Q14E To show,(a) 唯1(x,t)=Aeikx鈭抜蝇... [FREE SOLUTION]

Chapter 12: Q14E (page 556)

To show,
(a) $唯_{1} (x, t) = A e^{i k x 鈭� i 蝇 t}$ is the solution of both Klein-Gordon and the Schrodinger equations.
(b) $唯_{2} (x, t) = A e^{i k x} c o s 蝇 t$ is the solution of both Klein-Gordon but not the Schrodinger equations.
(c) The $唯_{2}$ is a combination of positive and negative energy solutions of the Klein-Gordon equation.
(d) To compare the time dependence of $唯^{2}$ for $唯_{1}$ and $唯_{2}$ .

Short Answer

Expert verified

(a) The solution of both Klein-Gordon and the Schrodinger equations is $唯_{1} (x, t) = A e^{i k x 鈭� i 蝇 t}$ .

(b) The solution of both Klein-Gordon but not the Schrodinger equations $唯_{2} (x, t) = A e^{i k x} \cos 蝇 t$ .

(c) The first part $\frac{1}{2} A e^{i k x + i 蝇 t}$ is the negative energy solution, and the second part $\frac{1}{2} A e^{i k x 鈭� i 蝇 t}$ is the positive energy.

(d) The function ${| 唯_{1} |}^{2}$ doesn't have the time dependence but ${| 唯_{2} |}^{2}$ has the time dependence.

Step by step solution

Given data

The given functions are $唯_{1}$ and $唯_{2}$ .

Concept of the Wave equation

The Klein-Gordon equation is given as,

$鈭� c^{2} {鈩�}^{2} \frac{{鈭�}^{2}}{鈭� x^{2}} 唯 (x, t) + m^{2} c^{4} 唯 (x, t) = 鈭� {鈩�}^{2} \frac{{鈭�}^{2}}{鈭� t^{2}} 唯 (x, t)$

The Schr枚dinger Equation is given as,

$鈭� \frac{{鈩�}^{2}}{2 m} \frac{{鈭�}^{2}}{鈭� x^{2}} 唯 (x, t) = i 鈩� \frac{鈭�}{鈭� t} 唯 (x, t)$

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:

$\frac{{鈭�}^{2}}{d x^{2}} 唯_{1} (x, t) = 鈭� k^{2} 唯_{1} (x, t)$ 鈥︹赌�(1)

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

$\frac{{鈭�}^{2}}{鈭� t^{2}} 唯_{1} (x, t) = 鈭� 蝇^{2} 唯_{1} (x, t)$ 鈥︹赌�.(2)

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

$c^{2} k^{2} {鈩�}^{2} 唯_{1} (x, t) + m^{2} c^{4} 唯_{1} (x, t) = 蝇^{2} {鈩�}^{2} 唯_{1} (x, t)$

Substitute $p = k 鈩�$ and $E = 蝇鈩�$ in the above equation and simplify as:

$c^{2} p^{2} + m^{2} c^{4} = E^{2}$

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

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:

$\begin{matrix} \frac{鈭�}{鈭� t} 唯_{1} (x, t) = \frac{鈭�}{鈭� t} (A e^{i k x 鈭� i 蝇 t}) \\ = 鈭� i 蝇唯_{1} (x, t) \end{matrix}$

鈥︹赌�(3)

After substitution equations (1) and (3) in the Schr枚dinger equation, obtain:

$\frac{{鈩�}^{2} k^{2}}{2 m} 唯_{1} (x, t) = \frac{p^{2}}{2 m} 唯_{1} (x, t)$

On the right, we have,

. $鈩� 蝇唯_{1} (x, t) = E 唯_{1} (x, t)$

For nonrelativistic particles,

. $E = \frac{p^{2}}{2 m}$

The wave function is proved to be a solution.

$唯_{1} (x, t) = A e^{i k x 鈭� i 蝇 t}$ is the solution of both Klein-Gordon and the Schrodinger equations.

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

(b)

The Candidate wave function is given as,

$唯_{2} (x, t) = A e^{i k x} \cos 蝇 t$

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

$\begin{matrix} \frac{{鈭�}^{2}}{鈭� x^{2}} 唯_{2} (x, t) = A e^{i k x} t \cos 蝇 t {(卤 i K)}^{2} \\ = 鈭� k^{2} 唯_{2} (x, t) \end{matrix}$

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

$\begin{matrix} \frac{{鈭�}^{2}}{鈭� t^{2}} 唯_{2} (x, t) = A e^{i k x} \cos 蝇 t (鈭� 蝇^{2}) \\ = 鈭� 蝇^{2} 唯_{2} (x, t) \end{matrix}$

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:

$\begin{matrix} \frac{鈭�}{鈭� t} 唯_{2} (x, t) = A e^{i k x} \sin 蝇 t (鈭� 蝇) \\ 鈮� 鈭� 蝇唯_{2} (x, t) \end{matrix}$

The wave function is proved not to be a solution.

$唯_{2} (x, t) = A e^{i k x} \cos 蝇 t$ is the solution of both Klein-Gordon but not the Schrodinger equations.

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

(c)

The Candidate wave function is given, as shown below:

$\begin{matrix} 唯_{2} (x, t) = A e^{i k x} \cos 蝇 t \\ \cos 胃 = \frac{e^{i x} + e^{鈭� i x}}{2} \end{matrix}$

The Candidate wave function is given as,

$唯_{2} (x, t) = A e^{i k x} \cos 蝇 t$

After the expansion of $\cos 蝇 t$ in exponential terms, obtain:

$\begin{array}{l} 唯_{2} (x, t) = A e^{i k x} \cos 蝇 t \\ 唯_{2} (x, t) = A e^{i k x} \frac{e^{i 蝇 t} + e^{鈭� i a x}}{2} \\ 唯_{2} (x, t) = \frac{1}{2} A e^{i k x + i 蝇 t} + \frac{1}{2} A e^{i k x 鈭� i 蝇 t} \end{array}$

The first part $\frac{1}{2} A e^{i k x + i 蝇 t}$ is the negative energy solution, and the second part $\frac{1}{2} A e^{i k x 鈭� i 蝇 t}$ is a positive energy.

Comparison of the time dependence of Ψ2forΨ1 and Ψ2

(d)

The Candidate wave function is given as,

. $唯_{1} (x, t) = A e^{i k x 鈭� i 蝇 t}$

The probability density is given as follows:

$\begin{array}{l} {| 蠄_{1} (x, t) |}^{2} = {(A e^{i k x 鈭� i 蝇 t})}^{2} \\ {| 蠄_{1} (x, t) |}^{2} = A^{2} \end{array}$

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

${| 蠄_{1} (x, t) |}^{2} = A^{2}$

So, it does not have time dependence.

The Candidate wave function is given as,

. $唯_{2} (x, t) = A e^{i k x} \cos 蝇 t$

The probability density is given as below:

$\begin{array}{l} {| 蠄_{2} (x, t) |}^{2} = {(A e^{i k x} \cos 蝇 t)}^{2} \\ {| 蠄_{2} (x, t) |}^{2} = A^{2} \cos^{2} 蝇 t \end{array}$

So, it does have time dependence.

The function ${| 唯_{1} |}^{2}$ doesn't have the time dependence but ${| 唯_{2} |}^{2}$ has the time dependence.

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Short Answer

Step by step solution

Given data

Concept of the Wave equation

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

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

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

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

Comparison of the time dependence of Ψ2forΨ1 and Ψ2

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