Question about the superposition of energy states

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Homework Statement


The three lowest energy states of an infinitely deep square well (of width L, between x=0 and x=L) are:

Ψ1(x,t) = N sin(πx/L) e-iω1t
Ψ2(x,t) = N sin(2πx/L) e-iω2t
Ψ3(x,t) = N sin(3πx/L) e-iω3t
  • N = sqrt(2/L) is the normalization, to make the total probability = 1.
  • Each wave function oscillates with a different frequency, ωi = 2πfi = Ei/ħ. This relation between energy and frequency is the same as for photons.
1)

Suppose a particle in the well is described by the wave function Ψ2. If you measure its energy, what result will you obtain, as a multiple of the ground-state energy E1?
E2 is a factor of (2)2 or 4 larger than the ground-state energy E1

2)

Suppose the particle's wave function isΨ = 0.616Ψ1 + 0.7Ψ2 + 0.361Ψ3. If you measure the energy of the particle, what is the probability that you will obtain these results:


P(E1) =

Homework Equations


P(x) = |ψ(x,t)|2

The Attempt at a Solution


I tried computing |ψ(x,t)|2 but the expression is nasty and I don't see how I can shrink this expression down. I honestly have no clue how to interpret most of these symbols or even how to start the problem itself.
 

Answers and Replies

  • #2
Simon Bridge
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Anyway, ##|\Psi(x,t)|^2=1##... (or 1/N if not normalized) so it does not help you.

Recall:
A measurement of energy involves operating on the state by the Hamiltonian operator.
##c_n = \int \psi_n^* \psi## where ##H\psi_n = E_n\psi_i## and ##\Psi(x,t)=\psi(x)\chi(t)##
... this stuff should be in your notes, or in a text.
This integration is easier than the one you attempted bc you can exploit ##\int \psi_i^*\psi_j = \delta_{ij}##

If ##\psi## is normalized, and you know ##\psi = \sum c_i\psi_i##
Then ##|\psi|^2 = \sum c_i^*c_i = 1##
Then ##P(E_n)=c_n^*c_n##
 
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  • #3
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I'm sorry, I have no idea what half of this means... we talked about normalization ( my professor wrote down the equation for the normalization condition in class and mentioned that it means there's 100% probability that the particle exists somewhere in this well ) but that's as far as he went. We aren't going in depth with the mathematical formalism of QM if at all, so solving the schrodinger equation for only space as a second order differential equation was as far as he went. Of course I would love to know what you mean by "operating on the state by the Hamiltonian operator" but I'm afraid at this time I just don't.

In regards to the last line you wrote, ci* is the conjugate of ci but in this case aren't the coefficients ci just real numbers? so the probability of finding the particle at it's ground state energy is |ci*ci|2?
 
  • #4
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solving the schrodinger equation for the wave function dependent ONLY on position is what I meant to say.
 
  • #5
Simon Bridge
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OK... that means you are doing it by rule then.
The rule is in the last line I wrote... well done... it should be in your class notes or you cannot do the problem.
But be careful... if the coefficients are all real, then ##c_n^*c_n = c_n^2##
 
  • #6
259
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I'm sorry, I have no idea what half of this means... we talked about normalization ( my professor wrote down the equation for the normalization condition in class and mentioned that it means there's 100% probability that the particle exists somewhere in this well ) but that's as far as he went. We aren't going in depth with the mathematical formalism of QM if at all, so solving the schrodinger equation for only space as a second order differential equation was as far as he went. Of course I would love to know what you mean by "operating on the state by the Hamiltonian operator" but I'm afraid at this time I just don't.

In regards to the last line you wrote, ci* is the conjugate of ci but in this case aren't the coefficients ci just real numbers? so the probability of finding the particle at it's ground state energy is |ci*ci|2?
The text our class is using is Tipler's 6th Edition, the mathematics surrounding this stuff doesn't involve Hamiltonian Mechanics.
 
  • #7
259
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OK... that means you are doing it by rule then.
The rule is in the last line I wrote... well done... it should be in your class notes or you cannot do the problem.
But be careful... if the coefficients are all real, then ##c_n^*c_n = c_n^2##
okay, I think it was written a bit differently. We sort of rushed through this stuff in class, I'll re-check my notes and the book. Thanks for your help i'll post back when more questions involving this problem arise.
 
  • #8
259
8
Anyway, ##|\Psi(x,t)|^2=1##... (or 1/N if not normalized) so it does not help you.

Recall:
A measurement of energy involves operating on the state by the Hamiltonian operator.
##c_n = \int \psi_n^* \psi## where ##H\psi_n = E_n\psi_i## and ##\Psi(x,t)=\psi(x)\chi(t)##
... this stuff should be in your notes, or in a text.
This integration is easier than the one you attempted bc you can exploit ##\int \psi_i^*\psi_j = \delta_{ij}##

If ##\psi## is normalized, and you know ##\psi = \sum c_i\psi_i##
Then ##|\psi|^2 = \sum c_i^*c_i = 1##
Then ##P(E_n)=c_n^*c_n##
is \delta_{ij}## the Kronecker Delta here?
 
  • #9
Simon Bridge
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Yes.

Note. If ##c## is real, then ##c^* =c## so ##c^*c = (c^*)c = (c)c = c^2## ...
 
  • #10
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Yes.

Note. If ##c## is real, then ##c^* =c## so ##c^*c = (c^*)c = (c)c = c^2## ...
This works out well and produces the correct answer, although I don't feel I understand what's truly going on here i'll run with it and maybe it will just fill itself in.....:confused:
 
  • #11
Simon Bridge
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The text our class is using is Tipler's 6th Edition, the mathematics surrounding this stuff doesn't involve Hamiltonian Mechanics.
Since you brought it up: it kinda does, but only indirectly.
In QM, the "Hamiltonian" is the energy operator
... the time independant schrodinger equation is ##\text H \psi = E\psi##
... solutions to this equation are called "energy states" and H is called "the Hamiltonian" for the system by analogy with Hamiltons formulation of classical mechanics.
I think you may find the word "operator" in the index in Tipler, or see:
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hamil.html
 
  • #12
Simon Bridge
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This works out well and produces the correct answer, although I don't feel I understand what's truly going on here i'll run with it and maybe it will just fill itself in.....:confused:
Thats the trouble with learning just the rules... you don't get to understand much.
OTOH, nobody actually understands QM, and trying to talk about what is "truly" going on can start arguments.
Enjoy.
 

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