6

Note that $\hat{\mathbf{P}}^n = \hat{\mathbf{P}}$ only holds if $n\neq 0$, so your manipulation is incorrect. You need to extract the $n=0$ case before you replace $\hat{\mathbf P}^n$ with $\hat{\mathbf P}$, which is why the additional steps in the correct solution are required.


5

Consider a Lagrangian system with $n$ DOF. In the case where the Hessian matrix $\frac{\partial^2 L}{\partial v^i \partial v^j}$ has constant rank $r$, it is possible to replace $r$ velocities with $r$ momenta in the definition of the Hamiltonian. It is proven in theorem 2 of my Phys.SE answer here, that this Hamiltonian will not depend on the remaining $n-r$...


4

I think that the problem is that the square root of the Laplacian is a non-local opertor and non-locality is usully regarded as a bad thing in physics. The long range nature shows up in the general expression $$ (-\nabla^2)^s f(x)\equiv \frac{4^s}{\pi^{n/2}}\frac{\Gamma(s+n/2)}{\Gamma(-s)} \int_{{\mathbb R}^n} d^ny \frac{\left\{f(x)-f(y)\right\}}{|x-y|^{...


4

To be precise, this is not a potential, but the potential energy of two particles interacting. Let us take for example two charged particles with charges $q_1,q_2$, separated by vector $\mathbf{r}_{2} - \mathbf{r}_{1}$. The potential created by the first particle is $$\varphi_1(\mathbf{r}) = \frac{q_1}{|\mathbf{r} -\mathbf{r}_1|},$$ whereas the potential ...


3

It is a different application of the word determinate. Griffiths simply means that the result of a measurement is determinate if the state is an eigenfunction. That is not the same as determinism in time evolution.


3

If you're looking for a formulation of QFT that resembles Schrödinger's equation in single-particle QM and that can be solved on a (infinitely fast) computer, here it is: Scalar fields For a single scalar field with Hamiltonian $$ \newcommand{\pl}{\partial} H \sim \int d^3x \Big[\big(\dot\phi(x)\big)^2 + \big(\nabla\phi(x)\big)^2 + V\big(\phi(x)\big)\...


2

I am not experienced with cubic PDEs... only Infinite dimensional such. Your system looks like the first quantum correction in Wigner flows in phase space, but I cannot get much help from that magnificent formulation, off the cuff... I could only formulate your problem in 19th century language. You are seeking the Green's function for $$ \left (\partial_t ...


2

if you have this kind of differential equations: $$\vec{\ddot{r}}=-\vec{F}(\vec{r})\tag 1$$ you can get the Hamiltonian. multiply equation (1) from the left with $\vec{\dot{r}}$ $$\vec{\dot{r}}\cdot \vec{\ddot{r}}=-\vec{\dot{r}}\cdot\vec{F}(\vec{r})$$ thus: $$\frac{1}{2}\frac{d}{dt}(\vec{\dot{r}}\cdot \vec{\dot{r}})= -\vec{\dot{r}}\cdot\vec{F}(\vec{r})$$...


2

Indeed, the eigenstates of the unperturbed Hamiltonian ($\hat{H}_0$) are not the eigenstates of the perturbation ($\hat{H}_1$) - otherwise one wouldn't need the perturbation theory, since the perturbing Hamiltonian is diagonal: it does not mix the states and merely adds a correction to the energy levels of $\hat{H}_0$. Let me further note that frequently ...


2

A more geometric approach is to consider the $2n+1$-dimensional contact manifold ${\cal M}$ with coordinates $(q^i,p_j,t)$. The Hamiltonian action functional is $$S_H[\gamma]~=~\int_I \gamma^{\ast} \Theta, \qquad \Theta~=~p_j \mathrm{d}q^j -H \mathrm{d}t, \tag{1}$$ where $\gamma:I\to {\cal M}$ is a curve. This action formulation (1) is world-line (WL) ...


2

Yes, QFT books are unfortunately often vague about the connection to reality until the chapters where they discuss scattering, in my experience. The measurement-related postulates in QFT are the same as in QM but with some difficulties. $| \langle p | \psi \rangle |^2 $ is the probability density of getting a momentum $p$ in an experiment measuring momentum; ...


2

Why don't they commute? You can work out the maths. It does not have anything to do with the factor but the form. $$ \begin{align} [\mathbf L \cdot \mathbf S, \mathbf L_x] & = [L_xS_x+L_yS_y +L_zS_z, L_x] \\ & = S_x[L_x,L_x] + S_y[L_y,L_x] + S_z[L_z,L_x] \\ & = 0 - S_y (i \hbar L_z) + S_z (i \hbar) L_y \\ & = i\hbar (\mathbf L \times \...


1

Generally, the non-diagonal elements are complex numbers, i.e. they may have a phase: \begin{equation} \hat{H} = \begin{bmatrix} E_1 & \Delta e^{i\phi}\\ \Delta e^{-i\phi} & E_2\\ \end{bmatrix}. \end{equation} In your case the phase is $\phi = \pi$. From the point of view of the energy levels the phase is irrelevant, as they are determined in terms ...


1

In a canonical transformation, the new hamiltonian could have nothing to do with the initial hamiltonian, it just have to preserve Hamilton's equations. So in the new variables $(Q,P,t)$ you have to have that $$\dot{Q} = \frac{\partial K}{\partial P} \qquad \dot{P} = -\frac{\partial K}{\partial Q}$$ where $K$ is the new Hamiltonian. Whenever this happens ...


1

If $H= \sigma_z P$, where $$ P= \left[\matrix{0&0\cr 0&1}\right] $$ is the projector on spin down, then $$ H= \left[\matrix{1&0\cr 0 &-1}\right]\left[\matrix{0&0\cr 0&1}\right]= \left[\matrix{0&0\cr 0 &-1}\right]. $$ so the two eigenstates are $(1,0)^T$ with energy zero and $(0,1)^T$ with energy $-1$ and $$ \exp\{-iHt/\hbar\...


1

If you change the generalized coordinate the eigenvalues of your new equation of motion must be the same otherwise the dynamic of your new system is changed. to obtain the eigenvalues you must linearize your equation of motions. Example: $$\underbrace{\begin{bmatrix} \ddot{x} \\ \ddot{y} \\ \end{bmatrix}}_{\vec{\ddot{q}}}+\underbrace{\left[ \begin {...


1

Rearranging the radial equation, we can write : $\frac{d^2}{dt^2}u-[{\frac{l(l+1)}{r^2}}-\frac{\alpha}{r}]u -\kappa^2u = 0$ , where $\kappa = \sqrt{\frac{-2mE}{\hbar^2}}$ , and $\alpha = \frac{2me^2}{4\pi\epsilon_0\hbar^2}$ As $r \to \infty$ , The the effect of the term in the $[$ $]$ (which is the "effective Potential" , $V_{eff}$) becomes ...


1

In $\left[b_{\vec{p}}^{s \dagger} b_{\vec{p}}^{s}+c_{\vec{p}}^{s \dagger} c_{\vec{p}}^{s}\,\,,\,\,b^r_{\vec{q}}\right] = \left[b_{\vec{p}}^{s \dagger}\,\,,\,\,b^r_{\vec{q}}\right] b_{\vec{p}}^{s}$, you assumed that $b^r_{\vec{q}}$ commutes with all the other operators. However, it anti-commutes (and therefore does not commute). Using the relation $[AB,C] = ...


1

We start with a function $L(q, v) $, and use it to define a function $$p_q(v) := \frac{\partial L} {\partial v}(q, v). $$ Then obviously we can define another function $$H'(q, v) = p_q(v)v - L(q, v).$$ All of this is trivial. The point is that we want to invert $v \rightarrow p_q(v)$. Suppose we have obtained such a map, for each fixed $q$, that is, we have ...


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