A particle of mass m, in an infinite potential well of length a, has the following initial wave function at t = 0: y (x, 0) = ih 2m 5a sin (3²) + √a sin (Sax). a (3.247) and an energy spectrum En = -ħ²² n²/(2ma²). Find y(x, t) at any later time t, then calculate and the probability current density vector J(x, t) and verify that + V · J(x, t) = 0. Recall that p = y*(x, t) y(x, t) and J(x, t) = (y(x, t) Vy* (x, t) — y* (x, t) V y (x, t)).

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Exercise 3.6 and please show each step.
ability density, p(x, t), and the current density, J(x, t).
(d) Verify that the probability is conserved; that is, show that ap/at + V. J(x, t) = 0.
Exercise 3.6
A particle of mass m, in an infinite potential well of length a, has the following initial wave
function at t = 0:
3π.χ
y (x, 0) = √
√ sin (³x) + √ sin (Sax).
5a
a
ih
2m
and an energy spectrum En -ħ²²n²/(2ma²).
Find y(x, t) at any later time t, then calculate and the probability current density vector
J(x, t) and verify that + V · J(x, t) = 0. Recall that p = y*(x, t) y (x, t) and J(x, t) =
(v(x, t)Vy*(x, t)-w*(x, t) Vy(x, t)).
(3.247)
Exercise 3.7
Consider a system whose initial state at t = 0 is given in terms of a complete and orthonormal
set of three vectors: 1), 12), 103) as follows: 1 (0)) = 1/√√3|61) + A\¢2) + 1/√6103),
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40
2:10 AM
Transcribed Image Text:ability density, p(x, t), and the current density, J(x, t). (d) Verify that the probability is conserved; that is, show that ap/at + V. J(x, t) = 0. Exercise 3.6 A particle of mass m, in an infinite potential well of length a, has the following initial wave function at t = 0: 3π.χ y (x, 0) = √ √ sin (³x) + √ sin (Sax). 5a a ih 2m and an energy spectrum En -ħ²²n²/(2ma²). Find y(x, t) at any later time t, then calculate and the probability current density vector J(x, t) and verify that + V · J(x, t) = 0. Recall that p = y*(x, t) y (x, t) and J(x, t) = (v(x, t)Vy*(x, t)-w*(x, t) Vy(x, t)). (3.247) Exercise 3.7 Consider a system whose initial state at t = 0 is given in terms of a complete and orthonormal set of three vectors: 1), 12), 103) as follows: 1 (0)) = 1/√√3|61) + A\¢2) + 1/√6103), Q Search 40 2:10 AM
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