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
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$$=\frac{-<T>}{\mu}$$
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.
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$$or,\; \frac{<P^2>}{\mu}= 2<^>$$
Which proves the virtual theorem.
- $$or,\;\; |\psi(0)|^2= \frac{-2\mu}{\hbar^2}<\frac{dv}{dr}>$$Whcih proves that \(|\psi(0)|^2\) is simply related to force \(<\frac{dv}{dr}>\)
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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:
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:
- Mathews, P.M and K Venkatesan.A Text Book of Quantum Mechanics.New Delhi: Tata McGraw Hill Publishing Co. Ltd, 1997.
- Merzbacher, E.Quantum Mechanics .New York: John Wiley, 1969.
- Prakash, S and S Salauja.Quantum Mechanics.Kedar Nath Ram Nath Publishing Co, 2002.
- 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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