15
votes
Exponential operator approximation: Suzuki-Trotter Expansion
Hint: Use the BCH formula multiple times:
$$ e^{tA}e^{tB}~\stackrel{\text{BCH}}{=}~e^{tA+tB+[tA,tB]/2 +{\cal O}(t^3)}~\stackrel{\text{BCH}}{=}~e^{tA+tB}e^{[tA,tB]/2}+{\cal O}(t^3),\tag{1}$$
$$ e^{tB}e^...
8
votes
Accepted
Exponential operator approximation: Suzuki-Trotter Expansion
First (as I encourage you to explicitly check), expand the left hand side for small epsilon, keeping track of the ordering since (as you said) $[H_0, V] \neq 0$ in general
\begin{eqnarray}
e^{-i \...
5
votes
Accepted
Basis of infinite dimension Hilbert spaces in quantum mechanics
In mathematics, one usually defines the basis of an infinite dimension vector space as a set of vectors such that any other vector can be written as a finite linear combination of the vectors in the ...
5
votes
Accepted
Are loops allowed for paths in the path integral formulation?
Yes, loops are perfectly fine. But we never draw a "typical" path from the path integral anyway.
Formally, the path integral is over the Wiener space of continuous paths (see also this ...
3
votes
Accepted
Confusion on Hamiltonian unbounded from below and Ostrogradsky Instability
What is important is that the spectrum of the Hamiltonian operator $\hat{H}$ is bounded from below, and that it has a ground state; not that the classical Hamiltonian $H$ is bounded from below in the ...
2
votes
Accepted
Confusion with Hund's rule
Hund's first rule is based on non-relativistic quantum mechanics without any spin-orbit interactions. Since the non-relativistic Hamiltonian (in the absence of magnetic fields) is spin-independent, ...
2
votes
Assigning Measurement Operators to POVM Operators
One can consider the POVM operator with the sum as a coarse graining operation that takes each measurement result labeled by both $n$ and $k$ and labels them all by $n$ alone.
For example, if $n$ ...
2
votes
Variance/Standard deviation of an observable on a state that is a linear combination of eigenvectors of that observable
The oscillator Hamiltonian is non-degenerate, but this property is not necessary. Without loss of generality, to be self-evident later, stick to a two-dimensional Hilbert space, assuming
$$
A|1\rangle ...
2
votes
Is superposition a real quantum state or just an unknown
Is superposition a real quantum state or just an unknown ?
Theories that posit that quantum superposition is a byproduct of our ignorance of (possibly inaccessible) attributes with deterministic ...
1
vote
Long term dispersion of the wavepacket
It’s actually exactly like optics. This is because the paraxial Helmholtz equation is the same as the time dependent Schrödinger equation. There are different ways to get it.
One method is to use the ...
1
vote
Finding error in Choi matrix calculation which suggests a CP map is not CP
When you show that the map is CPTP, you write $p_1 = tr[ (1-\Pi) \rho ]$, but in the definition of your map $\Phi$, you write $1 - tr[\Pi \rho]$. The former is correct. Both expressions are equal for ...
1
vote
Accepted
Finding error in Choi matrix calculation which suggests a CP map is not CP
Your map is not CPTP. In fact, it is not even linear, so it is unclear what the Choi state even means.
The correct map would be
$$\Phi(\sigma) = \Pi_R \sigma \Pi_R + (\mathrm{tr}(\sigma) - \mathrm{tr}(...
1
vote
Is the position wavefunction a fundamental property of quantum object or is it an emergent property of the combination of more fundamental properties?
The standard frame of quantum theories are:
Large quantum systems behave classically and can be desribed as individual objects like a golf ball, an experimenter with a hammer, setting up an experiment,...
1
vote
Should I partial trace the hamiltonian or partition function for a reduced system?
No, the first approach is not correct.
E.g., take a 3-site Heisenberg chain with open boundary conditions, where you project the middle spin onto $\lvert\uparrow\rangle$. Then, you can easily check ...
1
vote
If an operator $A$ commutes with the Hamiltonian $H$ do they have common eigenstates?
Basically, what they are trying to say is that the "full" Hilbert space (the direct product space) contains unphysical states and you have to restrict yourself to physical states. We often ...
1
vote
Does a particle still behave as a wave after being detected as a particle?
...and leaves a pattern of a particle on the screen (the pattern being two humps that represent the two ranges of positions the photon land on).
You should be a little careful about how you describes ...
1
vote
Perturbation theory for an asymmetric top
Don't you need only $l^{\prime} = l$, since the angular momentum does not mix together subspaces with different values of $l$?
1
vote
Accepted
Are there any known bulk superinsulators?
Superinsulators were originally proposed in two dimensions, and it appears that the experimental efforts have focused on this case. However, it has been argued theoretically that 3D superinsulators ...
1
vote
Is superposition a real quantum state or just an unknown
A (pure) state is a state: there is nothing unknown about it.
The best possible example to clarify this is polarized light. Suppose you have light with diagonal polarization: $\vert \nearrow\rangle$. ...
1
vote
What happens when a photon interacts with a free electron?
In the classical theory, for a linearly polarized plane EM wave, the electron oscillates in the polarization direction, perpendicular to the propagation direction. An oscillating charge emits ...
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