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For solutions of Klein-Gordon equation, the quantity,

ititis interpreted as charge density. Show that for a positive-energy plane-wavesolution. It is a real constant, and for negative-energy solution.It is a negative of that constant.

Short Answer

Expert verified

The quantity is a real constant for positive-energy plane-wave solution, and negative to that for negative-energy solution is proved.

Step by step solution

01

Given data

Charge density is itit.

02

Concept of the Wave equation

The Klein-Gordon equation is given as,

c222x2(x,t)+m2c4(x,t)=22t2(x,t)

The Schr枚dinger Equation is given as,

22m2x2(x,t)=it(x,t)

Charge density is given as,

itit

03

Calculation for positive wave function

The positive wave functions are given as follows:

1(x,t)=Aeikxit 鈥︹赌(1)

1*(x,t)=Aeikx+it 鈥︹赌(2)

Differentiate the equation (1) partially with respect totas:

t1(x,t)=Aeikxit(i)t1(x,t)=i1(x,t)

鈥︹赌︹赌(3)

Differentiate the equation (2) partially with respect to t:

t1*(x,t)=Aeikx+it(i)t1*(x,t)=i1*(x,t)

颁丑补谤驳别鈥塪别苍蝉颈迟测=i1*t1i1t1*

Substitute t1(x,t)=i1(x,t) andt1*(x,t)=i1*(x,t) in the equationand simplify as:

颁丑补谤驳别鈥塪别苍蝉颈迟测=i1*(i1)i1i1*(x,t)=i21*1i21*1=i2((Aeikx+it)(Aeikxit))1i2((Aeikx+it)(Aeikxit))=22A2

04

Calculation for negative wave function 

The negative energy wave functions are:

1*(x,t)=Aeikxit 鈥︹赌(4)

1*(x,t)=Aeikxit 鈥︹赌(5)

Differentiate the equation (4) partially with respect to t as:

t1(x,t)=Aeikx+it(i)t1(x,t)=i1(x,t)

Differentiate the equation (5) partially with respect to tas:

t1*(x,t)=Aeikxit(i)t1*(x,t)=i1*(x,t)

Substitute t1(x,t)=i1(x,t)and t1*(x,t)=i1*(x,t) in the equation and simplify as:

颁丑补谤驳别鈥塪别苍蝉颈迟测=i1*(i1)i1i1*(x,t)=i21*1+i21*1=i2((Aeikx+it)(Aeikxit))1+i2((Aeikx+it)(Aeikxit))=22A2

This is exactly the negative of the constant for this part.

The quantity is a real constant for positive-energy plane-wave solution, and negative to that for negative-energy solution is proved.

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