Tag Info

New answers tagged

1

You can think of the continuos formalism as being the limiting case of the discrete-momentum one: if the momentum is taken as a discrete variable (which amounts to constraining the particles to be in some finite volume $V$) the fourier expansion of the (real, scalar) field is: $$ \tag{1} \varphi(x) = \sum_{\textbf{k}} \frac{1}{\sqrt{2V \omega_{\textbf{k}}}} ...


0

$$i {{\partial}\over{\partial t}}\pi=[\pi,\int d^3x\tfrac{1}{2}\pi^2+\tfrac{1}{2}\phi()\phi]$$ $$=[\pi,\int d^3x\tfrac{1}{2}\phi()\phi]$$ $$=\tfrac{1}{2}\int d^3x[\pi,\phi()\phi]$$ $$=\tfrac{1}{2}\int d^3x[\pi,\phi]()\phi+\phi[\pi,()\phi]$$ $$=\tfrac{1}{2}\int d^3x[\pi,\phi]()\phi+\phi[\pi,()]\phi+\phi()[\pi,\phi]$$ $$=\tfrac{1}{2}\int ...


0

Yes, if $\hat{O}$ commutes with $\hat{H}_0$, $\hat{O}$ is diagonalized in the base of eigenvectors of $\hat{H}_0$. Let's show this. If the operator $\hat{O}$ commutes with $\hat{H}_0$, we have $$\hat{O} \hat{H}_0 = \hat{H}_0 \hat{O}. \qquad (1)$$ Let's assume for simplicity that the spectrum of $\hat{H}_0$ is non-degenerate. Now, let $V_1$ be an ...


1

Your claim that the derivative is in the expansion of $\langle x'|\hat{p}\hat{x}|\psi\rangle$ acts on everythin on the right is correct. Realize that $$\langle x|\hat{p}|\psi\rangle=\frac{\hbar}{i}\frac{d}{dx}\langle x|\psi\rangle=\frac{\hbar}{i}\frac{d}{dx}\psi(x)$$ So $$\langle x'|\hat{p}\hat{x}|\psi\rangle=\frac{\hbar}{i}\frac{d}{dx'}\langle ...


3

The classical poisson bracket with the generator of any transformation gives the infinitesimal evolution with respect to that transformation. The familiar $$ \partial_t f = \{H,f\}$$ means nothing else than the time evolution of any observable is given by its Poisson bracket with the Hamiltonian, which is the generator of time translation. More generally, ...


1

Quantization is often treated as a mystery, yet it can be seen to arise naturally from classical Hamiltonian mechanics.1 This yields the quantization prescription $$ \{\dot{},\dot{}\} \mapsto \frac{1}{{\mathrm{i}\hbar}}[\dot{},\dot{}]$$ for all classical observables on the phase space that are to be turned into quantum operators.2 Therefore, if we want to ...


0

Quantization, in the Heisenberg picture is very much related to establishing quantum non-trivial commutators between the main observables of the theory. Originally this is linked to the classical Poisson bracket, which in the Hamiltonian formalism includes the cannonical momentum. Hope I gave some insight, as to why we are led to define the conjugade ...


0

In Hamiltonian formalism, operators such as $a(k)$ and $a^{\dagger}(k)$ can depend on time (in Heisenberg picture), or they can not (in Schroedinger picture). They don't depend on space in sense that they are given for any 3-momentum $k$. But there is a Fourier transform $\tilde{a}(x)$ and $\tilde{a}^{\dagger}(x)$, which is given for any spacial point $x$. ...



Top 50 recent answers are included