- #1
tanaygupta2000
- 208
- 14
- 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 !
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 !