QM: Harmonic Oscillator wave function

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Homework Statement


For the n = 1 harmonic oscillator wave function, find the probability p that, in an experiment which measures position, the particle will be found within a distance d = (mk)-1/4√ħ/2 of the origin. (Hint: Assume that the value of the integral α = ∫01/2 x2e-x2/2 dx is known and express your result as a function of α)

Homework Equations


distance from 0 to d; d = (mk)-1/4√ħ/2

Normalize condition: Cn = 1/ (π2√2nn!)

Harmonic Oscillator wave function for n = 1 ψ1 = C1(2s)e-s2/2

Probability density
∫ψn(x)*ψ(x)

s = (km)1/41/2 x

The Attempt at a Solution


I first plugged in s and normalized condition into the harmonic oscillator wave function.
ψ(s) = (km/π)1/42x/√2ħ e-√(km)x2/2ħ

∫ψ(s)*ψ(s) ?
I'm not sure if this is the right approach to tackling this problem.
 

Answers and Replies

  • #2
fzero
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Probability density
∫ψn(x)*ψ(x)

##\psi^*(x)\psi(x)## is the probability density. This means that it is the probability to find the particle in the interval from ##x## to ##x+dx##. Can you think of a way to use this to determine the probability to find the particle in a certain interval?
 
  • #3
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the intervals would be from ∫-dd. Is there a way to simplify down the
distance given?
d = (mk)-1/4√h-bar/2

s = [(km)1/4/√h-bar] x

d = x/s√2 ; -d = -x/s√2 this doesn't seem right.

the integral, α = ∫-dd x2e-x2/2
 
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  • #4
fzero
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the intervals would be from ∫-dd. Is there a way to simplify down the
distance given?
d = (mk)-1/4√h-bar/2

s = [(km)1/4/√h-bar] x

d = x/s√2 ; -d = -x/s√2 this doesn't seem right.

the integral, α = ∫-dd x2e-x2/2

I think you're confusing yourself by having too many variables defined. The distance d is given in the units of the x variable, so set the probability up as an integral over x. Then you can make a change of variable to one defined like the s variable. Your limits of integration will become very simple and you should be able to express the result in terms of the ##\alpha## integral.
 
  • #5
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-ss ψ*(x)ψ(x) dx ?

P = C24s2e-s2/2 + eis2/2

x = [√h-bar/mk1/4] * 1/√2
s = [(km)1/4/√h-bar] x
s = 1/√2
so...
this doesn't seem right, am I suppose to get a function out of probability density?
 
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  • #6
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I am looking for the answer for this one too, I had it on my exam few days ago
 
  • #7
fzero
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-ss ψ*(x)ψ(x) dx ?

P = C24s2e-s2/2 + eis2/2

x = [√h-bar/mk1/4] * 1/√2
s = [(km)1/4/√h-bar] x
s = 1/√2
so...
this doesn't seem right, am I suppose to get a function out of probability density?

Why does your first line have s as the limit of integration? Where did P in the 2nd line come from?

As you deduced in an earlier post, the limits of integration are given by d. Then we can use the probability density to determine the desired probability:

$$ P = \int_{-d}^d \psi^*(x) \psi(x) dx.$$

You should rewrite this expression using the normalized ##\psi(x)##. It should be possible to find a change of variable such that

$$ P = A \int_{-1/2}^{1/2} y^2 e^{-y^2} dy,$$

where ##A## is a constant that we can express in terms of ##m,k,\hbar##.
 
  • #8
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-So we have ψ1 = C1 2se-s2/2
as our wave function of this specific harmonic oscillator n = 1

- Probability density is given as ψ*ψ and within the intervals of -d to d
giving us ∫d-d ψ*(x)ψ(x)dx

ψ*ψ = (C1(2s)es2/2)(C1(2s)e-s2/2) = C2(2s)2

I'm still not getting this problem..and how the probability P is rewritten in the form of
A∫1/2-1/2y2e-y2dy
 
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  • #9
fzero
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-So we have ψ1 = C1 2se-s2/2
as our wave function of this specific harmonic oscillator n = 1

- Probability density is given as ψ*ψ and within the intervals of -d to d
giving us ∫d-d ψ*(x)ψ(x)dx

ψ*ψ = (C1(2s)es2/2)(C1(2s)e-s2/2) = C2(2s)2

This wavefunction is real. In particular

$$ \left( e^{-s^2/2} \right)^* = e^{-s^2/2} $$

so the exponentials do not cancel out.

I'm still not getting this problem..and how the probability P is rewritten in the form of
A∫1/2-1/2y2e-y2dy

Go back and try this again with the correct value of ##\psi^* \psi##. I meant what I said when I suggested to write out the expression in terms of ##x##. It seems pointless for me to keep offering suggestions when you won't bother following any of them.
 
  • #10
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ψ(x) = C2xe-x2/2
ψ*ψ = |C2xe-x2/2|2

∫|C2xe-x2/2|2 dx
= C2∫4x2ex4/4 is this correct, all terms in x

so only when an exponential expression is cancelled out using pd is when it function is imaginary :ex: e-ix

Now by doing what I did above is the pdf corrected? I squared the function and set it under the integral of x.
 
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  • #11
vela
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No, it's not correct. You have a definition for s in terms of x. You need to use that. You don't simply replace s by x. You're making it way too complicated. This is just basic algebra.
 
  • #12
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I will try this from scratch
 
  • #13
fzero
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Hopefully it will help to note the following. When you wrote

The Attempt at a Solution


I first plugged in s and normalized condition into the harmonic oscillator wave function.
ψ(s) = (km/π)1/42x/√2ħ e-√(km)x2/2ħ

back in the OP, you were already confusing the s and x variables. The 1st excited state wavefunction is given in terms of x by (I am assuming that your normalization and the rest of your formula is correct, please check with your text to make sure.)

$$\psi(x) = \left(\frac{km}{\pi}\right)^{1/4} \frac{2x}{\sqrt{2}\hbar} \exp\left[ - \frac{\sqrt{km} x^2 }{2\hbar} \right].$$

This is the expression you want to start with. The change of variable to s is defined to make the argument of the exponential in the probability density come out as ##-s^2##, but you can verify this yourself.
 
  • #14
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Hopefully it will help to note the following. When you wrote



back in the OP, you were already confusing the s and x variables. The 1st excited state wavefunction is given in terms of x by (I am assuming that your normalization and the rest of your formula is correct, please check with your text to make sure.)

$$\psi(x) = \left(\frac{km}{\pi}\right)^{1/4} \frac{2x}{\sqrt{2}\hbar} \exp\left[ - \frac{\sqrt{km} x^2 }{2\hbar} \right].$$

This is the expression you want to start with. The change of variable to s is defined to make the argument of the exponential in the probability density come out as ##-s^2##, but you can verify this yourself.

Now with this expression ψ(x) I can find the probability density (ψ*(x)ψ(x)) = √(km/π) 2x2/hbar2 exp[-(km)x4/4hbar2
 
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