# Homework Help: Calculating expectation value of U

1. Feb 5, 2016

1. The problem statement, all variables and given/known data

$H$ is the Hamiltonian of an electron and is a Hermitian operator. It satisfies the following equation:

$H |\phi_n\rangle = E_n |\phi_n\rangle$

Let $U = e^{\frac {iHt}{\hbar}}$. Find the expectation value of U in state $|\phi_n\rangle$

2. Relevant equations

$\langle U\rangle = \langle\phi_n| U |\phi_n\rangle$

3. The attempt at a solution

I have tried to use the general expression for H, which is: $H = i\hbar \frac{\partial}{\partial t}$

Applying this to U, I have: $U = e^{\frac {iH\frac{\partial t}{\partial t}}{\hbar}} = e^{-1}$.

This implies:

$\langle U\rangle = \langle\phi_n| e^{-1} |\phi_n\rangle = e^{-1} \langle\phi_n| \phi_n \rangle = e^{-1}$

But I've been told this is wrong, although I have not been told exactly why. Instead I have been told the solution is:

$\langle U\rangle= \langle\phi_n| e^{\frac {iHt}{\hbar}} |\phi_n\rangle = \langle\phi_n| e^{\frac {iE_nt}{\hbar}} \phi_n \rangle = e^{\frac {iE_nt}{\hbar}} \langle\phi_n |\phi_n \rangle= e^{\frac {iE_nt}{\hbar}}$

Although the explanation for this alternate solution was lacking. Namely, why is it true that:

$e^{\frac {iHt}{\hbar}} | \phi_n \rangle = e^{\frac {iE_nt}{\hbar}} |\phi_n\rangle$ ?

Isn't it only true if the Hamiltonian, H, is directly acting on $|\phi_n\rangle$? Why is it that when H is in the exponent of the operator that it is still acting on the state function to yield $E_n$ as the eigenvalue? Any explanation on this and why my approach is wrong would be great!

Last edited: Feb 5, 2016
2. Feb 5, 2016

### blue_leaf77

Try expanding $U$ in power series.

3. Feb 5, 2016

Okay, I think I see where this is going. But when doing the expansion and trying to carry terms out, is $\frac {iHt}{\hbar} | \phi_n \rangle = \frac {iE_nt}{\hbar} | \phi_n \rangle$? Doesn't the operator H have to be applied to both t and $| \phi_n \rangle$ (i.e. $\frac{i}{\hbar} H( t | \phi_n \rangle)$) since t is an independent variable as opposed to constant, no? Then is $\frac{i}{\hbar} H( t | \phi_n \rangle) = \frac {i}{\hbar} E_n( t | \phi_n \rangle)$ necessarily?

4. Feb 5, 2016

### blue_leaf77

In quantum mechanics, time is not an operator, instead it's just parameter. Therefore, an operator (like in this case $H$) being applied to $t$ is not defined. You can always place $t$ wherever you want (of course, so long as there is no derivative of $t$ present in your equation).

5. Feb 5, 2016

Doesn't H have a time derivative? Isn't it: $H = i\hbar \frac{\partial}{\partial t}$? This isn't explicitly stated in the question, but isn't this the general form for the Hamiltonian given by Schrödinger's equation and shouldn't this be assumed?

6. Feb 5, 2016

### blue_leaf77

Actually, the relation between $H$ and the partial derivative you are talking about is what gives the form of $U$ being $e^{iHt/\hbar}$ for the case of time-independent Hamiltonian. The general equation for $U(t)$ is
$$i\hbar\frac{\partial}{\partial t}U(t) = HU(t)$$
from which you can see clearly that if $H$ is time independent, the solution for $U$ is indeed $e^{-iHt/\hbar}$.

One cannot really say that $i\hbar\frac{\partial}{\partial t} = H$. For instance, try sandwhiching this equation with $|u_n\rangle$ from left and right, where $|u_n\rangle$ satisfies $H|u_n\rangle = E_n|u_n\rangle$. In the RHS, you will get $E_n$ but in the LHS you will get zero because $|u_n\rangle$ is not a function of time.

Last edited: Feb 5, 2016
7. Feb 5, 2016

I was under the assumption all observables are given by operators. Does this imply time is not an observable in QM or am I overlooking something?

8. Feb 5, 2016

### blue_leaf77

Yes, time is an exception and is not an observable in QM; there is no operator associated with time.

9. Feb 5, 2016

Okay, that makes sense. But this would then also imply that $|u_n\rangle$ does not necessarily satisfy Schrödinger's equation, correct? But aren't all states in QM describable by Schrödinger's equation? So the problem I've stated here would be more of a mathematical example of operators as opposed to actually modelling an allowed system in QM? That is, the functions $\phi_n$ don't necessarily satisfy the time-dependent Schrödinger equation.

Last edited: Feb 5, 2016
10. Feb 5, 2016

Well, it's nice to finally find out. Thank you! I'm sure this has been discussed in other works....would you happen to have any references that expand on time not being an observable in QM?

11. Feb 5, 2016

### blue_leaf77

The eigenstate of $H$ must satisfy the time-independent Schroedinger equation as it was derived from this equation to begin with. What makes you think otherwise?
I would suggest subchapter 2.1 in Modern Quantum Mechanics by Sakurai.

12. Feb 5, 2016

Shouldn't the solution to U in this case be $U = e^{-iHt/\hbar}$. Therefore my form of U is not necessarily the same. I guess I was just never told that H is necessarily independent of time, or that $|\phi_n\rangle$ is the time-independent component of the state function. This is what is leading me to still not quite understand why the Hamiltonian, H, does not act on both t and $|\phi_n\rangle$ when I perform the series expansion.

Edit: typos

13. Feb 5, 2016

### blue_leaf77

I have edited my previous comment, yes the time evolution operator should be $e^{-iHt/\hbar}$.
$H$ can be time dependent, the most common example is the interaction between an atom with electromagnetic field. In this example, I have never seen the exact solution for the energy eigenkets and if they should exist, they will most likely be a function of time - the Hamiltonian is always changing, so one can expect that the eigenket at a given instant of time be different from another time. Also, for time dependent Hamiltonian, the form of $U$ will not be simply $e^{-iHt/\hbar}$ - its form depends on whether or not the Hamiltonian at different times commute (a short discussion on this subject is also presented in the book I mentioned in my previous comment).
It seems that part of you is still clinging with the fixation that $i\hbar\frac{\partial}{\partial t} = H$. This is not the correct starting point, instead you should start from
$$i\hbar\frac{\partial}{\partial t}U = HU$$
For now let's assume that $H$ is time independent, therefore $U=e^{-iHt/\hbar}$. Then apply the above equation to an energy eigenket $|u_n\rangle$,
$$i\hbar\frac{\partial}{\partial t}U|u_n\rangle = HU|u_n\rangle$$
and substitute for $U$ the given form before
$$i\hbar\frac{\partial}{\partial t}e^{-iHt/\hbar}|u_n\rangle = He^{-iHt/\hbar}|u_n\rangle \\ i\hbar\frac{\partial}{\partial t}e^{-iE_nt/\hbar}|u_n\rangle = e^{-iE_nt/\hbar}H|u_n\rangle \\ E_n|u_n\rangle = H|u_n\rangle$$
That's how you arrive at the time-independent Schroedinger equation.

$H$ does not act on $t$ because first, $t$ is not an operator and second, it's not correct to write $i\hbar\frac{\partial}{\partial t} = H$

14. Feb 5, 2016

Okay, I think I'm starting to follow you better. That seems to make a lot of sense. And just to clarify, the time derivative is only acting on the $e^{-iHt/\hbar}$ and not the $|u_n\rangle$, too, correct? So by taking the series expansion, the exponential term becomes a constant since the H from the series acting on the eigenket becomes the eigenvalue, and then the H outside of the sum operates on the eigenket to yield the eigenvalue once again, correct?

15. Feb 6, 2016

### blue_leaf77

Yes.
Yes, exactly.

16. Feb 6, 2016

### PeroK

I just wanted to add one thing. Trust the mathematics! It makes no difference what $H, \phi_n, E_n, i, \hbar, t$ are in this case. The way I would look at the time issue is:

At any time $t$, $\frac{it}{\hbar}$ is just some constant - it's just a complex number. Let's call it $\alpha$. For any operator, $H$ and any function $\phi$:

$e^{\alpha H} \phi = e^{\alpha H \phi}$

And, if $H\phi = E\phi$ then:

$e^{\alpha H} \phi = e^{\alpha H \phi} = e^{\alpha E \phi} = e^{\alpha E} \phi$

That's essentially what you are trying to prove. It doesn't actually depend on anything to do with QM at all! It's a property of any operator and its eigenfunction. H doesn't need to be a Hamiltonian. Just any old operator with one of its eigenfunctions will do. (That's perhaps a mathematician's view!)