Finding expectation values for given operators

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SUMMARY

The forum discussion centers on calculating expectation values for operators U, G, and S defined in terms of the Hamiltonian H of an electron in solids. The operators are expressed as U = e^(iHt/ħ), S = sin(Ht/ħ), and G = (ε - H)^(-1). The user successfully simplifies the expectation values for U and S in the eigenstate |Φn> to 1/e and sin(i), respectively, while seeking clarification on the correct approach for G and the state |Ψ> = e^(-α) ∑ |Φn>. The professor advises using power series expansions for the operators and emphasizes the importance of maintaining distinct indices in summations.

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Homework Statement


The Hamiltonian of an electron in solids is given by H. We know that H is an Hermitian operator, it satisfies the following eigenvalue equation:

H|Φn> = εnn>

Let us define the following operators in terms of H as:

U = e^[(iHt)/ħ] , S = sin[(Ht)/ħ] , G = (ε - H)-1

a) Find the expectation value of U, G and S operators in state |Φn> as:
n|U|Φn>
n|S|Φn>
n|G|Φn>

b) Now considr that electron is in state |Ψ> = e ∑ |Φn> (where ∑ denotes an infinite sum from n = 0 to n = ∞), where α is a complex variable. Calculate the expectation value of U, G, and S operators in state |Ψ>.

Homework Equations


H = (-ħ/2m)*(d2/dx2)
H = iħ*(d/dt)
Hint from professor: ex = ∑xn/n!

The Attempt at a Solution


It's been a long time since I took the pre-requisite courses for this quantum course so I am having a lot of problems figuring out how to do this question. I saw online that H could be expressed as iħ*(d/dt), plugging that into the equations for U and S simplifies them to 1/e and sin(i) respectively. After being simplified they can be removed from the brackets leaving me with:

(1/e)*<Φnn> and sin(i)<Φnn>

For both of these expressions the term in brackets is equal to 1, leaving me with an expectation value of 1/e for U and sin(i) for S. Can someone tell me if my approach is correct? The textbook for the course only expresses H in position dependent form (first relevant equation) and not in the time dependent form I have used, so I feel as though I'm "cheating" somehow and I won't get full marks. We probably have to solve it using the hint (3rd relevant equation) somehow but I don't understand how. Aside from that I have no idea how to find the expectation value for G and I haven't even gotten to part b yet. Any help would be appreciated.
 
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phys-student said:
|Ψ> = e-α ∑ |Φn>
The question asks you to calculate the expectation values with respect to ##|\phi_n\rangle##, which is known to be an eigenstate of ##H##, therefore you don't need to define your own state.
phys-student said:
Hint from professor: ex = ∑xn/n!
You should consider this hint instead. This hint tells you that you can expand a smooth function of x in terms of its powers. Do the same with ##U##, ##G##, and ##S##, namely expand each of these function operators in terms of powers of ##H##.
 
I've tried expressing U in terms of powers of H and got the following:

e(iHt/ħ) = [e(it/ħ)]H = ∑(H^n)/n! = 1 + H + H2/2 + H3/6 +...

What exactly am I supposed to do with this now?... I think my main source of confusion is that I'm unfamiliar with Dirac notation (which unfortunately seems to be a big part of the course). Is the following statement valid?

<U> = <[e(it/ħ)]H>
= <∑Hn/n!>
= <Φn|∑εn/n!|Φn>
= ∑εn/n!<Φnn> = ∑εn/n!

Am I allowed to pull an infinite sum outside the brackets like that? Am I allowed to simply substitute ε for H like I did in the 3rd line? Is it okay to leave the expectation value as an infinite sum? Sorry if these are stupid questions I'm just having a lot of trouble with this question
 
phys-student said:
[e(it/ħ)]H = ∑(H^n)/n!
More correctly, it should be
$$
e^{iHt/\hbar} = \sum_{n=0} \frac{1}{n!} \left(\frac{iHt}{\hbar}\right)^n
$$
phys-student said:
<Φn|∑εn/n!|Φn>
You must not use the same index between the states and the summation, make them different. Also don't forget to include the factor ##\frac{it}{\hbar}##.
phys-student said:
Am I allowed to pull an infinite sum outside the brackets like that?
As I have mentioned above, the kets should actually have different index from that of the sum. In other words, these kets were not originally part of the summation. Therefore you can bring them inside the summation sign.
phys-student said:
Am I allowed to simply substitute ε for H like I did in the 3rd line?
That "substitution" actually follows from
$$
H^n|\phi_m\rangle = \epsilon_m^n |\phi_m\rangle
$$
phys-student said:
Is it okay to leave the expectation value as an infinite sum?
Pay attention in the form of the series at the end of your calculation. Does it look similar to the series you have in the beginning when you expand the exponential operator?
 
Thank you, I've been able to get through all of part a. I'm on part b now, and I'm not sure how to express <Ψ|U|Ψ>...

Given that |Ψ> = e∑|Φn>

Can <Ψ|U|Ψ> be expressed as:

<Ψ|U|Ψ> = e∑<Φn|U|Φn>

So then the expectation value of U in state Ψ, is the infinite sum of the expectation value found in part a (in state Φn) multiplied by the constant e?
 
phys-student said:
|Ψ> = e-α ∑ |Φn>
The form of ##|\psi\rangle## given above actually looks fishy to me. All basis in the series have equal probability, I'm not sure if such form is normalizable.
Despite of that, when calculating the expectation value of an operator with respect to states which are written as linear combination of some basis, you have to retain both sums. In this problem you want to calculate ##\langle \psi |U| \psi \rangle##, with ##|\psi \rangle = e^{-\alpha} \sum_n |\phi_n\rangle##. Therefore
$$
\langle \psi |U| \psi \rangle = e^{-\alpha^*} \sum_n \langle \phi_n| U e^{-\alpha} \sum_m |\phi_m\rangle = e^{-2\Re[\alpha]}\sum_m \sum_n \langle \phi_n |U| \phi_m \rangle
$$
Since you know how ##U## acts on ##|\phi_m\rangle##, seeing what the next step is should be easy for you.
 

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