New answers tagged path-integral
2
The prefactor $F(t_f,t_i)$ is given in eq.(3-50) of Ref. 1 as
$$ F(t_f,t_i)~=~\qquad $$
$$\tag{3-50'} \int_{y(t_i)=0}^{{y(t_f)=0}}\!\!\! {\cal D}y~\exp\left\{\frac{i}{\hbar} \int_{t_i}^{t_f} \!\! dt[a(t) \dot{y}(t)^2+ b(t) y(t)\dot{y}(t)+c(t) y(t)^2 ] \right\}. $$
I doubt that there exists a closed formula for the path integral (3-50') for arbitrary ...
0
I have not the Feynman and Hibbs book, but I think, that, the action being quadratic, you will have to use the square root of the determinant of the second (coordinates) derivative matrix of the classical action.
You will confirm this by testing that, when $t_b \rightarrow t_a$, then $K(b, a, t_b, t_a) \rightarrow \delta(b - a)$
0
There is a difference between the mathematical treatment of Quantum Mechanics, and the pratical job of an experimenter.
Quantum mechanics says that the outcome of a measure is a particular eigenvalue of an operator:
$$ Particle \, Position : X^i(t)|\psi> = x^i(t)|\psi>$$
$$ Particle \, Spin : S_z\psi> = s_z|\psi>$$
$$ Fields : ...
0
It might be true for those systems whose only degrees of freedom are time and position. However, there are other internal components such as spin that do not directly reduce to position or time. So this is one example where this statement fails.
Having said that, I would not discount it entirely. Spin can certainly be measured by observing trajectories of ...
0
One point to consider, although not a definitive answer, is the following. The validity of the pilot-wave theory (Bohmian mechanics) relies on the truth of Feynman & Hibbs' postulate (F&H). This is because the pilot-wave theory only makes predictions about the positions of all particles, which along with the unobservable wave function constitute a ...
4
For simplicity, let's restrict the discussion to that of a single particle moving in one dimension. Path integrals can be performed in much broader contexts like quantum field theory, but I think that would conceptually obscure the issue at this point.
Let $H$ denote the (time-independent) quantum hamiltonian. Then the time evolution on the system is ...
0
What your friend actually meant is that you can obtain all desirable correlation functions. Assuming you're talking about a non-relativistic electron, consider a source term added to your action
$$
S'[x] = S[x] + \int dt \, J(t) x(t)\,.
$$
Now you can write any correlation function as a derivative of $\ln Z$ calculated at $J = 0$, i.e.
$$
\langle B| ...
2
I don't have the book with me right now, but from the passage you highlighted I gather that he's talking about his famous Feynman path integral formula. The partition function for a particle going from point A to B is given by
\begin{align}
Z = \int_{A \to B} [\mathcal{D}x]~ e^{iS[x]},
\end{align}
where $\mathcal{D}x$ is the measure that sums up over all ...
0
I am not sure if you are looking for this, but you can define a Lagrangian in such a way that the L-EOM (equation of motion) is the Schrödinger equation.
$\cal{L}=\Psi^{t}(i\frac{\partial}{\partial t}+\nabla^2/2m)\Psi$
$\frac{\partial\cal{L}}{\partial\Psi^t}=0$
The second term of the Lagrange-equation (derivative with respect to $\partial_{\mu}\Psi^t$) is ...
2
Recall the definition for a derivative is
$$\frac{\mathrm{d}x(t)}{\mathrm{d}t}=\lim_{\Delta t\to 0}\frac{x(t+\Delta t)-x(t)}{\Delta t}$$
So if we insist on having discrete values of time, we get
$$\frac{\mathrm{d}x}{\mathrm{d}t}\approx\frac{x(t+\Delta t)-x(t)}{\Delta t}$$
Or if everything is in integer multiples of some interval, we have
$$\dot{x}\approx ...
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