# I Describing Energy for Time-Dependent States

1. Mar 23, 2016

### Isaac0427

Hi all! Sorry if this question is stupid. The the time-independent schrodinger equation describes energy for a time-independent system, and the time-dependent schrodinger equation describes the time evolution of the wavefunction. So, how would you describe the energy for a time-dependent system? Thanks!

2. Mar 23, 2016

### blue_leaf77

It's in general very difficult to describe the exact time evolution of the wavefunction in the case of time-dependent Hamiltonian. However, there is one special case, which is the two-level system interacting with a sinusoidal electromagnetic field, which can be solved exactly. In this system, the system's state alternates between the two levels in a period determined by the EM field's oscillation. This oscillation of states is also sometimes called Rabi flopping.

3. Mar 23, 2016

### Isaac0427

Ok, but how would you describe the energy of a time-dependent wavefunction (please excuse my bad terminology).

4. Mar 23, 2016

### Isaac0427

Hold on, would iħ∂/∂t characterize time evolution and energy?

5. Mar 23, 2016

### atyy

Within the orthodox interpretation, quantum systems do not have properties until they are measured. So a quantum system does not have an energy until its energy is measured. When the energy is measured, the outcome is random and distributed according to the Born rule.

6. Mar 24, 2016

### Isaac0427

Ok, so then what does the time-independent schrodinger equation describe?

EDIT:
Also, particles don't have defined properties, which I know, but I also know that the wavefunction can describe the expectation value of a specific property. Using the hamiltonian operator, you can find the expectation value of energy, correct?

7. Mar 24, 2016

### Staff: Mentor

yes.

8. Mar 24, 2016

### Staff: Mentor

Its eigenvalues are the values that you might find for the energy when and if you measure it.

9. Mar 24, 2016

### Isaac0427

Could you say for a time-dependent system that $\hat H \psi = \hat E \psi = E \psi$ where $\hat H = \frac{- \hbar ^2}{2m} \frac{\partial ^2}{\partial x^2} + V$ and $\hat E = i \hbar \frac{\partial}{\partial t}$?

10. Mar 24, 2016

### kith

This needs some elaboration.

In fact, I think that the very name "time-independent Schrödinger equation" is bad terminology. A better name is "eigenvalue equation for the Hamilton operator". As Nugatory wrote, its solutions are the possible energies and associated wavefunctions which you get if you perform an energy measurement on your system or if you prepare your system in a state of definite energy. So the Hamilton operator is the operator which is associated with energy in QM.

If your system is in such an energy eigenstate, the (time-dependent) Schrödinger equation tells you that it doesn't change with time.

11. Mar 24, 2016

### kith

The last identity is false in general, i.e. it only holds if $\psi$ is a state of definite energy.

12. Mar 24, 2016

### Isaac0427

Why? If $\hat H \psi = E \psi$ (time independent Schrodinger equation) and $\hat H \psi = \hat E \psi$ (time dependent Schrodinger equation), then why is $\hat E \psi = E \psi$ not true for time-dependent systems? Is it because $\hat H \psi = E \psi$ does not apply to time-dependent systems? Does it?

13. Mar 25, 2016

### kith

Using your terminology, the answer is yes. But dividing systems into time-dependent and time-independent is problematic. It is the wavefunction (or speaking in more general terms, the state) of a system which does or doesn't change over time.

As I tried to tell you in my post #10, the (time-dependent) Schrödinger equation and the eigenvalue equation for the Hamiltonian operator are conceptually very different.

The (time-dependent) Schrödinger equation tells you how the wavefunction of your system changes over time. This equation is always valid (even in relativistic quantum theory, if we use the appropriate generalization). If you have an arbitrary system with an arbitrary initial wavefunction, the (time-dependent) Schrödinger equation tells you -in principle- how it will evolve in time.

The eigenvalue equation for the Hamiltonian on the other hand doesn't correspond to a certain physical situation. It tells you which definite energies are possible for your system and how the corresponding wavefunctions look like. In QM, such an equation exists for for every observable quantity (like momentum, angular momentum, spin, etc.) and the Hamiltonian corresponds to the observable "energy".

Last edited: Mar 25, 2016
14. Mar 26, 2016

### blue_leaf77

By calculating the expectation value of the Hamiltonian. I think you have noticed this fact and also Nugatory has confirmed this for you.
I don't think you can find $\psi$ such that $\hat{H}(t) \psi = E\psi$. In the case of time-independent Hamiltonian, the time evolution for an energy eigenstate $u_n(t)$ is described by $e^{-iE_n t/\hbar}$. Thus, $\hat{E} u_n(t)$ reduces to $E_n u_n(t)$. But for time-dependent system, the term describing the time evolution is no longer as simple-looking as $e^{-iE_n t/\hbar}$.

15. Mar 26, 2016

### PeroK

Leaving aside the physics, you are not using the mathematics carefully enough.

The time-independent equation emerges from the technique of separation of variables, where you look for $\Psi(x, t) = \psi(x) f(t)$. If the potential, hence the Hamiltonian, is time-independent, then the equation separates and the spatial equation is the "time independent" equation. In this case, $\psi$ is a function of the spatial coordinates and you get an eigenvalue equation involving this function as the eigenfunction:

$\hat {H} \psi(x) = E \psi(x)$

If the potential is time-dependent, then the Schroedinger equation does not separate and you don't get a time-independent equation. Moreover, the concept of eigenvalues doesn't make any mathematical sense. You have an operator equation that does not correspond to eigenvalues of any function:

$ih \hat{D}_t \Psi(x, t) = \hat {H}(t) \Psi(x, t)$

This is a more general operator equation on functions of two variables $(x, t)$ and there is no way to introduce an eigenvalue here at all.

16. Mar 26, 2016

### Isaac0427

Ok, but now I am getting very different answers. Does the equation $\hat H \psi = E \psi$ hold for time dependent systems? If not, how would you describe the expectation eigenalue of energy $E$?

17. Mar 26, 2016

### blue_leaf77

Take a look at the last paragraph of post #14 as well as the second half of post #15.

18. Mar 26, 2016

### Isaac0427

Ok, so you are saying that you can't do it with a time dependent system. However, in an earlier post you said
As an answer to a question about describing the energy of a time dependent wavefunction. Also, if you can't have an energy eigenvalue for time dependent systems, yet energy is an observable quantity, wouldn't the operator postulate (with every observable quantity q there is an operator Q...) be violated?

19. Mar 26, 2016

### blue_leaf77

What's wrong with calculating an expectation value of energy? You can do it regardless of whether the Hamiltonian depends on time.
The eigenvalue problem can still be applicable in the case of time-dependent Hamiltonian, but it will be time-specific, meaning that at each instant of time $t'$, the Hamiltonian $H(t')$ will have a specific spatial arrangement, and you can therefore write an eigenvalue problem exclusively for this time $H(t') u_n^{t'} = E_n^{t'} u_n^{t'}$. As soon as, an infinitesimal change in time occurs, the Hamiltonian will change and therefore another different eigenvalue problem applies.

20. Mar 27, 2016

### vanhees71

No, energy eigenstates represent stationary states. Note that it is not the Hilbert space vector $|\psi \rangle$ that represents the state but the ray it represents or, equivalently, the projector $\hat{\rho}_{\psi} = |\psi \rangle \langle \psi |$ as the statistical operator.

For a time-dependent state, the energy is indetermined, because it's not an energy eigenstate.