Quantum Harmonic Oscillator problem

[Caulfield]

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 = ∫0^1/2 x^2e^(-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

s=[km^(1/4) /ħ^2]*x

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

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

The Attempt at a Solution

Is the formula that I want to use here p=ψ∗(x)ψ(x) or
p=∫ψ∗(x)ψ(x)dx? (p stands for probability.)

s=[km^(1/4) /ħ^2]* (mk)^(-1/4) *√ħ/2 = 1/2

C1=1/π^2√2 so ψ1 = (1/π^2√2)e^-1/2

so, now, do I use:

p=ψ∗(x)ψ(x)= [(1/π^(2)√2)e^-1/2]^2= 1/2π^4 * e^-1

I believe that this is the right way, but the hint confuses me.

 

[vela]

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 = ∫0^1/2 x^2e^(-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

s=[km^(1/4) /ħ^2]*x

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

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

The Attempt at a Solution

Is the formula that I want to use here p=ψ∗(x)ψ(x) or
p=∫ψ∗(x)ψ(x)dx? (p stands for probability.)

s=[km^(1/4) /ħ^2]* (mk)^(-1/4) *√ħ/2 = 1/2

C1=1/π^2√2 so ψ1 = (1/π^2√2)e^-1/2

so, now, do I use:

p=ψ∗(x)ψ(x)= [(1/π^(2)√2)e^-1/2]^2= 1/2π^4 * e^-1

I believe that this is the right way, but the hint confuses me.

I don't see how you got that expression for $p$. What happened to the rest of the wave function?

 

[Caulfield]

I changed C1 when n=1 and s(d) when d = (mk)^(-1/4) *√ħ/2.

then ψ∗ψ=ψ^2 (since there is no i, ψ is real)

 

[Caulfield]

I made i typing mistake, I am sorry.

so ψ1 = C1(2s)e^-((s^2)/2)

Changing C1 and s gives ψ1 =(1/π^2√2)*(2*1/2)e^-((1/2^2)/2)

which gives ψ1 =(1/π^2√2)e^-1/8

and ψ1*ψ1=(1/2π^4)e^-1/4

 

[vela]

Where's $x$?

 

[Caulfield]

x is d.

 

[vela]

Why are you plugging $d$ in for $x$?

The problem asks you to find the probability the particle is found within a distance $d$ of the origin. What if it were to ask you find the probability the particle was found within a distance $d/2$ and $d$ of the origin? Would you still just plug in $d$ for $x$? How would the $d/2$ fit in?

 

[Caulfield]

right... so now i see it makes more sense to use ∫ψ∗(x)ψ(x)dx... ok, i got the result. thanks