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Consider a state|ε> that is in a quantum superposition of two particle-in-a-box energy eigenstates corresponding to n=2,3, i.e.: |ε> = ,[1/(2^.5)][|2> + |3>], or equivalently:

ε(x) = [1/(2^.5)][ψ_{2}(x) + ψ_{3}. Compute the expectation value of momentum: <p> = <ε|[itex]\widehat{}p[/itex]|ε>.

Relavent equations:

[itex]\widehat{}p[/itex] = -i[STRIKE]h[/STRIKE](∂ψ/∂x)

for n=2,...; ψ = sqrt(2/L) sin(2πx/L)

Attempt at solution:

I've computed <ε|[itex]\widehat{}p[/itex]|ε>

= .5[ <2| + <3| ] [itex]\widehat{}p[/itex] [ |2> + |3> ]

= .5 [<2|[itex]\widehat{}p[/itex]|2> + <3|[itex]\widehat{}p[/itex]|2> + <2|[itex]\widehat{}p[/itex]|3> + <3[itex]\widehat{}p[/itex]|3>]

= -i[STRIKE]h[/STRIKE]/L [ ∫^{L}_{0}sin(2πx/L)(∂ψ/∂x)sin(2πx/L) + sin(3πx/L)(∂ψ/∂x)sin(2πx/L) + sin(2πx/L)(∂ψ/∂x)sin(3πx/L) +sin(3πx/L)(∂ψ/∂x)sin(3πx/L)

= -i[STRIKE]h[/STRIKE]/L [ ∫^{L}_{0}(L/2π)sin(2πx/L)cos(2πx/L) + (L/2π)sin(3πx/L)cos(2πx/L) + (L/3π)sin(2πx/L)cos(3πx/L) + (L/3π)sin(3πx/L)cos(3πx/L)

I've tried to:

1) convert the trigs to exponentials and work through exponentials

2) directly integrate all of these

3) convert the trigs into different trig identities

I've been told that the answer = 0 but the closest I got when I converted to exponentials was 1.

Sorry if the text doesn't look right, this is my first time. the "h" with the cross through it is plancks constant divided by 2π

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# Expectation value for momentum operator using Dirac Notation

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