Question about the superposition of energy states

In summary, the conversation discusses the three lowest energy states of an infinitely deep square well, their corresponding wave functions and frequencies, and the relation between energy and frequency. The probability of obtaining a certain energy when measuring a particle in a specific wave function is given by the coefficient of that wave function squared. The Hamiltonian operator is also mentioned as it relates to the energy of the system.
  • #1
icesalmon
270
13

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.
 
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  • #2
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##
 
Last edited:
  • #3
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
solving the schrodinger equation for the wave function dependent ONLY on position is what I meant to say.
 
  • #5
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
icesalmon said:
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
Simon Bridge said:
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
Simon Bridge said:
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
Yes.

Note. If ##c## is real, then ##c^* =c## so ##c^*c = (c^*)c = (c)c = c^2## ...
 
  • #10
Simon Bridge said:
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
icesalmon said:
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
icesalmon said:
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.
 

1. What is the superposition of energy states?

The superposition of energy states is a concept in quantum mechanics where a quantum system can exist in multiple energy states simultaneously. This means that the system is in a combination of all possible energy states, rather than being in a single definite state.

2. How is the superposition of energy states different from classical physics?

In classical physics, a system can only exist in one energy state at a time. However, in quantum mechanics, the superposition of energy states allows a system to exist in a combination of energy states. This is one of the fundamental differences between classical and quantum physics.

3. What causes the superposition of energy states?

The superposition of energy states is caused by quantum superposition, which is a fundamental principle in quantum mechanics. It states that a quantum system can exist in multiple states simultaneously until it is observed or measured.

4. How is the superposition of energy states observed or measured?

The superposition of energy states can be observed or measured through experiments such as the double-slit experiment or the Stern-Gerlach experiment. These experiments use tools such as detectors or screens to observe the behavior of a quantum system and determine its energy states.

5. What are the practical applications of the superposition of energy states?

The superposition of energy states has many practical applications, including in quantum computing, quantum cryptography, and precision measurement tools. It also plays a crucial role in understanding and developing technologies based on quantum mechanics.

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