Understanding the General Hamiltonian from the Schrödinger Equation

In summary, we have discussed the possibility of obtaining the general Hamiltonian operator for a general Schrödinger equation, and clarified that the equation given in the conversation is actually the energy operator. We also explored the significance of the Hamiltonian as the time evolution operator in finding the state vector at any given time, and the differences between time-independent and time-dependent Hamiltonians. Additionally, we touched on the idea of expressing the Hamiltonian in different bases and the general nature of Dirac notation.
  • #1
joarb
2
0
Is it possible to say that for a general Schrödinger equation

[tex]i \hbar \frac{\partial}{\partial t} | \psi (t) \rangle = \hat{H} | \psi (t) \rangle [/tex]

one may obtain the general Hamiltonian operator

[tex] \hat{H} = i \hbar \frac{\partial}{\partial t} [/tex]

Thanks!
 
Physics news on Phys.org
  • #2
joarb said:
Is it possible to say that for a general Schrödinger equation

[tex]i \hbar \frac{\partial}{\partial t} | \psi (t) \rangle = \hat{H} | \psi (t) \rangle [/tex]

one may obtain the general Hamiltonian operator

[tex] \hat{H} = i \hbar \frac{\partial}{\partial t} [/tex]

Thanks!

not quite, what you have written is actually the energy operator:

[tex]\hat{E} = i \hbar \frac{\partial}{\partial t}[/tex]

the hamiltonian operator is:

[tex]\hat{H} = \frac{\hat{p}^2}{2m} + \hat{V}[/tex]
 
  • #3
joarb said:
Is it possible to say that for a general Schrödinger equation

[tex]i \hbar \frac{\partial}{\partial t} | \psi (t) \rangle = \hat{H} | \psi (t) \rangle [/tex]

one may obtain the general Hamiltonian operator

[tex] \hat{H} = i \hbar \frac{\partial}{\partial t} [/tex]

Thanks!

This observation is correct, but not very helpful.

Note that Schroedinger equation can be rewritten in an equivalent form

[tex]| \psi (t) \rangle = \exp(-\frac{i}{\hbar}\hat{H}t) | \psi (0) \rangle [/tex]

where the time evolution operator [itex] \exp(-\frac{i}{\hbar}\hat{H}t) [/itex] allows you to find the state vector [itex] |\psi (t) \rangle [/itex] at any time if you know the state vector [itex] \psi (0) \rangle [/itex] at time 0 and the Hamiltonian. This is the true significance of the Hamiltonian.

It is true that "operator" [itex]\frac{\partial}{\partial t} [/itex] also formally describes the infinitesimal time evolution. However, in order to apply it to [itex] | \psi (0) \rangle [/itex] you must already know the state vector at a later time [itex] | \psi (\Delta t) \rangle [/itex]. So, you must know the solution of the (time evolution) problem, which you are trying to solve in the first place.

Eugene.
 
Last edited:
  • #4
Hi Eugene,

you missed the [itex]t[/itex] in your time evolution operator. It should be

[tex]| \psi (t) \rangle = \exp(-\frac{i}{\hbar}\,t\,\hat{H}) | \psi (0) \rangle \;,[/tex]

where we assume, that the Hamiltonian is time-independent. In case of a time-dependent Hamiltonian, things get more complicated and the time evolution operater becomes (formally)

[tex]| \psi (t) \rangle = T\left(\exp(-\frac{i}{\hbar}\,\int_0^t\,\hat{H}(t')dt')\right) | \psi (0) \rangle \;,[/tex]

where [itex]T[/itex] is the time odering operator...
 
  • #5
quidamschwarz said:
Hi Eugene,

you missed the [itex]t[/itex] in your time evolution operator. It should be

[tex]| \psi (t) \rangle = \exp(-\frac{i}{\hbar}\,t\,\hat{H}) | \psi (0) \rangle \;,[/tex]

where we assume, that the Hamiltonian is time-independent.

You are right about that. I corrected my formulas.


quidamschwarz said:
In case of a time-dependent Hamiltonian, things get more complicated and the time evolution operater becomes (formally)

[tex]| \psi (t) \rangle = T\left(\exp(-\frac{i}{\hbar}\,\int_0^t\,\hat{H}(t')dt')\right) | \psi (0) \rangle \;,[/tex]

where [itex]T[/itex] is the time odering operator...

You are right about that as well. I was thinking about isolated systems for which the Hamiltonian is always time-independent.

Eugene.
 
  • #6
Isn't it also true that if you express the Hamiltonian in terms of the time derivative you are choosing a particular basis in which to express it? You could also choose to write it, for e.g., in the momentum basis. For that reason, I'd say that the Dirac notation is more general, being independent of a choice of basis.
 
  • #7
belliott4488 said:
Isn't it also true that if you express the Hamiltonian in terms of the time derivative you are choosing a particular basis in which to express it?

I don't think so. As far as I understand is the time parameter [itex]t[/itex] in non-relativistic QM (what we are discussing here) something different as position and momentum. Well, physically this is trivial. What I mean is the following:

A quantum state is given as a vector in a Hilbert space. Which Hilbert space this is depends on your chosen perspective. In position representation the quantum state is represented by a function [itex]\psi:\mathbb{R}^3\to\mathbb{C}[/itex].

The time evolution of [itex]\psi[/itex] is mathematically realized by a curve [itex]\psi_t[/itex] in this Hilbert space, so time is "just" the parameter of this curve. Since the Hamiltonian is an operator on the Hilbert space, it can include derivatives in [itex]x[/itex] (for position representation) but cannot include something like time-derivatives.
 
  • #8
I want to thank all contributors for an illuminating discussion. :)
 

1. What is a general Hamiltonian in the context of quantum mechanics?

A general Hamiltonian is a mathematical representation of the total energy of a quantum mechanical system. It is used to describe the time evolution of a quantum system and is essential in understanding many physical phenomena.

2. How is a general Hamiltonian derived from the Schrödinger equation?

The general Hamiltonian is derived from the Schrödinger equation by rearranging the equation to isolate the Hamiltonian operator on one side. This operator contains all the information about the system's energy and is defined by the system's potential and kinetic energy.

3. What is the significance of the general Hamiltonian in quantum mechanics?

The general Hamiltonian plays a crucial role in quantum mechanics as it allows us to calculate the behavior and properties of a quantum system. It is used to determine the system's energy levels, as well as the probabilities of different outcomes of measurements.

4. Can the general Hamiltonian be used for any quantum system?

Yes, the general Hamiltonian is a universal concept in quantum mechanics and can be applied to any quantum system, whether it is a single particle or a complex multi-particle system.

5. How does the general Hamiltonian relate to the Heisenberg uncertainty principle?

The Heisenberg uncertainty principle states that it is impossible to know the exact position and momentum of a particle simultaneously. The general Hamiltonian takes this into account and describes the energy of a system in terms of these uncertain variables, providing a more complete understanding of the system's behavior.

Similar threads

Replies
3
Views
756
Replies
7
Views
423
Replies
9
Views
303
Replies
17
Views
1K
Replies
5
Views
1K
Replies
2
Views
522
Replies
2
Views
180
Replies
5
Views
860
  • Quantum Physics
Replies
1
Views
629
Replies
5
Views
758
Back
Top