587 views

### Is there a prediction of quantum mechanics that could be construed as representing an “energy-time uncertainty relation?” [duplicate]

As the title suggests. Is there a prediction of quantum mechanics that could be construed as representing an "energy-time uncertainty relation?" Does there exist any reference to such a prediction, or ...
240 views

### Interpretation of the energy-time uncertainty [duplicate]

From the uncertainty relation it follows that: $$\Delta E \ \Delta t = \hbar$$ $\Delta E$ is the energy uncertainty of a state, $\Delta t$ should be the uncertainty of the lifetime $\tau_b$ of the ...
270 views

44 views

### Energy - Time uncertainty relation [duplicate]

I have a question regarding the interpretation of the relation $\Delta E \Delta t \ge 1$. First, what is the exact meaning of $\Delta t$? We know that $\Delta E$ is calculated as the standard ...
37 views

### Definite energy in Quantum Mechanics [duplicate]

According to Phillips' "Introduction to Quantum Mechanics", Chapter 4 "an eigenfunction of the Hamiltonian always describes a state of definite energy". But how can that be without ...
36 views

### Energy-Time Uncertainty Relation and Virtual Particles [duplicate]

I've come across a hole in my understanding. Heisenberg's Uncertainty Principle can be expressed in terms of energy and time as $$\Delta E \, \Delta t \geq \frac{\hbar}{2}$$ where $\Delta E$ is ...
26 views

### Does a quantum commutator exist for energy and time? [duplicate]

In quantum mechanics the position operator $\hat{x}$ and the momentum operator $\hat{p}$ have a commutator $$[\hat{x}, \hat{p}] = i\hbar$$ Does a similar commutator also exist for the uncertainty ...
18 views

### Applying uncertainty principle to energy states [duplicate]

Often for this I have heard, the longer the lifetime of the energy state, the uncertainty in the energy state decreases as a result of heisenberg's uncertainty principle. However doesn't that look at ...
### Why can't $i\hbar\frac{\partial}{\partial t}$ be considered the Hamiltonian operator?
In the time-dependent Schrodinger equation, $H\Psi = i\hbar\frac{\partial}{\partial t}\Psi,$ the Hamiltonian operator is given by $$\displaystyle H = -\frac{\hbar^2}{2m}\nabla^2+V.$$ Why can't we ...