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working on chap 5 homework
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notes/fourier
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notes/fourier
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general series
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∞
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f(x) ~ ∑ cₙ φₙ(x)
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ⁿ⁼⁰
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cₙ = ❬͟f͟,φ͟ₙ͟❭͟ᵥ where v is some modulo weighting
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‖φₙ‖ᵥ²
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@ -47,21 +47,23 @@ Energy eigenstates obey the following properties:
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Bra-ket Notation Wavefunction Notation
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Bra-ket Notation Wavefunction Notation
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Normalization
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Normalization
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⌠∞
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⌠∞
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❬Eₙ❙Eₙ❭ = 1 ⎮ │φₙ(x)│² dx = 1
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❬Eₙ❙Eₙ❭ = 1 ⎮ │φₙ(x)│² dx = 1
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⌡-∞
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⌡-∞
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Orthogonality
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Orthogonality
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⌠∞
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⌠∞
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❬Eₙ❙Eₘ❭ = δₙₘ ⎮ φₙ⃰(x) φₘ(x) dx = δₙₘ
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❬Eₙ❙Eₘ❭ = δₙₘ ⎮ φₙ⃰(x) φₘ(x) dx = δₙₘ
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⌡-∞
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⌡-∞
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Completeness
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Completeness
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⌠∞
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❬Eₙ❙Eₘ❭ = δₙₘ ⎮ φₙ⃰(x) φₘ(x) dx = δₙₘ
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❙Ψ❭ = ∑ cₙ ❙Eₙ❭ Ψ(x) = ∑ cₙ φₙ(x)
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⌡-∞
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ⁿ ⁿ
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@ -1,17 +1,12 @@
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A particle in an infinite square well has an initial state vector
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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,
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❙Ψ(t=0)❭ = A(❙φ₁❭ - ❙φ₂❭ + ι❙φ₃❭).
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❙Ψ(t=0)❭ = A(❙φ₁❭ - ❙φ₂❭ + ι❙φ₃❭).
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where ❙φₙ❭ are the energy eigenstates. This also means
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where ❙φₙ❭ are the energy eigenstates. This also means
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❬Ψ(t=0)❙ = A⃰(❬φ₁❙ - ❬φ₂❙ + ι❬φ₃❙)
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❬Ψ(t=0)❙ = A(❬φ₁❙ - ❬φ₂❙ - ι❬φ₃❙)
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❙Ψ(t=0)❭ = _͟A͟ (αβ❙φ₁❭ - βγ❙φ₂❭ + αγι❙φ₃❭)
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αβγ
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In the energy basis,
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In the energy basis,
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❙φ₁❭ ≐ ⎛1⎞ ❙φ₂❭ ≐ ⎛0⎞ and ❙φ₃❭ ≐ ⎛0⎞
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❙φ₁❭ ≐ ⎛1⎞ ❙φ₂❭ ≐ ⎛0⎞ and ❙φ₃❭ ≐ ⎛0⎞
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@ -23,11 +18,94 @@ So,
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⎜-A ⎟
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⎜-A ⎟
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⎝ιA ⎠.
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⎝ιA ⎠.
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(𝐚) Multiplying the state vector by its magnitude normalizes it.
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(𝐚) The state vector is normalized by taking the quotient of the magnitude.
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❙Ψ′(t=0)❭ ≐ __͟A͟__ ⎛ 1 ⎞ = _͟1͟ ⎛ 1 ⎞
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√(3A²) ⎜-1 ⎟ √3 ⎜-1 ⎟
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⎝ ι ⎠ ⎝ ι ⎠.
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When the Hamiltonian operates on ❙φₙ❭ with results according to the general eigenvalue equation, with Eₙ the measured energy of state n,
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Ĥ❙φₙ❭ = Eₙ❙φₙ❭.
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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,
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Eₙ = n͟²͟π͟²͟ħ͟².
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2mL²
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So,
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Ĥ❙Ψ′(t=0)❭ = _͟1͟ (E₁❙φ₁❭ - E₂❙φ₂❭ + ιE₃❙φ₃❭).
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√3
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(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.
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Energy Probability
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π͟²͟ħ͟². 𝓟₁=¹/₃
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2mL²
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4͟π͟²͟ħ͟². 𝓟₂=¹/₃
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2mL²
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9͟π͟²͟ħ͟². 𝓟₃=¹/₃
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2mL²
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The average value of the energy, or the expectation value, is given by
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│❬Ψ′❙Ĥ❙Ψ′❭│² = │_͟1͟ (❬φ₁❙ - ❬φ₂❙ - ι❬φ₃❙)_͟1͟ (E₁❙φ₁❭ - E₂❙φ₂❭ + ιE₃❙φ₃❭)│²
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│√3 √3 │.
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❬Ψ′❙Ĥ❙Ψ′❭ = _͟1͟ (❬φ₁❙ - ❬φ₂❙ - ι❬φ₃❙)_͟1͟ (E₁❙φ₁❭ - E₂❙φ₂❭ + ιE₃❙φ₃❭)
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√3 √3
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= ¹/₃ (❬φ₁❙ - ❬φ₂❙ - ι❬φ₃❙)(E₁❙φ₁❭ - E₂❙φ₂❭ + ιE₃❙φ₃❭)
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= ¹/₃ (❬φ₁❙E₁❙φ₁❭ - ❬φ₁❙E₂❙φ₂❭ + ❬φ₁❙ιE₃❙φ₃❭) (-❬φ₂❙E₁❙φ₁❭ + ❬φ₂❙E₂❙φ₂❭ - ❬φ₂❙ιE₃❙φ₃❭)
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(-ι❬φ₃❙E₁❙φ₁❭ + ι❬φ₃❙E₂❙φ₂❭ - ι❬φ₃❙ιE₃❙φ₃❭).
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Orthogonality is a property of the energy eigenstates:
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⎧ 0 if n≠m
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❬Eₙ❙Eₘ❭ = δₙₘ, where δₙₘ = ⎨ , so
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⎩ 1 if n=m
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❬Ψ′❙Ĥ❙Ψ′❭ = ¹/₃ (❬φ₁❙E₁❙φ₁❭) (❬φ₂❙E₂❙φ₂❭) (-ι❬φ₃❙ιE₃❙φ₃❭)
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= ¹/₃ E₁ E₂ E₃ = ¹/₃ π͟²͟ħ͟² 4͟π͟²͟ħ͟² 9͟π͟²͟ħ͟² = _͟3͟ ⎛π͟²͟ħ͟²⎞³
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2mL² 2mL² 2mL² 2 ⎝ mL²⎠.
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(𝐜) Therefore, the energy expectation value
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│❬Ψ′❙Ĥ❙Ψ′❭│² = _͟9͟ ⎛π͟²͟ħ͟²⎞⁶
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4 ⎝ mL²⎠.
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Because the hamiltonian is time-independent, the state vector progresses timewise according to,
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❙Ψ′(t)❭ = exp(-ι͟Ĥ͟t͟ )❙Ψ′(t=0)❭ = _͟1͟ exp(-ι͟Ĥ͟t͟ ) (❙φ₁❭ - ❙φ₂❭ + ι ❙φ₃❭).
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ħ √3 ħ
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_͟1͟ ⎛exp(-ι͟E͟₁͟t͟ )❙φ₁❭ - exp(-ι͟E͟₂͟t͟ )❙φ₂❭ + ι exp(-ι͟E͟₃͟t͟ )❙φ₃❭⎞.
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√3 ⎝ ħ ħ ħ ⎠
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The coefficients for the states have the same magnitudes, so the probability of measuring a particular state at time t=0 is 1/3.
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❙Ψ(t=0)❭ = _͟A͟ (αβ❙φ₁❭ - βγ❙φ₂❭ + αγι❙φ₃❭)
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αβγ
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❙Ψ′(t=0)❭ ≐ __͟A͟__ ⎛ 1 ⎞ = _͟1͟ ⎛ 1 ⎞
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√(3A²) ⎜-1 ⎟ √3 ⎜-1 ⎟.
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⎝ ι ⎠ ⎝ ι ⎠
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