Particle in a Box: Energy Measurements and Probabilities Explained

  • #1
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0

Homework Statement



A particle is in an infinite square well extending from x = 0 to x = a. Its state is a
linear combination of the two lowest energy states.
Psi(x, t) = A(2 Psi1(x, t) + Psi2(x, t))
a) If a measurement of energy is made what are the possible results of the measurement?
What is the probability associated with each? What is the average value of energy?


Homework Equations





The Attempt at a Solution



I know that when the Hamiltonian operates on Psi_n(x,t) it give E_n Psi_n(x,t), right? So are the possible results just Psi1(x,t) and Psi2(x,t) or would they be 2A E1 Psi1(x,t) and A E2 Psi2(x,t)? I am confused because I know quantum state vectors are supposed to be normalized but I am confused how that translates to wave functions?

And as for the probability, I really am not sure where to start can someone point me in the right direction?

Thanks for the help
 
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  • #2
The possible outcomes of a measurement are the eigenvalues of the corresponding observable. The coefficients of the wavefunction, written as a superposition of eigenstates, will tell you what those respective probabilities are.

This question is really just a straightforward application of the quantum mechanical rules (postulates) so it's hard to help without giving the answer.
 
  • #3
phrygian said:

Homework Statement



A particle is in an infinite square well extending from x = 0 to x = a. Its state is a
linear combination of the two lowest energy states.
Psi(x, t) = A(2 Psi1(x, t) + Psi2(x, t))
a) If a measurement of energy is made what are the possible results of the measurement?
What is the probability associated with each? What is the average value of energy?


Homework Equations





The Attempt at a Solution



I know that when the Hamiltonian operates on Psi_n(x,t) it give E_n Psi_n(x,t), right? So are the possible results just Psi1(x,t) and Psi2(x,t) or would they be 2A E1 Psi1(x,t) and A E2 Psi2(x,t)? I am confused because I know quantum state vectors are supposed to be normalized but I am confused how that translates to wave functions?

And as for the probability, I really am not sure where to start can someone point me in the right direction?

Thanks for the help


First of all, all this equation states: ''Psi(x, t) = A(2 Psi1(x, t) + Psi2(x, t))'' is two particles oscillating through a specific position and a time. You must state they are in a superpositioning, which is actually a type of interference. The total energy of one of these particles is:

[itex]E=\hbar \omega[/itex]. It also has a respective wave number [itex]k[/itex]: [itex]\frac{\hbar^{2}k^{2}}{2m}[/itex]. Now work it out for two particle systems.
 
  • #4
ManyNames said:
First of all, all this equation states: ''Psi(x, t) = A(2 Psi1(x, t) + Psi2(x, t))'' is two particles oscillating through a specific position and a time. You must state they are in a superpositioning, which is actually a type of interference. The total energy of one of these particles is:

[itex]E=\hbar \omega[/itex]. It also has a respective wave number [itex]k[/itex]: [itex]\frac{\hbar^{2}k^{2}}{2m}[/itex]. Now work it out for two particle systems.
The problem statement clearly says the state describes one particle, not two.
 
  • #5
vela said:
The problem statement clearly says the state describes one particle, not two.

Then the equation required, the one given is not the ordinary standard equation i have ever seen. This is the one which will calculate the question at hand.

[itex]\psi(x,t) = [A sin(k,t)+B cos(k,t)]e^{-i \omega t}[/itex]
 
  • #6
ManyNames said:
Then the equation [...] given is not the ordinary standard equation i have ever seen.
Really? This is a standard question in introductory quantum mechanics.
 
  • #7
vela said:
Really? This is a standard question in introductory quantum mechanics.

Then why is there a factor of 2 for both psi identities which appear inside the brackets?
 
  • #8
What's a psi identity?

[itex]\Psi(x,t)[/itex] isn't an energy eigenstate. It's a superposition of the energy eigenstates [itex]\psi_1(x,t)[/itex] and [itex]\psi_2(x,t)[/itex].
 

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