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11 votes
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If a unitary operator is close to the identity, will it leave any state it acts on unchanged?

I assume that the spectrum of $G=G^*$ stays in $[0,+\infty)$ and $\Lambda \leq +\infty$ is an upper bound of it. $$\langle \psi |e^{-i\delta G} \psi\rangle = \langle\psi|\psi\rangle+ \langle \psi |(e^...
Valter Moretti's user avatar
6 votes
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Why doesn't Car-Parrinello molecular dynamics require an SCF calculation?

It is not entirely true that Car-Parrinello (CP) Molecular Dynamics (MD) doesn't require a self-consistent field (SCF) calculation at all. The method must start close to the SCF solution (actually a ...
GiorgioP-DoomsdayClockIsAt-90's user avatar
5 votes
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Density operator commutator in master equation

In general, you will probably want to look into the Gaussian formalism for continuous-variable states and the rotating frame here. But we can solve directly for this particular problem. In the present ...
Quantum Mechanic's user avatar
4 votes
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Symbol denoting parity eigenvalue

In nuclear physics the symbol is usually $P$. In that context the parity is rarely abbreviated without also using the total angular momentum. For example, the deuteron has $J^P=1^+$. I am trying to ...
rob's user avatar
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3 votes

If a unitary operator is close to the identity, will it leave any state it acts on unchanged?

Here is a very simple lower bound for the fidelity, maybe this helps in addition to the existing answer: Let $U=\exp(-i\delta G)$, then using Dahumel's formula one can show the inequality $|| \exp(A) -...
Refik Mansuroglu's user avatar
3 votes

Parametric down-conversion - QFT necessary?

Your approach is correct. The only issue is: why can we do it? Standard practice in QFT is to quantize the fields as free fields and then add the interactions perturbatively, assuming the couplings ...
flippiefanus's user avatar
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3 votes

Non-locality of the wavefunction in QM and Twistor theory

The discussion in the comments seems to be a classical example of miscommunication. People don't exactly define what they means by the terms that are used. Before, attempting to answer the OP question,...
flippiefanus's user avatar
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2 votes

Time dependency of Wave function and its probability density function (PDF)

When we study the Schrodinger wave equation, we have a time dependent wave function $\Psi(x,t)$, and when we deduce its Probability Density function we come to know there is no time dependence in the ...
Thomas Fritsch's user avatar
2 votes

What, exactly, does Schrödinger's wave equation describe (just in plain English, without any of the math please)

Unfortunately, in a specific sense your description is quite wrong. Many of the aspects that you describe are interpretations that arose years after the framework was first published. You are ...
Cleonis's user avatar
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2 votes

Is Bell's theorem wrong?

People tend to overcomplicate this. When we have two entangled polarised photons and pass them through polarising analysers (A and B) the correlation probability is given by $$p = \cos^2(\theta_A - \...
KDP's user avatar
  • 6,217
2 votes

Wave packet as a field configuration acting like a particle's wave function?

There is a mathematical equivalence between two different physical situations. The first situation is a wavepacket in a classical field. This describes some localized packet of energy in an ordinary ...
Andrew's user avatar
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1 vote

When does electrostatic effect failed and needs quantum mechanics?

Accelerated charges radiate energy, and a circular orbit implies constant acceleration and constant energy. Thus, a stable circular orbit for a charged particle is incompatible with classical physics....
ZeroTheHero's user avatar
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1 vote

A question about Quantum entanglement, 45° Light Polarization, and faster than light communications

When the photons leave the crystal, their polarization is in superposition. They have no definite polarization. When they hit a polarizer, 50% of them pass through, regardless of any entanglement. ...
Jakob Heitz's user avatar
1 vote
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Many-Worlds Interpretation Question

In the MWI different versions of a system can interfere with one another and when interference effects are strong it doesn't make much sense to say they're in different universes because they affect ...
alanf's user avatar
  • 8,797
1 vote

How do operators on kets and wavefunctions correspond?

Is it by showing that this holds for position and momentum operators, and then have the result follow from any observable having to be a function of these? Yes, naturally. Your text or instructor ...
Cosmas Zachos's user avatar
1 vote

When is the temperature relevant for a quantum many body system?

If the system is at $T=0$, all particles should be at the ground state and no excited state is possible. However, this is clearly not true for an arbitrary quantum system, given that if we solve ...
Roger V.'s user avatar
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1 vote

The usage of temperature in quantum mechanics

Many-body quantum mechanics is also often referred to as Quantum statistical physics, which is a more telling name in the sense that it points out that we are considering systems where one has to take ...
Roger V.'s user avatar
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