phy-4600/solutions/chap5/prob2

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    

    2͟π͟²͟ħ͟².          𝓟₂=¹/₃
     mL²                         

    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    ⎠

(𝐝)
    = _͟1͟ ⎛exp(-ι π͟²͟ħ͟t͟ )❙φ₁❭ - exp(-ι 2͟π͟²͟ħ͟t͟ )❙φ₂❭ + ι exp(-ι 9͟π͟²͟ħ͟t͟ )❙φ₃❭⎞
      √3 ⎝       2mL²                 mL²                    2L²       ⎠.


What are the possible measurements at time t = ħ/E₁?

(𝐞) The same value of energy will be measured for each state. Since there has been no change to the coefficients besides a change in phase, and the phase term goes to 1 under the modulus, the probabilities remain the same. 

❙Ψ′(t=ħ/E₁)❭ = _͟1͟ ⎛exp(-ι)❙φ₁❭ - exp(-ι E͟₂͟ )❙φ₂❭ + ι exp(-ι E͟₃͟ )❙φ₃❭⎞ 
               √3 ⎝                     E₁                  E₁      ⎠

    Energy          Probability     

     π͟²͟ħ͟².          𝓟₁=¹/₃
     2mL²    

    2͟π͟²͟ħ͟².          𝓟₂=¹/₃
     mL²                         

    9͟π͟²͟ħ͟².          𝓟₃=¹/₃
     2mL²