Answer to Question #207304 in Quantum Mechanics for shambhu

Question #207304

The Potential Step (E > U0):

(a) Express the reflection and transmission probabilities in terms of E and U0. What is the probability

that the incident particle will be reflected for the case U0 = 0.7 E?

(b) Obtain the probability density and the real and imaginary parts of the wavefunction for t = Ο€/2 Ο‰ .

Also, plot them.


1
Expert's answer
2021-06-16T14:22:12-0400

Part a

The physics of a potential step for the case "E>U_0"

"U(x)=[_{+U_0}^0 \\space _{x>0}^{x<0}"

"u(x)=e^{ikx}+Re^{-ikx}"

"k=\\sqrt{\\frac{2mE}{\\hbar^2}}"

"u'(x)=Te^{-ik'x}"

"k'=\\sqrt{\\frac{2mE-U_0}{\\hbar^2}}"

Continuity wave function "=x=0" implies

"1+R=T"

"[ike^{ikx}-ikRe^{-ikx}]_{x=0}=[ik'Te^{ik'x}]_{x=0}"

"k(1-R)=k'(1+R)"

"(k+k')R=(k-k')"

"R= \\frac{k-k'}{k+k'}"

"T=1-R= \\frac{2k}{k+k'}"

"u(x)=[^{e^{ikx}} {\\frac{k-k'}{k+k'}}e^{-ikx}" "x<0"

"\\frac{2k}{k+k'}e^{ik'x}" "x>0"

"P_{reflection}=|R|^2=(\\frac{k-k'}{k+k'})^2"

"P_{transmission}=1-P_{reflection}=\\frac{4kk'}{(k+k')^2}"

The transmission probability goes to 1 "k=k'" (since there is no step)

The transmission probability goes for 0 "k'=0" (since the kinetic energy is zero)


Part b

"P(x,x+\u0394x)\u2248|\u03a8(x,t)|^2\u03b4x."

"P(x,x+\u0394x)=\u222b^{x+\u0394x}_x|\u03a8(x,t)|^2dx"

"P(\u2212\u221e,+\u221e)=\u222b^\u221e_{\u2212\u221e}|\u03a8(x,t)|^2dx=1."

"P(\u221e,+\u221e)=\u222b^\u221e_{\u2212\u221e}|C|2dx=1."

"P(x=0,L)=\u222b^L_0|C|^2dx=1."

"C=\\sqrt\\frac{1}{L}" where "L= \\frac{\\pi}{2}"

"P(x=0,L\/2)=\u222b^{L\/2}_0 |\\sqrt{\\frac{1}{\\frac{\\pi}{2}}}|^2"

"=(\\frac{1}{0.5 \\pi}) \\frac{0.5 \\pi}{2} = 0.25"


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