Question 3.1: (a) Classically (i.e, from Maxwell’s equations), change in c......
(a) Classically (i.e, from Maxwell’s equations), change in charge density, ρ, is related to divergence of current density, J. Use this fact and the time-dependent Schrödinger wave equation for particles of mass m and charge e to derive the current density
J=-\frac{ie\hbar }{2m} \left(\psi ^{*} \nabla \psi -\psi \nabla \psi ^{*} \right)
(b) If a wave function can be expressed as \psi \left(x,t\right) =Ae^{i(kx-\omega t)} +Be^{i(-kx-\omega t)}, show that particle flux is proportional to A²−B².
(c) For current to flow through a tunnel barrier, the wave function must contain both left- and right-propagating exponentially decaying solutions with a phase difference. If \psi \left(x,t\right) =Ae^{kx-i\omega t} +Be^{-kx-i\omega t}, show that particle flux is proportional to Im(AB^{*}). Hence, show that if \psi \left(x,t\right) = Be^{-kx-i\omega t} particle flux is zero.
(d) Show that a particle of energy E and mass m moving from left to right in a onedimensional potential in such a way that V(x)=0 for x\lt 0 and V(x)=V_{0} for x\geq 0 has unity reflection probability if E\lt V_{0}.
(a) We will use the classical expression relating the divergence of charge density to current density to obtain an expression for the quantum mechanical current. The change in charge density ρ is related to the divergence of current density J through
\frac{\partial }{\partial t} \rho (r, t)=-\nabla \cdot J(r, t)In quantum mechanics, we assume that the charge density of a particle with charge e and wave function \psi can be written \rho =e\left|\psi \right| ^{2}, so that
\frac{\partial \rho }{\partial t} =e\frac{\partial }{\partial t} (\psi ^{*}\psi )=e\left(\psi ^{*}\frac{\partial \psi }{\partial t}+\psi \frac{\partial \psi ^{*} }{\partial t} \right)We now make use of the fact that the time-dependent Schrödinger equation for a particle of mass m moving in a real potential is
i\hbar \frac{\partial }{\partial t} \psi (r,t)=\left(-\frac{\hbar ^{2} }{2m}\nabla ^{2} +V(r) \right)\psi (r, t) =\hat{H}\psi (r,t)Multiplying both sides by \psi^{*} (r,t) gives
i\hbar \psi ^{*} (r, t)\frac{\partial }{\partial t} \psi (r, t)=-\frac{\hbar ^{2} }{2m} \psi ^{*} (r, t)\nabla ^{2} \psi (r, t)+\psi ^{*} (r, t)V(r)\psi (r, t)which, when multiplied by e/i\hbar, is the first term on the right-hand side in our expression for \partial \rho /\partial t. To find the second term, one takes the complex conjugate of the time-dependent Schrödinger equation and multiplies both sides by \psi (r, t). Because V(r) is taken to be real, we have \psi ^{*} (r, t)V(r)\psi (r, t)=\psi (r, t)V(r)\psi ^{*} (r, t), giving the rate of change of charge density
\frac{\partial \rho }{\partial t} =-\psi ^{*} \frac{e\hbar }{i2m} \nabla^2 \psi +\psi \frac{e\hbar }{i2m} \nabla^2 \psi ^{*} +\frac{e\psi ^{*}\psi }{i\hbar } (V(r)-V(r))\frac{\partial \rho }{\partial t}=\frac{ie\hbar }{2m} (\psi ^{*}\nabla^2 \psi -\psi \nabla^2 \psi ^{*} )
\frac{\partial \rho }{\partial t} =\frac{ie\hbar }{2m} \nabla \cdot (\psi ^{*}\nabla \psi -\psi \nabla \psi ^{*} )=-\nabla .J
Hence, an expression for the current density for a particle charge e described by state \psi is
J=-\frac{ie\hbar }{2m} (\psi ^{*}\nabla \psi -\psi \nabla \psi ^{*} )This expression has an obvious symmetry, some consequences of which we will now explore.
(b) In this exercise we will work with a particle of mass m and charge e that is described by a wave function \psi \left(x,t\right) =Ae^{i(kx-\omega t)} +Be^{i(-kx-\omega t)} , where k is real. The aim is to show that particle flux is proportional to A²−B².
Starting with the current density in one dimension and substituting in the expression for the wave function gives
J_{x} =-\frac{ie\hbar }{2m}\left(\psi ^{*} \frac{\partial }{\partial x}\psi -\psi \frac{\partial \psi ^{*} }{\partial x} \right)\psi ^{*} \frac{\partial \psi }{\partial x} =(A^{*}e^{-ikx}+B^{*} e^{ikx} )(ikAe^{ikx}-ikBe^{-ikx} )
\psi ^{*} \frac{\partial \psi }{\partial x} =ik\left|A\right| ^{2} -ikA^{*} Be^{-2ike} +ikB^{*}Ae^{2ikx} -ik\left|B\right|^{2}
and
\psi \frac{\partial \psi ^{*} }{\partial x} =(Ae^{ikx} +Be^{-ikx} ) (-ikA^{*}e^{-ikx} +ikB^{*} e^{ikx} )\psi \frac{\partial \psi ^{*} }{\partial x} = -ik\left|A\right|^{2} -ikBA^{*} e^{-2ikx} +ikAB^{*}e^{2ikx} +ik\left|B\right| ^{2}
so that, finally,
J_{x} =-\frac{ie\hbar }{2m}(ik2(\left|A\right|^{2} -\left|B\right|^{2} )) =\frac{e\hbar k}{m} (\left|A\right|^{2} -\left|B\right|^{2} )This is not an unexpected result, since the wave function \psi \left(x,t\right) is made up of two traveling waves. The first traveling wave Ae^{ikx} might consist of an electron probability density \rho =\left|A\right| ^{2} moving from left to right at velocity v=\hbar k/m carrying current density j_{+} =ev_{+} \rho =e\hbar k\left|A\right| ^{2} /m. The second traveling wave Be^{-ikx} consists of an electron probability \rho =\left|B\right| ^{2} moving from right-to-left at velocity v=-\hbar k/m carrying current density j_{-} =ev_{-} \rho =-e\hbar k\left|B\right| ^{2} /m. The net current density is just the sum of the currents J_{x}=j_{+} +j_{-} =e\hbar k(\left|A\right|^{2} -\left|B\right|^{2} )/m.
(c) A particle of mass m and charge e is described by the wave function \psi \left(x,t\right) =Ae^{kx-i\omega t} +Be^{-kx-i\omega t}, where \kappa is real. We wish to show that for current to flow through a tunnel barrier the wave function must contain both left- and right-propagating exponentially decaying solutions with a phase difference.
As in part (b) we start with the current density in one dimension and substitute in the expression for the wave function. This gives
J_{x} =-\frac{ie\hbar }{2m}\left(\psi ^{*} \frac{\partial }{\partial x}\psi -\psi \frac{\partial \psi ^{*} }{\partial x} \right)\psi ^{*} \frac{\partial \psi }{\partial x} = (A^{*}e^{\kappa x} +B^{*}e^{-\kappa x} )(A\kappa e^{\kappa x} -B\kappa e^{-\kappa x} )
\psi^{*} \frac{\partial \psi }{\partial x} =\kappa \left|A\right|^{2} e^{2\kappa x} +\kappa B^{*}A -\kappa A^{*}B -\kappa \left|B\right|^{2} e^{-2\kappa x}
and
\psi \frac{\partial \psi^{*} }{\partial x} =(Ae^{\kappa x}+Be^{-\kappa x} )(A^{*}\kappa e^{\kappa x} -B^{*} \kappa e^{-\kappa x})\psi \frac{\partial \psi^{*} }{\partial x} =\kappa \left|A\right|^{2} e^{2\kappa x} -\kappa AB^{*} +\kappa BA^{*} -\kappa \left|B\right|^{2}e^{-2\kappa x}
so that
J_{x} =-\frac{ie\hbar }{2m} 2\kappa (A^{*}B-B^{*} A )=-\frac{2e\hbar \kappa }{m} (Im(AB^{*} ))To show this last equality, one starts by explicitly writing the real and imaginary parts of the terms in the brackets
(A^{*} B-B^{*} A)=(A_{Re}-iA_{Im} )(B_{Re} +iB_{Im} )-(B_{Re} -iB_{Im})(A_{Re}+iA_{Im})Multiplying out the terms on the right-hand side gives
(A^{*} B-B^{*} A)=A_{Re}B_{Re} +iA_{Re} B_{Im} -iA_{Im} B_{Re} +A_{Im} B_{Im} -A_{Re} B_{Re} -iB_{Re} A_{Im} +iB_{Im} A_{Re} -A_{Im} B_{Im}This simplifies to
(A^{*} B-B^{*} A)=2iA_{Re} B_{Im} -2iA_{Im} B_{Re}=2i(A_{Re}B_{Im} -A_{Im} B_{Re} )Taking the imaginary part of the terms on the right-hand side allows one to add the real terms A_{Re}B_{Re} and A_{Im}B_{Im}, so that the expression may be factored
(A^{*}B-B^{*}A)=-2i(Im(A_{Re}B_{Re}-iA_{Re} B_{Im}+ iA_{Im}B_{Re} +A_{Im} B_{Im} ))(A^{*}B-B^{*}A)=-2i(Im((A_{Re}+iA_{Im} )(B_{Re}-iB_{Im} )))=-2i(Im(AB^{*} ))
J_{x} =-\frac{ie\hbar }{2m} 2\kappa (A^{*} B-B^{*}A )=-\frac{2e\hbar \kappa }{m} (Im(AB^{*} ))
The significance of this result is that current can only flow through a one dimensional tunnel barrier in the presence of both exponentially decaying terms in the wave function \psi \left(x,t\right) =Ae^{\kappa x} +Be^{-\kappa x}. In addition, the complex coefficients A and B must differ in phase. The right-propagating term Ae^{\kappa x} can only carry current if the left-propagating term Be^{-\kappa x} exists. This second term indicates that the tunnel barrier is not opaque and that transmission of current via tunneling is possible.
Suppose a particle of mass m and charge e is described by wave function \psi \left(x,t\right) = Be^{-\kappa x-i\omega t}. Substitution into the expression for the one-dimensional current density gives
J_{x} =-\frac{ie\hbar }{2m}\left(\psi ^{*}\frac{\partial }{\partial x} \psi -\psi \frac{\partial \psi ^{*} }{\partial x} \right) =-\frac{ie\hbar }{2m} (B^{*}e^{-\kappa x}(-B\kappa e^{-\kappa x} )-Be^{-\kappa x}(-B^{*}\kappa e^{-\kappa x} ) )J_{x} =-\frac{ie\hbar }{2m}(-B^{*}B\kappa e^{-2\kappa x} +BB^{*}\kappa e^{-2\kappa x} )=0
(d) To show that a particle of energy E and mass m moving in a one-dimensional potential in such a way that V(x) = 0 for x < 0 in region 1 and V(x) = V_{0} for x ≥ 0 in region 2 has unity reflection probability if E < V_{0}, we write down solutions of the time-independent Schrödinger equation \left(\frac{-\hbar ^{2} }{2m} \frac{d^{2} }{dx^{2} }+V(x) \right) \psi (x)=E\psi (x). The solutions that describe left- and right-traveling waves in the two regions are
\psi _{1} (x)=Ae^{ik_{1}x } +Be^{-ik_{1}x } for x < 0
and
\psi _{2} (x) =Ce^{ik_{2}x}+De^{-ik_{2}x} for x ≥ 0
When E < V_{0} , the value of k_{2} is imaginary, and so k_{2} → ik_{2} . The value of D is zero to avoid a wave function with infinite value as x\rightarrow \infty. The value of C is finite, but the contribution to the wave function \psi _{2} (x\rightarrow \infty )=Ce^{-k_{2}x}\mid _{x\rightarrow \infty } =0. Hence, we may conclude that the transmission probability in the limit x\rightarrow \infty must be zero, since \left|\psi _{2} (x\rightarrow \infty )\right| ^{2} = 0.
From this it follows that \left|A\right| ^{2} = \left|B\right| ^{2} = 1, which has the physical meaning that the particle is completely reflected at the potential step when E < V_{0}.
Alternatively, one could calculate the reflection probability from the square of the amplitude coefficient for a particle of energy E and mass m incident on a one-dimension potential step
B=\left(\frac{1 – k_{2}/k_{1} }{1 + k_{2}/k_{1} } \right)When particle energy E < V_{0}, the value of k_{2} is imaginary, so that k_{2} → ik_{2}, which gives
\left|B\right| ^{2} =B^{*}B=\left(\frac{1 + ik_{2}/k_{1} }{1 – ik_{2}/k_{1} }\right)\left(\frac{1 – ik_{2}/k_{1} }{1 + ik_{2}/k_{1} }\right) =1From this one may conclude that the particle has a unity reflection probability.