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2. A system at initial time t= 0 with wave function (x, 0) = Ae¬alxl propagates freely (no forces). Normalize p(x, 0), i.e., find A. Let k = and determine (x, 0) in momentum space, i.e., p(k,0). а. b.

2. A system at initial time t= 0 with wave function (x, 0) = Ae¬alxl propagates freely (no forces). Normalize p(x, 0), i.e., find A. Let k = and determine (x, 0) in momentum space, i.e., p(k,0). а. b.

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2. A system at initial time t= 0 with wave function (x, 0) = Ae¬alxl propagates freely (no forces). Normalize p(x, 0), i.e., find A. Let k = and determine (x, 0) in momentum space, i.e., p(k,0). а. b.
Transcribed Image Text:2. A system at initial time t= 0 with wave function (x, 0) = Ae¬alxl propagates freely (no forces). Normalize p(x, 0), i.e., find A. Let k = and determine (x, 0) in momentum space, i.e., p(k,0). а. b.
Expert Solution
GIVEN:

The initial time of the wave function is given as, t = 0

The wave function is given as, Ψ(x,0) = Ae-ax

The value of k is given as, k = ph

TO DETERMINE:

(a) Normalize Ψx,0 and find the value of A

(b) Ψx,0 in momentum space, that is, Ψpk,0

 

SOLUTION PART-1:

(a) Normalize Ψ(x,0) and find the value of A:

The normalization of wave function is given as,

∫-∞∞Ψx,02dx = 1 ---------------------- (1)

Substituting the values in equation (1), we have,

∫-∞∞Ae-ax2dx = 1

∫-∞∞A2e-2axdx = 1 (or)

A2∫-∞∞e-2axdx = 1 -------------------- (2)

The integral is an even function for the symmetric function. Therefore, equation (2) is given as,

2A2∫0∞e-2axdx = 1

Integrating the above equation, we get,

2A2e-2ax-2a0∞ = 1

Applying the limits, we get,

2A2-1-2a = 1

A2a = 1 (or) A2 = a

The value of A is calculated as,

A = a

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