Minimum average value of position

  • #1
tanaygupta2000
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Homework Statement:
Let ψ₀(x) and ψ₁(x) be the wave functions corresponding to the ground state and the first excited states of a one dimensional harmonic oscillator respectively. Consider the normalized state ϕ(x) = αψ₀(x) + βψ₁(x), where α and β are real numbers. The values of α and β for which <x>, the average value of the position is a minimum are:

(a) α = -β = 1/√2
(b) α = β = 1/√2
(c) α = 1/√3 and β = -√(2/3)
(d) α = 1/√3 and β = √(2/3)
Relevant Equations:
For one-dimensoinal harmonic oscillator,
Ground state wavefunction, ψ₀(x) = (mw/πℏ)^(1/4) exp[-(mw/2ℏ)x^2]
First excited state wavefunction, ψ₁(x) = (mw/πℏ)^(1/4) * x√(2mw/ℏ) * exp[-(mw/2ℏ)x^2]

Average value of position, <x> = ∫xϕ(x) dx
After getting the values of ψ₀(x) and ψ₁(x), I put them in the expression of ϕ(x) to get:
ϕ(x) = (mw/πℏ)^(1/4) * exp[-(mw/2ℏ)x^2] * [α + βx√(2mw/ℏ)]

Now when attempting to find the value of <x> by ∫xϕ(x) dx, I am having trouble determining the limits, as I am getting nothing useful by integrating from -∞ to ∞.
Also, do I even need to find <x>, can't I equate the derivative of simply ϕ(x) to 0 to get the minima, as <x> ∝ ϕ(x) which is relatively easier to differentiate?
Kindly help !
 

Answers and Replies

  • #3
PeroK
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Also, do I even need to find <x>, can't I equate the derivative of simply ϕ(x) to 0 to get the minima, as <x> ∝ ϕ(x) which is relatively easier to differentiate?
Kindly help !

By "minimum" value, I think they mean the greatest negative value of ##\langle x \rangle##. They do not mean ##\langle x \rangle = 0##. You could do that with ##a = 1, b = 0##, for example.

I think you have to calculate ##\langle x \rangle##, unless there is a clever trick I'm missing.
 
  • #4
Chandra Prayaga
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Also, do I even need to find <x>, can't I equate the derivative of simply ϕ(x) to 0 to get the minima, as <x> ∝ ϕ(x) which is relatively easier to differentiate?
As PeroK pointed out, your expression for <x> is wrong.
Also note that there is a difference between differentiating <x> and differentiating either ϕ(x) or its square which is the probability density: The probability density is a function of x, whereas <x> is not. <x> is a definite integral. <x> is now dependent on a and b. The statement of the problem requires you to find the values of a and b such that <x> is minimum.
 
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