General Result for two particle bound state

The hamiltonian for two body interacting through potential V(r) is given by, $$\hat H= \frac{-\hbar^2}{2\mu}\nabla^2+V(\vec r)\dotsm(1)$$ From the hamiltonian we derived the three important result or theorems as, 1. Energy level lie deeply as reduced mass \((\mu\)) increases. 2. Virial theorem is satisfied. 3. \(|\psi(0)|^2\) is simply related to force \(\biggl<\frac{\partial v}{\partial r}\biggr>\) In this chapter.

Summary

The hamiltonian for two body interacting through potential V(r) is given by, $$\hat H= \frac{-\hbar^2}{2\mu}\nabla^2+V(\vec r)\dotsm(1)$$ From the hamiltonian we derived the three important result or theorems as, 1. Energy level lie deeply as reduced mass \((\mu\)) increases. 2. Virial theorem is satisfied. 3. \(|\psi(0)|^2\) is simply related to force \(\biggl<\frac{\partial v}{\partial r}\biggr>\) In this chapter.

Things to Remember

  1. $$=\frac{-<T>}{\mu}$$

    Since, <T> is always positive, which show that \frac{\partial E}{\partial \mu}\) is always negative from this, we can conclude that energy level. i.e. deeply as reduced mass increases.

  2. $$or,\; \frac{<P^2>}{\mu}= 2<^>$$

    Which proves the virtual theorem.

  3. $$or,\;\; |\psi(0)|^2= \frac{-2\mu}{\hbar^2}<\frac{dv}{dr}>$$Whcih proves that \(|\psi(0)|^2\) is simply related to force \(<\frac{dv}{dr}>\)
  4. The hamiltonian for two body interacting through potential V(r) is given by,

    $$\hat H= \frac{-\hbar^2}{2\mu}\nabla^2+V(\vec r)$$

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General Result for two particle bound state

General Result for two particle bound state

General Result for two particle bound state:

The hamiltonian for two body interacting through potential V(r) is given by,

$$\hat H= \frac{-\hbar^2}{2\mu}\nabla^2+V(\vec r)\dotsm(1)$$

From the hamiltonian we can derive the three important result or theorems as,

1. Energy level lie deeply as reduced mass \((\mu\)) increases.

2. Virial theorem is satisfied.

3. \(|\psi(0)|^2\) is simply related to force \(\biggl<\frac{\partial v}{\partial r}\biggr>\)

Proof (1) :

The Hamiltonian can be expressed as,

$$H|\psi>= E|\psi>$$

$$or,\; E= <\psi|H>|psi>$$ and $$<\psi|psi>=1$$

So, $$\frac{\partial E}{\partial \mu}= <\psi|\frac{\partial H}{\partial \mu|>}\dotsm(2)$$

From equation (1)

$$\frac{\partial H}{\partial \mu}=\frac{-\hbar^2}{2}\biggl(-\frac{1}{\mu^2}\biggr)\nabla^2= \frac{\hbar^2}{2\mu}\nabla^2= \frac{-1}{\mu}\biggl(-\frac{\hbar^2}{2\mu}\nabla^2\biggr)$$

From equation (1) we have,

$$\frac{\partial H}{\partial \mu}= -\frac{1}{\mu}(H-V)$$

$$or,\;\; \frac{\partial H}{\partial \mu}= \frac{-T}{\mu}$$

So, from equation (2), we have,

$$\frac{\partial E}{\partial \mu}= <\psi|(\frac{-T}{\mu})|\psi>$$

$$=\frac{-1}{\mu}<\psi|T|\psi>$$

$$=\frac{-<T>}{\mu}$$

Since, <T> is always positive, which show that \frac{\partial E}{\partial \mu}\) is always negative from this, we can conclude that energy level. i.e. deeply as reduced mass increases.

Proof (2) :

We consider a term \( [ \hat r, \hat p , \hat H]\)

$$or, \; [ \hat r\; \hat P, \hat H] = [ \hat r_x\hat P_x+ \hat r_y\hat P_y+ \hat r_z\hat P_z, \frac{\hat P_x}{2\mu}+v]$$

$$=[\hat r_x\hat P_x+ \hat r_y \hat P_y+ \hat r_z\hat P_z, \hat v] + i\hbar \biggl(\frac{P_x^2}{2\mu}+ \frac{p_y^2}{2\mu}+\frac{P_y^2}{2\mu}+ \frac{P_z^2}{2\mu}\biggr)$$

$$i.e\;\; [ \hat r \hat P, \hat H]= [ \hat r \hat p , \hat v] + \frac{i\hbar \hat P^2}{2\mu}\dotsm(1)$$

Using Ehnerfest theorem we get,

$$\frac{d}{dT}<\psi|\vec r, \vec P|\psi>= \frac{1}{i\hbar}[<\psi|[\vec r\vec P, \hat H]|\psi>]\dotsm(2)$$

Since, \(\hat r\) and \(\hat p\) do not explictly depends on the time the LHS equation (2) must be zero. i.e.

$$<\psi|\hat r,\hat P:\hat H|\psi>=0$$

$$or,\;\; <\psi|\hat r,\hat P, V|\psi>, <\psi|\frac{i\hbar\hat P^2}{2\mu}| \psi>=0\dotsm(3)$$

Since first term give \(i\hbar < T>\) we have,

$$\frac{i\hbar}{2\mu}<P^2>= i\hbar <T>$$

$$or,\; \frac{<P^2>}{\mu}= 2<^>$$

Which proves the virtual theorem.

Proof (3):

Now the schrodinger equation can be written as,

$$[-\frac{\hbar^2}{2\mu}-\nabla^2+ V(r)]\psi=E\psi$$

Now the radiant part of above equation can be written as,

$$\frac{d^2X(r)}{dr^2}+ \frac{2\mu}{\hbar^2}(E- V(r)) X(r)=0\dotsm(1)$$

Where, X(r)= r, R(r)

Now, \(EX-VX= \frac{-\hbar^2}{2\mu}\frac{dX^2}{dr^2}\dotsm(2)\)

Differentiating equation (2) with respect to r, we get

$$Ex'- vx'-xv'=\frac{-\hbar^2}{2\mu}x'''$$

$$or,\;\; \frac{2\mu}{\hbar^2}(E-V)x'- \frac{2\mu}{\hbar^2}v'x= -x'''$$

$$or,\;\; \frac{2\mu}{\hbar^2}\frac{(EV)x'}{X}+\frac{x'''}{x}=\frac{2\mu}{\hbar^2}\frac{dv}{dr}$$

$$or,\;\; \frac{x''}{x}-\frac{x''}{x^2}x'= \frac{2\mu}{\hbar^2}\frac{dv}{dr}$$

$$or,\;\; \frac{d}{dr}\biggl(\frac{x''}{x}\biggr)= \frac{2\mu}{\hbar^2}\frac{dv}{dr}\dotsm(3)$$

On integrating we get,

$$\int_0^\infty dr x^2 \frac{d}{dr}\biggl(\frac{x''(r)}{x(r)}\biggr)=\int_0^\infty \frac{2\mu}{\hbar^2}(\frac{dv}{dr})dr$$

$$or,\; x^2\frac{x''}{x}\biggr]_0^\infty- \int_0^\infty 2xx'\frac{x''}{x'}= \frac{2\mu}{\hbar^2}<\frac{dv}{dr}>$$

$$or,\;\; -2\int_0^\infty x'' x' dr= \frac{2\mu}{\hbar^2}<\frac{dv}{dr}>$$

$$or,\;\; -\int_0^\infty \frac{d}{dr}[x]^2 dr= \frac{2\mu}{\hbar^2}<\frac{dv}{dr}>$$

$$or,\; -|x'(0)|^2= \frac{2\mu}{\hbar^2}<\frac{dv}{dr}>$$

$$or,\;\; -4\pi|R(0)|^2= \frac{2\mu}{\hbar^2}<\frac{dv}{dr}>$$

$$or,\;\; |\psi(0)|^2= \frac{-2\mu}{\hbar^2}<\frac{dv}{dr}>$$Whcih proves that \(|\psi(0)|^2\) is simply related to force \(<\frac{dv}{dr}>\)

Reference:

  1. Mathews, P.M and K Venkatesan.A Text Book of Quantum Mechanics.New Delhi: Tata McGraw Hill Publishing Co. Ltd, 1997.
  2. Merzbacher, E.Quantum Mechanics .New York: John Wiley, 1969.
  3. Prakash, S and S Salauja.Quantum Mechanics.Kedar Nath Ram Nath Publishing Co, 2002.
  4. Singh, S.P, M.K and K Singh.Quantum Mechanics.Chand & Company Ltd., 2002.

Lesson

Central Potential Problems

Subject

Physics

Grade

Bachelor of Science

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