.

Ψ2(x,t)=Aeikxcosωt

Second order partial derivative of the Candidate wave function with respect to x as:

2x2Ψ2(x,t)=Aeikxtcosωt±iK2=k2Ψ2(x,t)

Second order partial derivative of the Candidate wave function with respect t as:

2t2Ψ2(x,t)=Aeikxcosωt(ω2)=ω2Ψ2(x,t)

From special relativity, the wave function is proved to be a solution.

Similarly, takethe first order partial derivative of the Candidate wave function with respect to x as:

tΨ2(x,t)=Aeikxsinωt(ω)ωΨ2(x,t)

The wave function is proved not to be a solution.

Ψ2(x,t)=Aeikxcosωtis the solution of both Klein-Gordon but not the Schrodinger equations.

06

Calculation to show  Ψ2 is a combination of positive and negative energy solutions 

(c)

The Candidate wave function is given, as shown below:

Ψ2(x,t)=Aeikxcosωtcosθ=eix+eix2

The Candidate wave function is given as,

Ψ2(x,t)=Aeikxcosωt

After the expansion ofcosωt in exponential terms, obtain:

Ψ2(x,t)=AeikxcosωtΨ2(x,t)=Aeikxeiωt+eiax2Ψ2(x,t)=12Aeikx+iωt+12Aeikxiωt

The first part 12Aeikx+iωtis the negative energy solution, and the second part 12Aeikxiωtis a positive energy.

07

Comparison of the time dependence of   Ψ2forΨ1   and  Ψ2 

(d)

The Candidate wave function is given as,

.Ψ1(x,t)=Aeikxiωt

The probability density is given as follows:

|ψ1(x,t)|2=(Aeikxiωt)2|ψ1(x,t)|2=A2

For the first Candidate, since all the positive and time dependence is in terms of the exponential of an imaginary number,

|ψ1(x,t)|2=A2

So, it does not have time dependence.

The Candidate wave function is given as,

.Ψ2(x,t)=Aeikxcosωt

The probability density is given as below:

|ψ2(x,t)|2=(Aeikxcosωt)2|ψ2(x,t)|2=A2cos2ωt

So, it does have time dependence.

The function |Ψ1|2doesn't have the time dependence but|Ψ2|2 has the time dependence.

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