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visits member for 2 years, 9 months
seen Jul 29 at 7:51

Currently doing a PhD in Controlled Quantum Dynamics at Imperial College and the University of Oxford.


Jul
28
comment Is Hamiltonian a differential operator in second quantization?
@Minethlos Yes but you only determine the constants once you add both the homogeneous and particular solutions. So the relative weight of the homogeneous and particular solutions is determined by the boundary conditions or other constraints. The general solution is just that: the most general thing you can write down that solves the equation. Afraid I don't have any mathematical references. However one does frequently see Green's functions techniques used in finite-dimensional scattering problems, e.g. the Landauer-Buttiker transport formalism.
Jul
27
comment Is Hamiltonian a differential operator in second quantization?
@Minethlos Exactly the same question arises in the differential equation case. Strictly speaking you should add some undefined multiple $\alpha \lvert \phi\rangle$. The solution is then determined up to an unknown constant $\alpha$. It is never OK to just choose the prefactor ad hoc. Generally, the solution of such an equation is undetermined up to an arbitrary vector from the kernel of $(E - H_0)$. In the case of a differential equation this vector is determined by boundary conditions. For a scattering problem, this should probably be that the state is asymptotically an incoming plane wave.
Jul
27
comment Is Hamiltonian a differential operator in second quantization?
@Minethlos The homogeneous solution is just a zero eigenvector of the linear operator $(E - H_0)$. It is clear that one can always add such a zero eigenvector to the solution of $(E - H_0)\lvert \psi\rangle = V\lvert\psi\rangle$. So I don't see any need to restrict the conclusions to a statement about differential/infinite-dimensional operators.
Jul
27
comment Is Hamiltonian a differential operator in second quantization?
Note that the derivation in the linked answer does not need to make any assumptions about whether $H$ is a differential operator. Exactly the same manipulations could be performed for a finite-dimensional matrix: it is just matrix inversion, after all. In the case of an infinite-dimensional matrix the "inverse" is more frequently called the Green's function, but the principle is the same.
Jul
26
comment How is this possible that photons are absorbed?
To clarify: are you asking how it is possible that two different electrons can have exactly the same energy, so that a photon emitted by one can be absorbed by the other? The answer to this is indeed a sort of uncertainty principle argument: the photons emitted by atomic transitions do not have a perfectly sharp energy, but rather a range of energies. This is due to the finite lifetime of the atomic states which decay.
Jul
24
comment bose einstein phase transition
$\mu$ is implicitly a function of temperature. The phase transition occurs at $\mu = 0$ (for a non-interacting Bose gas).
Jul
24
comment How to Derive Atomic Hamiltonian and Cavity Hamiltonian?
You can use LaTeX style code pretty much everywhere. As far as I know \ket{} is not a standard LaTeX macro and you would need to define it yourself.
Jul
24
comment How to Derive Atomic Hamiltonian and Cavity Hamiltonian?
To typeset Dirac notation use, for example, \lvert a \rangle, which produces $\lvert a \rangle$. Derivation of the atomic Hamiltonian is trivial, it is just how one writes in Dirac notation a general $2\times 2$ matrix whose eigenvectors are $\lvert e\rangle$, $\lvert g\rangle$ with eigenvalues $\hbar \omega_{e,g}$. I'll write up a derivation of the field Hamiltonian later unless someone wants to jump in first.
Jul
21
comment Lennard Jones Total system energy
You are using a very high density of about $10^{28}~m^{-3}$, corresponding to an interparticle distance of 4 angstroms. Therefore there is a high probability of each particle being deep inside the repulsive core of another particle. Why don't you try decreasing the density? If your collisions are elastic then the energy should not rise, but I expect this is due to a numerical instability of your code.
Jul
20
comment Books on superconductivity and its relation to spontaneous symmetry breaking
You might like the textbooks by Altland & Simons and also Wen.
Jul
16
comment Why does not the bare interaction potential appear in the Bogoliubov theory?
Until the electron orbitals start to overlap, at which point the scattering problem is a many-electron problem and the concept of an atom-atom potential no longer makes sense. My point is that there does not exist any "bare" potential $V(r)$, there are only effective potentials valid beyond a certain radius. If you want to use the Lennard-Jones potential, you find that its Fourier transform is divergent as $k\to\infty$, due to the singularity as $r\to 0$. But if you don't care about physics at this scale, you replace it with a pseudo-potential whose Fourier transform is simple to work with.
Jul
16
comment Why does not the bare interaction potential appear in the Bogoliubov theory?
What is the "bare" atom-atom interaction? Of course at large distances it will look like $V(r)\sim 1/r^6$ for neutral atoms. But what about when $r$ is comparable to the Bohr radius?
Jul
15
answered What are typical error rates of quantum computers?
Jul
9
comment How is the degenerate electron gas state “degenerate”?
@JohnRennie I think the word degenerate in this context refers to the possibility of multiple particles occupying the same single-particle state (cf. multiple states with the same energy). Obviously in a Fermi system this possibility does not exist, whether or not the system is quantum degenerate. However the Pauli principle only becomes relevant at temperatures where fermions would "try" to occupy the same state in the absence of the exchange statistics.
Jul
9
comment Must we test whether e.g. $A=B$ and $A=C$ implies $B=C$ by experiment?
The transitive axiom of algebra relates symbols written down on a piece of paper. Feynman is talking about the outcomes of physical measurements. I think the broad point he is making is that one should not assume a priori that the abstract mathematics of algebra describes relationships between measurable quantities, but that it turns out experimentally that in this case (and in many others) it does.
Jul
2
comment Are the electrons at the centre of the Sun degenerate or not?
Could you specify exactly what it is you want to calculate? Whether or not you have a good approximation is always dependent on what you are trying to calculate in the first place.
Jul
2
comment Random walk recurrence term and the self-energy
Yes, I am pretty sure that there is an extensive mathematical literature on similar topics beyond Pierre-Louis' work, so you might have some luck mining that.
Jul
2
comment Random walk recurrence term and the self-energy
Sorry, the third link I posted was incorrect: here is the paper I meant. Its value is probably more in the slightly more physical discussion of what path-sums actually mean in the context of an example model. By the way, these papers are seriously, seriously mathematical. They will probably need some commitment if you want to understand them properly (I do not myself).
Jul
2
comment Random walk recurrence term and the self-energy
Feelings 100% mutual :) I do try to encourage more people with an interest in quantum stochastic problems to join SE whenever I can, it's a shame there are not more users here who share our interests.
Jul
2
comment Random walk recurrence term and the self-energy
An old colleague of mine wrote his PhD on a method to compute elements of the many-body perturbation expansion in terms of walks on graphs. Your question reminds me quite a lot of this. The basic idea is to represent quantum dynamics as a random walk on the graph whose adjacency matrix is the Hamiltonian. Unfortunately, his thesis is apparently not available online, but you might find some insights in his papers, here, here and here (I helped a bit with the last one, but just with numerics).