New answers tagged

1

This can be trivially achieved, if you are willing to go to extended phase space, that is add an extra degree of freedom to your system, represented by the momentum - position pair $(p_e, q_e)$. Then the dynamic system evolves under the effective Hamiltonian $$ H_\text{eff} = p_e + H(\mathbf{p}, \mathbf{q}, q_e) $$ The effective Hamiltonian $H_{\text{...


0

There is no "configuration manifold" in quantum mechanics. "Quantization" does not map points in the classical configuration manifold to anything in the quantum theory. Quantization produces a quantum theory from a classical theory, it is not a map that turns classical states into quantum states. "Quantization" is, in general, not a well-defined process, ...


0

All three coefficinets have dimensions of energy, so $J_2/J_1$ and $\lambda/J_1$ are dimensionless.


1

Normally the ("stationary") eigenstates of an operator (here, your nonhermitean hamiltonian) serve to find the time-dependent solution of the corresponding TDSE, by suitable multiplication by phases $e^{−itα}$, each. Why would you use a basis of eigenstates of another (hermitean oscillator H) operator? The coherent states $|\alpha\rangle$ you are ...


2

As far as I know, the classical and quantum Hamiltonian are related because of the Hamilton-Jacobi equation. The Hamilton Jacobi equation is a classical equation of motion that uses the Hamiltonian itself, and it reads $$\frac{\partial S}{\partial t}+H=0$$, and $S$ is termed Hamilton's principal function, together with the equation $p_j=\frac{\partial S}{\...


2

The relationship between classical and quantum mechanics is expressed the clearest in the process of Dirac's canonical quantisation. This consists of promoting replacing all Poisson brackets $$ \{A, B\} = \frac{\partial A}{\partial q} \frac{\partial B}{\partial p} - \frac{\partial B}{\partial q} \frac{\partial A}{\partial p} $$ by the corresponding ...


1

OP's Hamiltonian of the form $$ H~=~\frac{p^2+m^2}{2}, \qquad p^2~:=~p_{\mu}g^{\mu\nu}(x) p_{\nu}~\leq~0,\tag{1} $$ is called a super-Hamiltonian in e.g. MTW, cf. e.g. this Phys.SE post. (For a massless particle like the photon $m=0$.) The super-Hamiltonian (1) is the $e=1$ gauge of the Hamiltonian $$H~=~ \frac{e}{2}(p^2+m^2)\tag{2}$$ for a relativistic ...


3

The Hamiltonian is not always the sum of potential and kinetic energies. Like the Lagrangian, it is a theoretical construct that is not unique for any given physical system, and you can always construct a time-reparametrization invariant formulation of a system where the Hamiltonian vanishes. See this answer of mine for the construction and the first part of ...


-1

Maybe this is too naive a picture, but under your time-dependent Hamiltonian you can define a unitary time-evolution operator: $U(T,0) = {\cal T} \left( e^{-i \int_0^T H(t') dt'} \right)$ where $\cal T$ signifies the time-ordering operator. As $U$ is a unitary operator, you can express it in terms of a hermitian operator $H_{\mathrm eff}$ as $U(T,0) = e^{-...


2

No there are two in the alternative solution as well. You have one $N$ for each exponential, one with the plus sign and one with the minus sign. What you have is $$ \psi(\phi) = A \cos (k \phi) + B \sin (k\phi) = N_+ e^{i k \phi} + N_- e^{-i k\phi}. $$ To answer the comment below: In case you redo the calculation instead with the ansatz $\psi = N_+ e^{...


2

While total energy is conserved, like you say, the energy of an individual system need not be conserved if it is being manipulated from the outside. For Hamiltonians, this happens when you introduce a time-varying potential $V(x, t)$. Then $\frac{dH}{dt} = \{H,H\} + \frac{\partial H}{\partial t} = \frac{\partial V}{\partial t}$.


0

$\newcommand{\Ket}[1]{\left|#1\right>}$ $\newcommand{\Bra}[1]{\left<#1\right|}$ $\newcommand{\BK}[1]{\left<#1|#1\right>}$ The local energy can be used within an MCMC sampling scheme in order to optimize your variational neural network state $\Ket{\Psi}$ to the ground state of the applied Hamiltonian, $\mathcal{H}$. To optimize the network state ...


2

It is the equation of time evolution of the operator $H(\vec{x})$ in the Heisenberg picture. This is another formulation of QM where the Schrödinger equation is replaced with $$ \frac{dA(t)}{dt} = \frac{i}{\hbar} [H, A(t)] $$ for any operator $A$. In this case, I imagine that the total Hamiltonian would be given by $H = \int d^3 x\, H(\vec{x}, t)$, so you ...


24

The Hamiltonian for the He atom is: $$H = -\frac{\hbar^2}{2m_e}(\nabla_1^2 + \nabla_2^2) - \frac{2e^2}{4\pi\epsilon_0 r_1} - \frac{2e^2}{4\pi\epsilon_0 r_2} + \frac{e^2}{4\pi\epsilon_0 r_{12}}$$ where the electrons are denoted 1 and 2, and $r_i$ is the distance to the nucleus at the origin and $r_{12}$ the distance between the electrons. Since the ...


3

The reason that we take it to be the energy is that this is closely related to the classical energy, when one follows the standard rules and conventions of "quantizing" the classical quantities. In a similar manner, you can take any classical quantity $A$, and then from the classical concept of the Hamiltonian (which is the classical energy $E_K+V$ in ...


8

This is essentially the definition of energy in quantum mechanics -- it is $\hbar$ times the rate of change of phase. That's one of the fundamentally new ideas in quantum mechanics, so it can't be derived from anything you already know classically. If that's not very satisfying, we can say the same thing in more steps. The energy of an energy eigenstate is ...


1

You just need to keep scrolling down in the Wikipedia article, I will just add the mention of the Wick rotation. First of all this is a hypothesis, so you can understand it as a "well motivated guess", the reasoning coming from (as stated in the Wiki): $$\bar{A} = \lim_{T\rightarrow \infty} \frac{1}{T} \int_0^T {\rm d}t\, \langle \psi(t)|\hat{A}|\psi(t)\...


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