Question

Suppose you wanted to describe an unstable particle, that spontaneously disintegrates with a lifetime in that case the total probability of finding the particle somewhere should not be constant but should decrease at an exponential rate:

$\mathit{p}\left(t\right)\mathbf{=}{\mathbf{\int }}_{\mathbf{-}\mathbf{\infty }}^{\mathbf{\infty }}\left[\psi {\left(x,t\right)}^2\right]\mathit{d}\mathit{x}\mathbf{=}{\mathit{e}}^{\frac{\mathbf{-}\mathbf{t}}{\mathbf{t}}}$

A crude way of achieving this result is as follows. in equation 1.24 we tightly assumed that is real. That is certainly responsible, but it leads to the conservation of probability enshrined in equation 1.27. What if we assign to in imaginary part

$\mathit{V}\mathbf{=}{\mathit{V}}_{\mathbf{0}}\mathbf{=}\mathit{i}\mathit{\Gamma }$

Where is the true potential energy and is a positive real constant?

1. Show that now we get

$\frac{\mathbf{d}\mathbf{p}}{\mathbf{d}\mathbf{t}}\mathbf{=}\frac{\mathbf{2}\mathbf{\Gamma }}{\mathbf{h}}\mathit{p}$.

Solve for and find the lifetime of the particle in terms of$\mathit{\Gamma }$

Short answer

(a)The given equation is proved $\frac{dp}{dt}=-\frac{2\Gamma }{ħ}p$

(b)The lifetime of the particle in terms of t is$\tau =\frac{ħ}{2\Gamma }$

Step-by-step solution

Step 1: Define the Schrodinger equation

The Schrödinger equation, or Schrödinger's wave equation, is a partial differential equation that utilizes the wave function to describe the behavior of quantum mechanical systems. The Schrödinger equation can be used to determine the trajectory, location, and energy of these systems.

Prove the given equation  dpdt=2Γhp 

(a)

For an unstable particle

$P\left(t\right)={\int }_{\infty }^{\infty }{\left|\psi \left(x,t\right)\right|}^{2}dx=e^{-t/z}$

…(1)

Where for a stable particle (as indicated in eq.(1.27) in the book)

$P\left(t\right)={\int }_{\infty }^{\infty }{\left|\psi \left(x,t\right)\right|}^{2}dx=0$ …(2)

Since the potential energy for an unstable particle become $V=V_0-i\Gamma$, Schrodinger equation become

$Iħ\frac{\partial \psi }{\partial T}=\frac{{ħ}^2{\partial }^2\psi }{2m\partial x^2}+\left(V_0-i\Gamma \right)\psi$ …(3)

Where its conjugate is

$-iħ\frac{\partial \psi }{\partial t}=-\frac{{ħ}^2}{2m}\frac{{\partial }^2\psi }{\partial x^2}+\left(V_0+i\Gamma \right)\psi$ …(4)

Now,

$$\begin{gathered} \frac{dp}{dt}=\frac{d}{dt}{\int }_{-\infty }^{\infty }{\left|\psi \right|}^{2}dx \\ ={\int }_{-\infty }^{\infty }\frac{\partial }{\partial t}{\psi }^{*} \\ \frac{dp}{dt}={\int }_{-\infty }^{\infty }\left(\psi \frac{\partial {\psi }^{*}}{\partial t}+{\psi }^{*}\frac{\partial \psi }{\partial t}\right)dx \end{gathered}$$ …(5)

Substitute from Schrodinger equation and its conjugate, thus, the integrand become

$\left(\psi \left[\frac{-iħ}{2m}\frac{{\partial }^2{\psi }^{*}}{\partial x^2}-\frac{1}{ħ}{V}_0{\psi }^{*}\frac{\Gamma }{ħ}{\psi }^{*}\right]+{\psi }^{*}\left[\frac{-iħ}{2m}\frac{{\partial }^2{\psi }^{*}}{\partial x^2}+\frac{1}{ħ}{V}_0\psi -\frac{\Gamma }{ħ}{\psi }^{*}\right]\right)$

Therefore,

$$\begin{gathered} \frac{dp}{dt}={\int }_{\infty }^{\infty }\left(\frac{-iħ}{2m}\left[\psi \frac{{\partial }^2{\psi }^{*}}{\partial x^2}+{\psi }^{*}\frac{{\partial }^2\psi }{\partial x^2}\right]-2\frac{\Gamma }{ħ}\psi {\psi }^{*}\right)dx \\ =\frac{-iħ}{2m}{\int }_{-\infty }^{\infty }\left(\psi \frac{{\partial }^2{\psi }^{*}}{\partial x^2}+{\psi }^{*}\frac{{\partial }^2\psi }{\partial x^2}\right)dx-{\int }_{\infty }^{\infty }2\frac{\Gamma }{ħ}\psi {\psi }^{*}dx \\ \frac{dp}{dt}={\int }_{\infty }^{\infty }\frac{\partial }{\partial x}\left(\psi \frac{\partial {\psi }^{*}}{\partial x}+{\psi }^{*}\frac{\partial \psi }{\partial x^2}\right)dx-2\frac{\Gamma }{ħ}{\int }_{-\infty }^{\infty }{\left|\psi \right|}^{2}dx \end{gathered}$$ …(6)

Where, the first integral is zero from eq.(6), and the second integral equal to the probability, so,

$\frac{dp}{dt}=-\frac{2\Gamma }{ħ}p$

Hence ,its proved.

Determine the lifetime of the particle

(b)

If we derive eq.(1) with respect to time we can get

$$\begin{gathered} \frac{dp}{dt}=\frac{d}{dt}{e}^{-t/z} \\ =\frac{-1}{\tau }{e}^{t/z} \end{gathered}$$

So, from $\frac{dp}{dt}=-\frac{2\Gamma }{ħ}p$ we can get by substituting $\frac{dp}{dt}$ inside it

$$\begin{gathered} \frac{-e^{-t/z}}{\tau }=\frac{-2\Gamma }{ħ}{e}^{-t/z} \\ \tau =\frac{ħ}{2\Gamma } \end{gathered}$$

Therefore, the lifetime of the particle in terms of$\tau$is$\tau =\frac{ħ}{2\Gamma }$