Chapter 2: Q. 11P (page 50)

a) Compute $⟨ x ⟩, ⟨ p ⟩, <x^{2}>, <p^{2}>$ , for the states $ψ_{0}$ and $ψ_{1}$ , by explicit integration. Comment; In this and other problems involving the harmonic oscillator it simplifies matters if you introduce the variable $ξ \equiv \sqrt{\frac{m ω}{□}} x$ and the constant $α \equiv {(\frac{m ω}{π □})}^{1 / 4}$ .
b) Check the uncertainty principle for these states.
c) Compute $⟨ T ⟩$ (the average kinetic energy) and $⟨ V ⟩$ (the average potential energy) for these states. (No new integration allowed). Is their sum what you would expect?

Short Answer

Expert verified

Answer

(a) The value of is $\int x | ψ |^{2} d x - 0$ and the value of $⟨ p ⟩$ is

$m^{d ⟨ x ⟩} / d t = 0$ and the value of $⟨ x ⟩^{2}$ is $\frac{3 □}{2 m ω}$ and the value of $⟨ P ⟩^{2}$ is $\frac{3 m □ ω}{2}$ .

(b)The uncertainty principal for these states is greater than $\frac{□}{2}$

(c) The value of the average kinetic energy is $\{\begin{array}{l} \frac{1}{4} ω □ (n = 0) \\ \frac{3}{4} ω □ (n = 1) \end{array}\}$ and the average potential energy is $\{\begin{array}{l} \frac{1}{4} ω □ (n = 0) \\ \frac{3}{4} ω □ (n = 1) \end{array}\}$ .The sum of these two is same as expected value.

Step by step solution

- Define explicit integration

In this method, the accelerations and velocities at a specific time point are taken to be constant during a time interval and are utilised to find the location of the following time point.

- The value of x and P

(a)

The function $ψ_{0}$ is even, and $ψ_{1}$ is odd. In either case $∣ ψ^{2}$ is even, so the value of $⟨ x ⟩ = \int x | ψ |^{2} d x = 0$ . Therefore $⟨ p ⟩ = \frac{m d ⟨ x ⟩}{d t} = 0$

Thus, the value of $<x>$ is $\int x | ψ |^{2} d x - 0$ and the value of $⟨ p ⟩$ is $\frac{m d ⟨ x ⟩}{d t} = 0$

Now, evaluate the value of $<x^{2}>$ and $<p^{2}>$ .

The below values are known that

$ψ_{0} = α e^{- ξ^{2} / 2}$

$ψ_{1} = \sqrt{2} α ξ e^{- ξ^{2} / 2}$

For $n = 0$ , the value of $<x^{2}>$ will be

$<x^{2}> = α^{2} \int_{- \infty}^{\infty} x^{2} e^{- ξ^{2} / 2} d x$ ....................(1)

For given equation-

$ξ = \sqrt{\frac{m ω}{□}} x x = ξ \sqrt{\frac{□}{m ω}} x^{2} = ξ^{2} (\frac{□}{m ω})$

And

$x = ξ \sqrt{\frac{□}{m ω}} d x = d ξ \times \sqrt{\frac{□}{m ω}}$

Substitute the value of $x^{2}$ and $d x$ in equation (1)

$<x^{2}> = α^{2} \int_{- \infty}^{\infty} (ξ^{2} (\frac{□}{m ω})) \times e^{- ξ^{2}} \times d ξ \times \sqrt{\frac{□}{m ω}} <x^{2}> = α^{2} {(\frac{□}{m ω})}^{3 / 2} \int_{- \infty}^{\infty} ξ^{2} e^{- ξ^{2}} d ξ = \frac{1}{\sqrt{π}} (\frac{□}{m ω}) \frac{\sqrt{π}}{2} = \frac{□}{2 m ω}$

And the value of $<P^{2}>$ is

$<P^{2}> = \int ψ_{0} {(\begin{matrix} □ & d \\ i & d x \end{matrix})}^{2} ψ_{0} d x = □^{2} α^{2} \sqrt{\frac{m ω}{□}} \int_{- \infty}^{\infty} e^{- ξ^{2}} (\frac{d^{2}}{d ξ^{2}} e^{- ξ^{2}}) d ξ = - \frac{m □ ω}{\sqrt{π}} (\frac{\sqrt{π}}{2} - \sqrt{π}) = \frac{m □ ω}{2}$

For $n = 1$ , the value of $<x^{2}>$ will be

$<x^{2}> = 2 α^{2} \int_{- \infty}^{\infty} x^{2} ξ^{2} e^{- ξ^{2}} d x$

Substitute the value of and in the above equation. So,

$<x^{2}> = 2 α^{2} {(\frac{□}{m ω})}^{3 / 2} \int_{- \infty}^{\infty} ξ^{4} e^{- ξ^{2}} d ξ = \frac{3}{\sqrt{π}} \frac{2 □}{m ω} \frac{\sqrt{π}}{4} = \frac{3 □}{2 m ω}$

And the value of $<P^{2}>$ is

$<P^{2}> = - □^{2} 2 α^{2} \sqrt{\frac{m ω}{□}} \int_{- \infty}^{\infty} ξ e^{- ξ^{2} / 2} [\frac{d^{2}}{d ξ^{2}} (ξ e^{- ξ^{2}} / 2)] d ξ = - \frac{2 m □ ω}{\sqrt{π}} (\frac{3 \sqrt{π}}{4} - \frac{3}{2} \sqrt{π}) = \frac{3 m □ ω}{2}$

Therefore, the value of $<x^{2}>$ is $\frac{3 □}{2 m ω}$ and $<P^{2}>$ is $\frac{3 m □ ω}{2}$

- The uncertainty principal of above states

(b)

The value of will be for :

$σ_{x} = \sqrt{<x^{2}> - {<x>}^{2}} = \sqrt{\frac{□}{2 m ω}} σ_{P} = \sqrt{<P^{2}> - {<P>}^{2}} = \sqrt{\frac{m ω □}{2}}$

$σ_{x} σ_{P} = \sqrt{\frac{□}{2 m ω}} \sqrt{\frac{m ω □}{2}} = \frac{□}{2}$

(Right t the uncertainty moment)

For $n = 1$ , the value will be :

$σ_{x} = \sqrt{\frac{3 □}{2 m ω}} σ_{P} = \sqrt{\frac{3 m ω □}{2}} σ_{x} σ_{P} = 3 \frac{□}{2} > \frac{□}{2}$

The uncertainty principal for these states is greater than $\frac{□}{2}$

- The value of the average kinetic energy and the average potential energy.

(c)

The average kinetic energy is:

$<T> = \frac{1}{2 m} <P^{2}>$

$= \{\begin{array}{l} \frac{1}{4} ω □ (n = 0) \\ \frac{3}{4} ω □ (n = 1) \end{array}\}$

The average potential energy is

$<V> = \frac{1}{2} m ω^{2} <x^{2}>$

$= \{\begin{array}{l} \frac{1}{4} ω □ (n = 0) \\ \frac{3}{4} ω □ (n = 1) \end{array}\}$

$<T> + <V> = <H> = \{\begin{matrix} \frac{1}{2} ω □ (n = 0) = E_{0} \\ \frac{3}{2} ω □ (n = 1) = E_{1} \end{matrix}\}$ , as expected.

So, the value of the average kinetic energy is $\{\begin{matrix} \frac{1}{4} ω □ (n = 0) \\ \frac{3}{4} ω □ (n = 1) \end{matrix}\}$ and the average potential energy is $\{\begin{matrix} \frac{1}{4} ω □ (n = 0) \\ \frac{3}{4} ω □ (n = 1) \end{matrix}\}$ . The sum of these two is same as expected value.