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Infinite Square Well - Particle in linear combination of states

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A particle of mass m is trapped in a one-dimensional infinite square well running from x= -L/2 to L/2. The particle is in a linear combination of its ground state and first excited state such that its expectation value of momentum takes on its largest possible value at t=0.


I know the process of solving PDE's clear as day, thats not the issue. The problem is that I'm tripping myself out on how to write [itex]\Psi(x,t)[/itex] as a linear combination of its ground state + first excited state.

My hunch is to approach the problem like this :

[itex]\Psi(x,t)[/itex]=c[itex]_{0}[/itex][itex]\Psi_{0}(x,t)[/itex]+c[itex]_{1}[/itex][itex]\Psi_{1}(x,t)[/itex]

where 0 and 1 represent the ground state and first state, respectively

I'm confusing myself on what needs to fill in the equations! Any help would be appreciated.
 
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  • #2
vanhees71
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The clever answer is that this problem is ill-posed, because strictly speaking a particle in an infinite-square well has no momentum observable, but I'm pretty sure that this is not what the professor wants to know ;-).

So guess you are supposed to find [itex]c_0[/itex] and [itex]c_1[/itex] such that the naive "momentum"-expectation value
[tex]\int_{-L/2}^{L/2} \mathrm{d} x \; \psi^*(x,t) \left (-\hbar \frac{\mathrm{d}}{\mathrm{d} x} \psi(x,t) \right )=\text{max},[/tex]
where [itex]\psi[/itex] must be properly normalized, i.e.,
[tex]\int_{-L/2}^{L/2} \mathrm{d} x \; |\psi(x,t)|^2=0.[/tex]
 
  • #3
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The clever answer is that this problem is ill-posed, because strictly speaking a particle in an infinite-square well has no momentum observable, but I'm pretty sure that this is not what the professor wants to know ;-).

So guess you are supposed to find [itex]c_0[/itex] and [itex]c_1[/itex] such that the naive "momentum"-expectation value
[tex]\int_{-L/2}^{L/2} \mathrm{d} x \; \psi^*(x,t) \left (-\hbar \frac{\mathrm{d}}{\mathrm{d} x} \psi(x,t) \right )=\text{max},[/tex]
where [itex]\psi[/itex] must be properly normalized, i.e.,
[tex]\int_{-L/2}^{L/2} \mathrm{d} x \; |\psi(x,t)|^2=0.[/tex]
thanks! that makes sense. By the way, the problem wants me to find the nomalized ψ(x,t) and the usual dynamical quantities (<x>, <p>, check m(d/dx)<x>=p, Δx, Δp, check ΔxΔp satisfies uncertainty principle)

how do we write ψ(x,t) as a linear combination of its ground and first excited state like the problem asks??
 
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  • #4
vela
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Just like you did in the original post.
 
  • #5
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I'm stuck on the fact that were given the phrase "expectation value of the momentum is a maximum when t=0."

the energy of the second state is higher than that of the first, which implies the momentum of the second state which is more than that in the first (ps- I'm just talking about expectation values)

Does anyone know what this means ?
 
  • #6
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Still confused about that phrase, if anyone has ideas please let me know ! Thanks
 

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