Expectation Value of Hamiltonian with Superposition

In summary: So the calculations done by the shortcut method are correct.I have found that ##\langle \hat{H} \rangle = \frac{2 \hbar^2 \pi^2}{m L^2}## which is the same expression as for ##E_2##.I think it makes sense that my expression lies between ##E_1## and ##E_3## (afterall, it contains components of both, so isn't purely one or the other), and since the energy is quantised, it can only be... in between! :)So the calculations done by the shortcut method are correct.Yes, this is all correct. Good job!
  • #1
BOAS
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Homework Statement


[/B]
Particle in one dimensional box, with potential ##V(x) = 0 , 0 \leq x \leq L## and infinity outside.

##\psi (x,t) = \frac{1}{\sqrt{8}} (\sqrt{5} \psi_1 (x,t) + i \sqrt{3} \psi_3 (x,t))##

Calculate the expectation value of the Hamilton operator ##\hat{H}## . Compare it with the energy eigenvalues ##E_1##, ##E_2##, and ##E_3##.

Homework Equations

The Attempt at a Solution


[/B]
The subscript refers to the different eigenvalue solutions.

##\psi_1 = e^{-i \frac{E_1 t}{\hbar} } \sqrt{\frac{2}{L}} \sin \frac{\pi}{L} x##

and

##\psi_3 = e^{-i \frac{E_3 t}{\hbar} } \sqrt{\frac{2}{L}} \sin \frac{3 \pi}{L} x##

Using the fact that ##\hat{H} \phi = E \phi## I find that

##\hat{H} \phi = -\frac{\hbar^2}{2m} \frac{\partial^2 \phi}{\partial x^2}## and cancelling out the stationary wave equation, I get that

##E = \frac{\hbar^2 \pi^2}{2m L^2 \sqrt{8}} (\sqrt{5} + 9 \sqrt{3})##

and since the expectation value of the hamiltonian is the total energy, that should be what I'm looking for.

I am confused about how I compare this value to ##E_n## when I have a superposition of stationary states. Does the expression ##E_n = \frac{\hbar^2}{2m} (\frac{n \pi}{L})## apply to all stationary states?

i.e is it literally a case of plugging in ##n = 1,2,3## ?

I'm not sure if I'm explaining myself clearly. How does the case where ##n = 1## effect my ##\psi_3##?
 
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  • #2
BOAS said:
##
E = \frac{\hbar^2 \pi^2}{2m L^2 \sqrt{8}} (\sqrt{5} + 9 \sqrt{3})
##
That's not correct, you should calculate ##\langle \psi | H| \psi \rangle## instead.
 
  • #3
blue_leaf77 said:
That's not correct, you should calculate ##\langle \psi | H| \psi \rangle## instead.

I don't understand your notation.

Is that equivalent to ##\langle \hat{H} \rangle = \int^{\infty}_{\infty} dx \psi* \hat{H} \psi## ?
 
  • #4
BOAS said:
Is that equivalent to ##\langle \hat{H} \rangle = \int^{\infty}_{\infty} dx \psi^* \hat{H} \psi## ?
Yes, it's equivalent to that integral form.
 
  • #5
BOAS said:
Does the expression ##E_n = \frac{\hbar^2}{2m} (\frac{n \pi}{L})## apply to all stationary states?

##E_n = \frac{\hbar^2}{2m} (\frac{n \pi}{L})^2##
 
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  • #6
blue_leaf77 said:
Yes, it's equivalent to that integral form.

Ok,

##\hat{H} \psi = - \frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2} + V \psi= \frac{\hbar^2}{2m} \sqrt{\frac{2}{L}} (\frac{\sqrt{5}}{\sqrt{8}} e^{\frac{-i E_1 t}{\hbar}} \frac{\pi^2}{L^2} \sin \frac{\pi}{L}x + i \frac{\sqrt{3}}{\sqrt{8}} e^{\frac{-i E_3 t}{\hbar}} \frac{9 \pi^2}{L^2} \sin \frac{3 \pi}{L} x)##

This simplifies a little

##\hat{H} \psi = \sqrt{\frac{2}{L}} ( E_1\frac{\sqrt{5}}{\sqrt{8}} e^{\frac{-i E_1 t}{\hbar}} \sin \frac{\pi}{L}x + i E_3 \frac{\sqrt{3}}{\sqrt{8}} e^{\frac{-i E_3 t}{\hbar}} \sin \frac{3 \pi}{L} x)##

##\hat{H} \psi = \frac{1}{2 \sqrt{L}} ( E_1 \sqrt{5} e^{\frac{-i E_1 t}{\hbar}} \sin \frac{\pi}{L}x + i E_3 \sqrt{3} e^{\frac{-i E_3 t}{\hbar}} \sin \frac{3 \pi}{L} x)##

##\phi * = \frac{1}{2 \sqrt{L}} (\sqrt{5} e^{\frac{i E_1 t}{\hbar}} \sin \frac{\pi}{L} x - i \sqrt{3} e^{\frac{i E_3 t}{\hbar}} \sin \frac{3 \pi}{L} x)##

##\phi * \hat{H} \phi = \frac{1}{4L} ( E_1 \sqrt{5} e^{\frac{-i E_1 t}{\hbar}} \sin \frac{\pi}{L}x + i E_3 \sqrt{3} e^{\frac{-i E_3 t}{\hbar}} \sin \frac{3 \pi}{L} x)(\sqrt{5} e^{\frac{i E_1 t}{\hbar}} \sin \frac{\pi}{L} x - i \sqrt{3} e^{\frac{i E_3 t}{\hbar}} \sin \frac{3 \pi}{L} x)##

##\phi * \hat{H} \phi = \frac{1}{4L} (5 E_1 \sin^2 \frac{\pi}{L}x + 3 E_3 \sin^2 \frac{3 \pi}{L}x + i \sqrt{3} \sqrt{5} E_3 e^{\frac{-i(E_3 - E_1)t}{\hbar}} \sin \frac{3 \pi}{L}x \sin \frac{\pi}{L} x - i \sqrt{3} \sqrt{5} E_1 e^{\frac{-i(E_1 - E_3)t}{\hbar}} \sin \frac{\pi}{L} x \sin \frac{3 \pi}{L} x)##

When I integrate this, am I correct in thinking I can apply the orthogonality theorem to get rid of all but the sine squared terms?
 
  • #7
BOAS said:
When I integrate this, am I correct in thinking I can apply the orthogonality theorem to get rid of all but the sine squared terms?
Yes you are, because the eigenstates of a Hamiltonian are orthogonal.
 
  • #8
BOAS said:
Ok,

##\hat{H} \psi = - \frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2} + V \psi= \frac{\hbar^2}{2m} \sqrt{\frac{2}{L}} (\frac{\sqrt{5}}{\sqrt{8}} e^{\frac{-i E_1 t}{\hbar}} \frac{\pi^2}{L^2} \sin \frac{\pi}{L}x + i \frac{\sqrt{3}}{\sqrt{8}} e^{\frac{-i E_3 t}{\hbar}} \frac{9 \pi^2}{L^2} \sin \frac{3 \pi}{L} x)##

This simplifies a little ...

There's no advantage in replacing ##\psi_n## with the appropriate ##sin## function (unless you were trying to prove orthogonality). It is better and simpler to keep it general for as long as possible.

Your original mistake was to use the coefficients ##a_1, a_3## rather than ##|a_1|^2, |a_3|^2## when you took your shortcut.

It's also simpler to keep it as ##E_1, E_3## for as long as possible. That last little part of the question is almost a hint to do this!
 
  • #9
PeroK said:
There's no advantage in replacing ##\psi_n## with the appropriate ##sin## function (unless you were trying to prove orthogonality). It is better and simpler to keep it general for as long as possible.

Your original mistake was to use the coefficients ##a_1, a_3## rather than ##|a_1|^2, |a_3|^2## when you took your shortcut.

It's also simpler to keep it as ##E_1, E_3## for as long as possible. That last little part of the question is almost a hint to do this!

I can definitely appreciate this after getting myself into plenty of trouble with accounting for all the terms.

Thanks for the tips.
 
  • #10
Isn't the integral over the sin^2 terms divergent??
 
  • #11
jimbododge said:
Isn't the integral over the sin^2 terms divergent??
The sine functions involved in this calculation are actually only within the interval ##0<x<L##. Outside this interval the wavefunctions vanish.
 
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  • #12
blue_leaf77 said:
Yes you are, because the eigenstates of a Hamiltonian are orthogonal.

I have found that ##\langle \hat{H} \rangle = \frac{2 \hbar^2 \pi^2}{m L^2}## which is the same expression as for ##E_2##.

I think it makes sense that my expression lies between ##E_1## and ##E_3## (afterall, it contains components of both, so isn't purely one or the other), and since the energy is quantised, it can only be ##E_2##.
 
  • #13
BOAS said:
I have found that ##\langle \hat{H} \rangle = \frac{2 \hbar^2 \pi^2}{m L^2}## which is the same expression as for ##E_2##.

I think it makes sense that my expression lies between ##E_1## and ##E_3## (afterall, it contains components of both, so isn't purely one or the other), and since the energy is quantised, it can only be ##E_2##.

Yes and no. That's the right answer, but if the coefficents were different, you could get any answer between ##E_1## and ##E_3##. There's no obligation on the expected energy to be precisely ##E_2##. Any specific measurement of energy will give either ##E_1## or ##E_3## but the average (expected) value doesn't have to be ##E_2##.
 
  • #14
PeroK said:
Yes and no. That's the right answer, but if the coefficents were different, you could get any answer between ##E_1## and ##E_3##. There's no obligation on the expected energy to be precisely ##E_2##. Any specific measurement of energy will give either ##E_1## or ##E_3## but the average (expected) value doesn't have to be ##E_2##.

Hmm, this is confusing.

##\langle \hat{H} \rangle = \Sigma_n E_n |c_n|^2## describes what you're saying about a specific measurement giving either ##E_1## or ##E_3##. Afterall. ##E_2## has no associated probability.

So the numbers of this question were chosen to give E_2. That seems somewhat misleading. Ok, so depending on the weighting given to each wavefunction, you might expect the expectation value to shift towards the 'heavier' one.
 
  • #15
BOAS said:
Hmm, this is confusing.

##\langle \hat{H} \rangle = \Sigma_n E_n |c_n|^2## describes what you're saying about a specific measurement giving either ##E_1## or ##E_3##. Afterall. ##E_2## has no associated probability.

So the numbers of this question were chosen to give E_2. That seems somewhat misleading. Ok, so depending on the weighting given to each wavefunction, you might expect the expectation value to shift towards the 'heavier' one.

Whoever set the question probably didn't anticipate your interpretation! Yes, weighting is exactly how to look at it. If ##a_1## were small and ##a_3## nearly one, then ##<H>## would be nearly ##E_3##; and vice versa.
 
  • #16
BOAS said:
##
\langle \hat{H} \rangle = \frac{2 \hbar^2 \pi^2}{m L^2}
##
Either there is a typo or you were calculating it wrong - the factor of two should be another number.
EDIT: Sorry, I forgot that you must have cancel 4 in the numerator with 2 in the denominator. You answer is right.
 
  • #17
blue_leaf77 said:
Either there is a typo or you were calculating it wrong - the factor of two should be another number.
EDIT: Sorry, I forgot that you must have cancel 4 in the numerator with 2 in the denominator. You answer is right.

No problem - I appreciate your help.
 
  • #18
Shouldn't the probability of state 2 be zero, seeing as it was not involved in the original sum of functions? How can the probabilities of each state vary between different functions if they're given by the equation for En? I found that the total energy is given by 5/8 E1 + 3/8 E3... Seeing as these probabilities add to 1, would these represent the relative probabilities of the two functions?

Thanks so much!
 
  • #19
jimbododge said:
How can the probabilities of each state vary between different functions if they're given by the equation for En?
I don't get what you intend to say there.
jimbododge said:
I found that the total energy is given by 5/8 E1 + 3/8 E3.
Average energy is a more correct term than the total energy. The numerical value above is correct though.
jimbododge said:
Seeing as these probabilities add to 1, would these represent the relative probabilities of the two functions?
Yes, the probabilities should indeed add up to one and they also represent the probabilities of finding the system in the corresponding states. What is it that confuses you?
 

Related to Expectation Value of Hamiltonian with Superposition

1. What is the expectation value of Hamiltonian with superposition?

The expectation value of Hamiltonian with superposition is a measure of the average energy of a quantum system in a specific state. It is calculated by taking the inner product of the state vector with the Hamiltonian operator, and then squaring the result.

2. How is the expectation value of Hamiltonian with superposition related to quantum mechanics?

The expectation value of Hamiltonian with superposition is a fundamental concept in quantum mechanics. It allows us to make predictions about the behavior of a quantum system and is used extensively in calculations and experiments.

3. Can the expectation value of Hamiltonian with superposition be negative?

Yes, the expectation value of Hamiltonian with superposition can be negative. This reflects the fact that quantum systems can have negative energy levels, unlike classical systems where energy is always positive.

4. How is the expectation value of Hamiltonian with superposition different from the classical concept of expectation value?

The expectation value of Hamiltonian with superposition is different from the classical concept of expectation value in that it takes into account the principles of quantum mechanics, such as superposition and uncertainty. In classical mechanics, the expectation value is simply the average of all possible outcomes, while in quantum mechanics, it is a complex mathematical calculation involving the state of the system and the Hamiltonian operator.

5. Why is the expectation value of Hamiltonian with superposition important in quantum computing?

The expectation value of Hamiltonian with superposition is important in quantum computing because it allows us to determine the energy of a quantum system, which is crucial for understanding the behavior and evolution of the system. It is also used in algorithms such as the variational method to optimize the energy of a quantum system and solve complex problems.

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