Tunneling Effect
Quantum tunnelling or tunneling refers to the quantum mechanical phenomenon where a particle tunnels through a barrier that it classically could not surmount. This plays an essential role in several physical phenomena, such as the nuclear fusion that occurs in main sequence stars like the Sun.[1] It has important applications to modern devices such as the tunnel diode,[2] quantum computing, and the scanning tunnelling microscope. The effect was predicted in the early 20th century and its acceptance as a general physical phenomenon came mid-century.[3] Tunnelling is often explained using the Heisenberg uncertainty principle and the wave–particle duality of matter. Pure quantum mechanical concepts are central to the phenomenon, so quantum tunnelling is one of the novel implications of quantum mechanics.
Summary
Quantum tunnelling or tunneling refers to the quantum mechanical phenomenon where a particle tunnels through a barrier that it classically could not surmount. This plays an essential role in several physical phenomena, such as the nuclear fusion that occurs in main sequence stars like the Sun.[1] It has important applications to modern devices such as the tunnel diode,[2] quantum computing, and the scanning tunnelling microscope. The effect was predicted in the early 20th century and its acceptance as a general physical phenomenon came mid-century.[3] Tunnelling is often explained using the Heisenberg uncertainty principle and the wave–particle duality of matter. Pure quantum mechanical concepts are central to the phenomenon, so quantum tunnelling is one of the novel implications of quantum mechanics.
Things to Remember
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The Quantum Mechanical behaviour of particle for finite value of \(V_0>E\) is obtained by solving one dimensional time independent Schrodinger equation,
$$\frac{-\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2}+ V(x)\psi(x)= E\psi(x)$$
- $$k_1= \sqrt{\frac{2mE}{\hbar^2}}$$
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Tunneling Effect
Tunneling effect: (\(E<V_0\))
fig: Potential Barrier ( case of \(E<V_0\))
Consider a flux of particles having same energy 'E' and mass 'm' be incident from left on the potential ( Barrier ) of the form,
$$V(x)= V_0\; for\; x\leq x\leq a$$
$$= 0 \; for \; x<0$$
$$=0\; for \; x>a$$
According to classical mechanics, for the energy of incident particle E less than height of potential barrier, all particle must be reflected in region (I). The probability of transmission from x>0 is zero. So x>0 is classically forbidden region.
The Quantum Mechanical behaviour of particle for finite value of \(V_0>E\) is obtained by solving one dimensional time independent Schrodinger equation,
$$\frac{-\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2}+ V(x)\psi(x)= E\psi(x)$$
$$or,\; \frac{d^2\psi(x)}{dx^2}+ \frac{2m}{\hbar^2}[E-V(x)]\psi(x)=0\dotsm(2)$$
In region (I) V(x)=0, \(\psi(x)= \psi_I(x)\) , equation (2) becomes,
$$\frac{d^2\psi_I(x)}{dx^2}+\frac{2mE}{\hbar^2}\psi_I(x)=0$$
$$or,,\; \frac{d^2\psi_I(x)}{dx^2}+K_1^2 \psi_I(x)=0$$
$$Where,\;\; k_1= \sqrt{\frac{2mE}{\hbar^2}}\dotsm(4)$$
The oscillatory solution of (3) is,
$$\psi_I(x)= Ae^{ik_1 x}+ B^{-ik_1 x}\dotsm(5)$$
In region (II)
$$\psi(x)=\psi_{II}(x), V(x)= V_0>E$$
Equation (2) become
$$\frac{d^2\psi_{II}(x)}{dx^2}-\frac{2m}{\hbar^2}[V_0-E]\psi_{II}(x)=0$$
$$or,\; \frac{d^2\psi_{II}(x)}{dx^2}- k_2^2 \psi_{II}(x)=0\dotsm(6)$$
$$Where,\; K_2= \sqrt{\frac{2m(V_0-E}{\hbar^2}}\dotsm(7)$$
The Non-oscillatory solution of equation (6) is,
$$\psi_{II}(x)= Ce^{k_2x}+ De^{-k_2x}\dotsm(8)$$
In region (III) V(x)=0
Equation (2) becomes,
$$\frac{d^2\psi_{III}(x)}{dx^2}+\frac{2mE}{\hbar^2}\psi_{III}(x)=0$$
$$or,\;\; \frac{d^2\psi_{III}(x)}{dx^2}+k_1^2 \psi_{III}(x)=0\dotsm(9)$$
Solution of equation (9) is oscillatory and given by,
$$\psi_{III}(x)= Fe^{ik_1 x}+ Ge^{-ik_1 x}$$
Since there is no barrier in region (III) i.e. x>a; so [G=0]
$$\psi_{III}(x)= Fe^{ik_1 x}\dotsm(10)$$
Using Boundary condition at x=0
$$\psi_I(x)|_{x=0}= \psi_{II}(x)|_{x=0}$$
$$A+B= C+D\dotsm(11)$$
and
$$\frac{d\psi_I(x)}{dx}\biggl|_{x=0}= \frac{d\psi_{II}(x)}{dx}\biggr|_{x=0}$$
$$\Rightarrow A-B= \frac{k_2}{ik_1}(C-D)\dotsm(12)$$
Adding (11) and (12) we get,
$$A= \biggl[1+\frac{k_2}{ik_1}\biggr]\frac c2 + \frac D2 \biggl[1- \frac{k_2}{|k_1}\biggr]\dotsm(13)$$
Subtracting (12) from (11)
$$B=\biggl[1-\frac{k_2}{|k_1}\biggr]\frac C2 + \frac D2 \biggl[1+ \frac{k_2}{ik_1}\biggr]\dotsm(14)$$
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
One Dimensional Quantum Mechanical Problems
Subject
Physics
Grade
Bachelor of Science
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