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A particle of mass 'm' is initially in a ground state of 1- D Harmonic oscillator potential V(x)...

  • #1
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Homework Statement


[/B]
A particle of mass 'm' is initially in a ground state of 1- D Harmonic oscillator potential V(x) = (1/2) kx2 . If the spring constant of the oscillator is suddenly doubled, then the probability of finding the particle in ground state of new potential will be?
(A) 21/4/(1+ 21/2)
(B) 25/4/(1+21/2)
(C) 2/(1+21/2)
(D) 23/2/(1+21/2)


Homework Equations


I calculated state with the wave function of one dimentional harmonic oscillator given by
Ψ = (k/πħ)¼ exp (-kx2/2ħ)

When k was doubled, new wave function becomes

Ψ'= (2k/πħ)¼ exp (-2kx2/2ħ)


The Attempt at a Solution


I tried solving the question by calculating the probability by finding the inner product of the two but I cannot solve it
2yvqid1.jpg

 

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Answers and Replies

  • #2
vela
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Your expression for the ground-state wave function has errors as the argument of the exponential isn't unitless.

Your approach is fine, but we can't really help you if you don't show your calculations.
 
  • #3
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Your expression for the ground-state wave function has errors as the argument of the exponential isn't unitless.

Your approach is fine, but we can't really help you if you don't show your calculations.
I am attaching my solution in the attachment below.
2yvqid1.jpg
 

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  • #4
vela
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To calculate an inner product, you need to evaluate an integral.
 

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