Energy operator in Quantum Mechanics

  • #1
I learned that the energy operator is
##\hat{E} = i\hbar \frac{\partial}{\partial t} ##
and the Hamiltonian is
##\hat{H} = \frac{-\hbar^2}{2m}\nabla^2+V(r,t)##

If the Hamiltonian represents the total energy of the system. I expect the two should be the same. Did I misunderstand the concept of an operator?
 

Answers and Replies

  • #2
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
14,359
6,730
I learned that the energy operator is
##\hat{E} = i\hbar \frac{\partial}{\partial t} ##
and the Hamiltonian is
##\hat{H} = \frac{-\hbar^2}{2m}\nabla^2+V(r,t)##

If the Hamiltonian represents the total energy of the system. I expect the two should be the same. Did I misunderstand the concept of an operator?
They are equal through the Schroedinger equation:
$$i\hbar \frac{\partial}{\partial t} \Psi = \hat H \Psi = (\frac{-\hbar^2}{2m}\nabla^2+V)\Psi$$
 
  • Like
Likes etotheipi
  • #3
48
5
They are equal through the Schroedinger equation:
$$i\hbar \frac{\partial}{\partial t} \Psi = \hat H \Psi = (\frac{-\hbar^2}{2m}\nabla^2+V)\Psi$$
You may be interested in this paper
https://arxiv.org/abs/1407.6762
(look a formula 5)
 
  • #4
686
141
I learned that the energy operator is
##\hat{E} = i\hbar \frac{\partial}{\partial t} ##
and the Hamiltonian is
##\hat{H} = \frac{-\hbar^2}{2m}\nabla^2+V(r,t)##

If the Hamiltonian represents the total energy of the system. I expect the two should be the same. Did I misunderstand the concept of an operator?
The Hamiltonian operator IS the energy operator. The LHS of the time-dependent Schrodinger equation (the RHS of your first expression) is not an operator on states in the usual sense, but rather indicates that the Hamiltonian is the infiniestimal generator of time translations.
 
  • #5
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
15,709
7,032
It is very important to understand that the operator representing energy is ##\hat{H}##, and time is NOT an observable but a parameter in quantum mechanics. That's so, because otherwise ##t## and ##\hat{H}## would be canonically conjugate observables and thus for any system ##\hat{H}## would have entire ##\mathbb{R}## as its spectrum, which is obviously wrong, because then we couldn't describe the discrete energy levels of, e.g., a hydrogen atom with quantum mechanics. Also there would be nothing stable, because the Hamiltonian were not bounded from below and there'd be no stable ground state.

The time-dependent Schrödinger equation,
$$\mathrm{i} \partial_t \psi(t,\vec{x})=\hat{H} \psi(t,\vec{x}),$$
describes the time evolution of the wave function, representing the position-probability amplitude, i.e., ##|\psi(t,\vec{x}|^2## represents the position-probability distribution or a particle prepared in the corresponding pure state.
 
  • Like
Likes Chan Pok Fung
  • #6
@vanhees71

I tried hard to understand your words. I raised this thread because I saw an exercise in the textbook asking to show ##\Delta E \Delta t \leq \frac{\hbar}{2}##. It gives the solution using the commutator ##[\hat{E}, t] = i\hbar##.

However, my professor said that ##t## is a parameter instead of a dynamic variable. It is not meaningful to talk about the commutator between time and energy. We may take this ##\Delta t## as the measurement time of a dynamic variable.

I guess this is related to what you are saying here. Could you explain more about this statement:
That's so, because otherwise ##t## and ##\hat{H}## would be canonically conjugate observables and thus for any system ##\hat{H}## would have entire ##\mathbb{R}## as its spectrum, which is obviously wrong, because then we couldn't describe the discrete energy levels of, e.g., a hydrogen atom with quantum mechanics. Also there would be nothing stable, because the Hamiltonian were not bounded from below and there'd be no stable ground state.
I only understand the mathematically meaning of the canonically conjugated variables. How shall we relate the math to physics? Why would this result in a world with no stable ground states?
 
  • #7
PeterDonis
Mentor
Insights Author
2019 Award
30,722
9,712
Why would this result in a world with no stable ground states?
For there to be a stable ground state, there must be a minimum value in the spectrum of the Hamiltonian, since a stable ground state must be a state of minimum possible energy. But if the spectrum of the Hamiltonian includes all real numbers, it cannot have a minimum value. Physically, this would mean that any state whatever would have an amplitude to emit energy in some form and transition to a lower energy state; no state could possibly be stable.
 
  • Like
Likes vanhees71
  • #8
@PeterDonis

It's much clearer. Thanks! But I still have some difficulties connecting the maths with the physics.

First, if ##t## and ##\hat{H}## are canonically conjugate observables. How would this result in ##\hat{H}## having only real number regime?

Second, if ##\hat{H}## can only be real, why can't it have a minimum value?

Thanks!
 
  • #9
591
413
Second, if can only be real, why can't it have a minimum value?
What's the smallest real number?
 
  • #10
Oh, I understand now. I thought it means that H can be imaginary.
 
  • #11
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
15,709
7,032
@vanhees71

I tried hard to understand your words. I raised this thread because I saw an exercise in the textbook asking to show ##\Delta E \Delta t \leq \frac{\hbar}{2}##. It gives the solution using the commutator ##[\hat{E}, t] = i\hbar##.

However, my professor said that ##t## is a parameter instead of a dynamic variable. It is not meaningful to talk about the commutator between time and energy. We may take this ##\Delta t## as the measurement time of a dynamic variable.

I guess this is related to what you are saying here. Could you explain more about this statement:

I only understand the mathematically meaning of the canonically conjugated variables. How shall we relate the math to physics? Why would this result in a world with no stable ground states?
It's, of course, ##\Delta t \Delta E \geq \hbar/2##, but this uncertainty relation is special. It's not one of the standard Heisenberg uncertainty relations, which refer to observables, and this one reads
$$\Delta A \Delta B \geq \frac{1}{2} |\langle \mathrm{i} [\hat{A},\hat{B}] \rangle|.$$
The energy-time uncertainty relation is a delicate issue. You have to clearly analyze in any specific case what ##\Delta t## precisely means.

E.g., there's a long-standing debate about the definition of "tunneling time", i.e., what does tunneling mean in a dynamical context and how to define the time a particle needs to "tunnel" through a potential barrier. I'm not sure, whether this age-old problem has been satisfactorily solved in the sense that physicists have reached a consensus about the meaning of "tunneling time".
 
  • Like
Likes Chan Pok Fung and PeroK
  • #12
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
15,709
7,032
For there to be a stable ground state, there must be a minimum value in the spectrum of the Hamiltonian, since a stable ground state must be a state of minimum possible energy. But if the spectrum of the Hamiltonian includes all real numbers, it cannot have a minimum value. Physically, this would mean that any state whatever would have an amplitude to emit energy in some form and transition to a lower energy state; no state could possibly be stable.
Yes, and in addition there'd be no discrete states of any quantum system, because as for ##\hat{x}## and ##\hat{p}## both ##\hat{H}## and ##\hat{t}## would have the entire real line as their spectra. That's clearly wrong since the entire development of quantum theory was pushed mostly by the ability to explain the discrete spectra of atoms and riddles like the anomalous Zeeman effect, which is not satisfactorily explainable in any classical way, because you need spin 1/2, which does not exist in classical mechanics. There were more and more ad hoc manipulations within the Bohr-Sommerfeld old quantum theory necessary to get close to such phenomena and to argue that all of a suden not only integer but half-integer quantum numbers were needed in some cases. Also multi-electron atoms were not understandable without such ad-hoc manipulations of the foundations of old quantum theory, and "multi-electron atoms" already included He with only 2 electrons!

So both matrix mechanics's (Born, Jordan, Heisenberg) and wave mechanics's (Schrödinger) greatest early triumph was the correct derivation of the hydrogen spectrum (solved by Pauli for matrix mechanics and Schrödinger for wave mechanics) and shortly thereafter also of multi-electron atoms, the discovery of half-integer angular momenta (spin) and the solution of the Zeeman riddle.

So the idea of time as an observable was considered wrong very early on. The argument concerning the continuous spectrum of the Hamiltonian which follws from such an assumption, was most clearly stated first by Pauli in his famous Encyclopedia article on wave mechanics, which still is among the best quantum mechanics text-book style writings ever!
 
  • Like
Likes Chan Pok Fung
  • #13
atyy
Science Advisor
13,998
2,276
As others have said, the Hamiltonian ##\hat{H} = \frac{-\hbar^2}{2m}\nabla^2+V(r,t)## is the energy operator.

The Hamiltonian has 2 major uses:

1) When you want to calculate the probability of obtaining various results when you measure a quantity, then you use the operator corresponding to the measured quantity and the Born rule. So when you want to calculate the probability of obtaining various values when you measure the energy, you use the Hamiltonian and the Born rule.

2) The Hamiltonian also govern how the quantum state evolves with time via the Schroedinger equation ##i\hbar \frac{\partial}{\partial t}\psi = \hat{H}\psi##.

I raised this thread because I saw an exercise in the textbook asking to show ##\Delta E \Delta t \leq \frac{\hbar}{2}##. It gives the solution using the commutator ##[\hat{E}, t] = i\hbar##.
That is not correct. One cannot use ##\hat{E} = i\hbar \frac{\partial}{\partial t} ## as the operator when calculating the variance of results obtained when measuring the energy. When calculating the probability of obtaining various results in a measurement of energy, one must use the Hamiltonian.

Take a look at https://arxiv.org/abs/quant-ph/0609163, where section 3 discusses the myth that there is a time-energy uncertainty relation.
 
  • Like
Likes Chan Pok Fung and vanhees71
  • #14
Take a look at https://arxiv.org/abs/quant-ph/0609163, where section 3 discusses the myth that there is a time-energy uncertainty relation.
Thanks for the suggestion. I spent a whole afternoon reading it. What a well-written pedagogical article!
 
  • Like
Likes vanhees71

Related Threads on Energy operator in Quantum Mechanics

Replies
33
Views
4K
  • Last Post
Replies
6
Views
3K
Replies
9
Views
1K
Top