phy-4600/solutions/chap5/prob2
2016-02-28 17:15:29 -05:00

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A particle in an infinite square well has an initial state vector, with A a real number and ι the imaginary unit,
❙Ψ(t=0)❭ = A(❙φ₁❭ - ❙φ₂❭ + ι❙φ₃❭).
where ❙φₙ❭ are the energy eigenstates. This also means
❬Ψ(t=0)❙ = A(❬φ₁❙ - ❬φ₂❙ - ι❬φ₃❙)
In the energy basis,
❙φ₁❭ ≐ ⎛1⎞ ❙φ₂❭ ≐ ⎛0⎞ and ❙φ₃❭ ≐ ⎛0⎞
⎜0⎟ ⎜1⎟ ⎜0⎟
⎝0⎠, ⎝0⎠, ⎝1⎠.
So,
❙Ψ(t=0)❭ ≐ ⎛ A ⎞
⎜-A ⎟
ιA ⎠.
(𝐚) The state vector is normalized by taking the quotient of the magnitude.
❙Ψ′(t=0)❭ ≐ __͟A͟__ ⎛ 1 ⎞ = _͟1͟ ⎛ 1 ⎞
√(3A²) ⎜-1 ⎟ √3 ⎜-1 ⎟
ι ⎠ ⎝ ι ⎠.
When the Hamiltonian operates on ❙φₙ❭ with results according to the general eigenvalue equation, with Eₙ the measured energy of state n,
Ĥ❙φₙ❭ = Eₙ❙φₙ❭.
The measured energies for a point particle in an infinite square well are given by, with L the x-width of the well, and m the particle mass,
Eₙ = n͟²͟π͟²͟ħ͟².
2mL²
So,
Ĥ❙Ψ(t=0)❭ = _͟1͟ (E₁❙φ₁❭ - E₂❙φ₂❭ + ιE₃❙φ₃❭).
√3
(𝐛) There is an equal chance of measuring any of the three values, so 𝓟ₙ=1/3. The measured enemies are given by the previous expression.
Energy Probability
π͟²͟ħ͟². 𝓟₁=¹/₃
2mL²
4͟π͟²͟ħ͟². 𝓟₂=¹/₃
2mL²
9͟π͟²͟ħ͟². 𝓟₃=¹/₃
2mL²
The average value of the energy, or the expectation value, is given by
│❬Ψ❙Ĥ❙Ψ❭│² = │_͟1͟ (❬φ₁❙ - ❬φ₂❙ - ι❬φ₃❙)_͟1͟ (E₁❙φ₁❭ - E₂❙φ₂❭ + ιE₃❙φ₃❭)│²
│√3 √3 │.
❬Ψ❙Ĥ❙Ψ❭ = _͟1͟ (❬φ₁❙ - ❬φ₂❙ - ι❬φ₃❙)_͟1͟ (E₁❙φ₁❭ - E₂❙φ₂❭ + ιE₃❙φ₃❭)
√3 √3
= ¹/₃ (❬φ₁❙ - ❬φ₂❙ - ι❬φ₃❙)(E₁❙φ₁❭ - E₂❙φ₂❭ + ιE₃❙φ₃❭)
= ¹/₃ (❬φ₁❙E₁❙φ₁❭ - ❬φ₁❙E₂❙φ₂❭ + ❬φ₁❙ιE₃❙φ₃❭) (-❬φ₂❙E₁❙φ₁❭ + ❬φ₂❙E₂❙φ₂❭ - ❬φ₂❙ιE₃❙φ₃❭)
(-ι❬φ₃❙E₁❙φ₁❭ + ι❬φ₃❙E₂❙φ₂❭ - ι❬φ₃❙ιE₃❙φ₃❭).
Orthogonality is a property of the energy eigenstates:
⎧ 0 if n≠m
❬Eₙ❙Eₘ❭ = δₙₘ, where δₙₘ = ⎨ , so
⎩ 1 if n=m
❬Ψ❙Ĥ❙Ψ❭ = ¹/₃ (❬φ₁❙E₁❙φ₁❭) (❬φ₂❙E₂❙φ₂❭) (-ι❬φ₃❙ιE₃❙φ₃❭)
= ¹/₃ E₁ E₂ E₃ = ¹/₃ π͟²͟ħ͟² 4͟π͟²͟ħ͟² 9͟π͟²͟ħ͟² = _͟3͟ ⎛π͟²͟ħ͟²⎞³
2mL² 2mL² 2mL² 2 ⎝ mL²⎠.
(𝐜) Therefore, the energy expectation value
│❬Ψ❙Ĥ❙Ψ❭│² = _͟9͟ ⎛π͟²͟ħ͟²⎞⁶
4 ⎝ mL²⎠.
Because the hamiltonian is time-independent, the state vector progresses timewise according to,
❙Ψ′(t)❭ = exp(-ι͟Ĥ͟t͟ )❙Ψ′(t=0)❭ = _͟1͟ exp(-ι͟Ĥ͟t͟ ) (❙φ₁❭ - ❙φ₂❭ + ι ❙φ₃❭).
ħ √3 ħ
_͟1͟ ⎛exp(-ι͟E͟₁͟t͟ )❙φ₁❭ - exp(-ι͟E͟₂͟t͟ )❙φ₂❭ + ι exp(-ι͟E͟₃͟t͟ )❙φ₃❭⎞.
√3 ⎝ ħ ħ ħ ⎠
The coefficients for the states have the same magnitudes, so the probability of measuring a particular state at time t=0 is 1/3.
❙Ψ(t=0)❭ = _͟A͟ (αβ❙φ₁❭ - βγ❙φ₂❭ + αγι❙φ₃❭)
αβγ