Understanding Energy Operator in Time-Dependent Schrodinger Equation

[good_phy]

Hi liboff proble 5.28 says

time dependent Schrödinger equation permits the identity such as $E = i\hbar \frac{\partial}{\partial x}$ (E is operator)

But i don't understand E( is operator in this problem) can be thought energy operator

Is energy operator only H, Hamiltonian?

If E is energy operator, We can find some uncertainty by using commute relation

$$\Delta E \Delta t = \frac{1}{2} \hbar$$

Considering this relation, We can think if we know current energy eigenstate, meaning we

know exact energy value, uncertainty of t,time is indefinity.

What does it means? we can't find exact time that state measured experienced?

what does it means?Please remove my confuse.

Thank you.

 

[Manilzin]

I'm not sure I understand you correctly, also, your second image didn't show. But I have never seen E being represented as an operator, it has always been the eigenvalue of the operator H. Also, the operator you wrote has the dimensions of momentum, not of energy, so I don't understand how this can relate to energy at all. Could you perhaps provide more details?

 

[Fredrik]

The Schrödinger equation is

[tex]i\hbar\frac{\partial}{\partial t}\psi=H\psi[/itex]<br /> <br /> That's why the operator on the left can be thought of as an energy operator. It <i>is</i> the hamiltonian. Note that it's d/dt, not d/dx. Also note that there's no time operator, so the energy-time "uncertainty relation" has nothing to do with commutators.[/tex]

 

[Manilzin]

Ah ok, with time derivatives it makes more sense. However, I still disagree that this is the "energy operator". Nor is it the Hamiltonian, any more than an eigenvalue of H is the Hamiltonian. Rather, it's just what gives the time evolution of a system.

 

[Fredrik]

In advanced texts (i.e. Weinberg) the Hamiltonian is actually defined as the generator of translations in time. That approach goes something like this:

There must be a unitary operator U(t) that translates a state a time t. The unitarity implies that its Taylor expansion takes the form U(t)=1-iHt+... where "1" is the unit operator, and H is a Hermitian operator. Let's call H "the Hamiltonian". The property U(t+t')=U(t)U(t') implies that U(t)=exp(-iHt). If you multiply by i and take the time derivative, you get idU(t)/dt=HU(t). So the time translation operator satisfies the Schrödinger equation (in units such that $\hbar=1$).

Given a state vector $|\psi\rangle$, you can define a time dependent state vector $|\psi;t\rangle=U(t)|\psi\rangle$. Since U(t) satisfies the Schrödinger equation, the time dependent state vector must satisfy it too.

 

[Manilzin]

I agree with all this, and the fact that the Hamiltonian generates time translation is easily seen from the Schrödinger equation itself. What I mean is that the Hamiltonian and the time derivative operator mentioned are not the same operators. They are related through the Schrödinger equation, but they are not the same. In classical qm, time is not an observable on the same footing as position and momentum. We can't act with the "time operator" on a state and find its eigenvalue or some probability for an eigenvalue. Rather, it is a parameter. How the quantum state evolves with this parameter, given the system Hamiltonian H, is what we get from the Schrödinger equation.