Expected energy and viability of a linear combination of wave functions

In summary, the conversation discusses proving the viability of a given wave function eigenfunction decomposition and finding its expected energy. It also touches on the concepts of probability, orthogonality, and the expected energy of a discrete random variable. The relationship between frequency and energy is also mentioned, with the conclusion that if the professor is using units of ##\hbar=1##, then ##\omega_n = E_n##.
  • #1
xicor
37
0

Homework Statement



Show that the following wave function eigenfunction decomposition is viable and find it's expected energy.

ψ(x, t) = [1/(1+i)]ψ1(x)e^(-iw1t) - (1/√2)ψ2(x)e^(-iw2t)

Homework Equations


∫ψ(x)*ψ(x)dx = 1 probability

∫ψm(x)*ψn(x)dx = 0 orthogonality

∫ψm(x)*ψn(x)dx = δmn

Ʃ|C[itex]_{n}[/itex]|[itex]^{2}[/itex] = 1

<H> = Ʃ|C[itex]_{n}[/itex]|[itex]^{2}[/itex]*E[itex]_{n}[/itex]

The Attempt at a Solution



From my understanding so far, δmn is value given when you combine the condition for both probability and orthogonality. However I don't think I fully understand why you can just then said that the sum of the constants squared of the wave functions is equal to 1. Using the equation I can find that the first part of wave function is 1/2 and the second part is 1/2 so the sum of that is 1 so the wave function is viable. However I'm not sure what to do with the expected energy. I think that ω = En/h but I am unsure as to how the linear combination effects the expression of expected energy because shouldn't <H> =En since the sum of the constants is just 1?

Thanks for everyone that helps.
 
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  • #2
I don't think I fully understand why you can just then said that the sum of the constants squared of the wave functions is equal to 1.
Well - if you made a measurement of the eigenvalue, what is the probability of finding the particle in one of the eigenstates?
I'm not sure what to do with the expected energy
How do you find the expectation value of a discrete random variable in normal statistics?
<H> =En since the sum of the constants is just 1?
But each value of the energy does not appear with equal probability does it?
 
  • #3
Alright, so the first part just requires some understanding of linear algebra. We are still dealing with time independent wave functions but I think we will be learning that notation very soon.

But each value of the energy does not appear with equal probability does it?

Well, I could see showing the different probability for each value of energy but in this case the expected energy would be <H> = (1/2)E1 + (1/2)E2 where En = ω[itex]_{n}[/itex]h
 
  • #4
Well, I could see showing the different probability for each value of energy but in this case the expected energy would be <H> = (1/2)E1 + (1/2)E2 where En = ωn
That last equation doesn't work - what is the relationship between frequency and energy?

Alright, so the first part just requires some understanding of linear algebra. We are still dealing with time independent wave functions but I think we will be learning that notation very soon.
... no idea what you are talking about.
 
  • #5
Wouldn't the relationship just be based off Planck's equation for quantized energy? That would show that En = hv where "h" is Planck's constant and "v" is the frequency. As far as I can see, the last equation is in that form unless I'm not understanding it correctly.
 
  • #6
xicor said:
Wouldn't the relationship just be based off Planck's equation for quantized energy? That would show that En = hv where "h" is Planck's constant and "v" is the frequency. As far as I can see, the last equation is in that form unless I'm not understanding it correctly.
##\omega = 2\pi\nu## ... for some reason I left the h off the end when I quoted you...

I'm sorry, the way you wrote
... but in this case the expected energy would be...
I inferred that you had some issue with the resulting relation.
I was trying to guess what it was instead of just asking. Now I'm asking :)
 
Last edited:
  • #7
That was just because I wasn't sure about using the Cn probabilities before where both E1 and E2 have an equivalent probability. I think the professor was just using ω as a substitute for En/h because he never brought up the term of 2*pi. The ω variable is taking the place of the wave function ψn(x,t) = Cn*ψn(x)e[itex]^{-iEnt/h}[/itex]. I was just uncertain if I was correctly using the formula for the expected energy <H>.
 
  • #8
Oh OK - well that was exactly how to use them.
##\langle H \rangle = c_1^\star c_1 E_1 + c_2^\star c_2 E_2##

I think the professor was just using ω as a substitute for En/h
But in the wave-function he(?) writes ##e^{-i\omega_n t}## as the time-dependency ... in that relation, ##\omega_n = E_n/\hbar##
If he is using ##\omega=E_n/h## then that would be out by a factor of ##2\pi##.

If he uses units of ##\hbar=1## then you can write ##\omega_n = E_n##.
 

Related to Expected energy and viability of a linear combination of wave functions

1. What is the expected energy of a linear combination of wave functions?

The expected energy of a linear combination of wave functions is the average energy value that is predicted by the Schrödinger equation. It is calculated by taking the weighted sum of the individual energy values of each wave function, where the weights are determined by the coefficients of the linear combination.

2. How is the expected energy related to the individual energy values of the wave functions?

The expected energy is related to the individual energy values of the wave functions through a weighted sum. The coefficients of the linear combination determine the weights of each individual energy value, and the expected energy is the average of these values.

3. Can the expected energy be higher or lower than the individual energy values?

Yes, the expected energy can be higher or lower than the individual energy values. This is because the coefficients of the linear combination can have a positive or negative effect on the final expected energy value, depending on their magnitude and sign.

4. How is the viability of a linear combination of wave functions determined?

The viability of a linear combination of wave functions is determined by the coefficients of the linear combination. If the coefficients are all non-zero and satisfy certain conditions, then the linear combination is considered viable and can be used to accurately predict the behavior of a quantum system.

5. Can a linear combination of wave functions have a negative viability?

No, a linear combination of wave functions cannot have a negative viability. The coefficients of the linear combination must be non-zero and satisfy certain conditions in order for the linear combination to be considered viable. If the coefficients are all positive, then the viability will also be positive.

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