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1

Are these transformations different to Lorentz and coordinate transformations when discussing symmetry? Yes and no. They have in common that we use them to look at the problem from another point of view. When one study wave phenomena it is very common (if not automatic) to use the Fourier transform (and sometimes the Laplace transform). For example the ...


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No. If $$a_p(t_0)\left | 0\right>=0$$ then $$a_p(t)\left|0\right>=e^{iH(t-t_0)}a_p(t_0)e^{-iH(t-t_0)}\left|0\right>=0$$ since we usually think of the vacuum as a energy eigenstate with energy zero (even if it's not zero, the above equation is still true).


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To keep things simple, let's talk about plane acoustic waves in one dimension. If we solve the wave equation in one dimension , we find that the acoustic pressure as a function of space and time is of the form $$P(x,t) = Ae^{i(kx -\omega t)}$$ where $A$ is the maximum amplitude, $x$ and $t$ are the displacement and time respectively, $\omega$ is the ...


4

The upshot is that we need one condition to specify how the operators in the Schrödinger and Heisenberg picture are connected. This is usually done by declaring that the two pictures agree at some fixed instant $t_0$. To summarize: The Schrödinger operator $\phi(\textbf{x},t_0)$ does not depend on time $t$, while the Heisenberg operator ...


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You may consider reading about Aharonov-Bohm effect. This is one of those cases, where the phase of the wave function, in sum with the electromagnetic 4-potential, is extremely important, as it gives different physical results. This effect was also checked experimentally, so it is not a pure theoretical abstraction.


2

The Fourier transform of a waveform has to contain an infinite range of frequencies in order to represent the full information content of the original. It is of course possible to create a waveform that only contains a finite range of frequencies (or rather - one for which the high frequencies vanish exponentially). The Gaussian is one such shape - its FT is ...


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The $p$ in the Fourier transform of the (free real scalar) field $$ \phi(x) = \int \left(a(\vec p)\exp(-\mathrm{i}px) + a^\dagger(\vec p)\exp(\mathrm{i}px)\right)\frac{\mathrm{d}^3p}{2p^0} $$ is a number, it is essentially a change in coordinates on $\mathbb{R}^n$ like every Fourier transform. The canonically conjugate momentum $\pi(x) := \frac{\partial ...


2

So, I suppose that $Φ(k)$ is the probability density of the momentum. Is this true? Almost. $\Phi(k)$ is the probability amplitude for the momentum of the particle. The probability density is obtained as usual by squaring the amplitude, giving $|\Phi(k)|^2$. For a free particle, all values of momentum are always allowed, which enables the superposition ...


2

First, realize we are doing an approximation when we are evaluating the coefficient $A_p$ in the Fourier series $$ \psi(x) = \sum_p A_p \psi_p(x)$$ by the integral $$ A_p = \int_{-\infty}^\infty \psi_p^*(x)\psi(x)\mathrm{d}x$$ since the limits should really be the boundary of the interval on which the Fourier series is periodic. Furthermore, $\psi(x) = ...


3

Because these are actually Fourier transform of the usual Green functions. Considere the Schrödinger equation : $$ \hat{\mathcal{H}}|\Psi(t)\rangle=\mathrm{i}\partial_t|\Psi(t)\rangle $$ The general solution $|\Psi(t)\rangle$ of such equation for a time-independant hamiltonian $\hat{\mathcal{H}}$ can be expressed in terms of Green function $G(x',x,t)$ : $$ ...


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Step 1: Write the function $\phi(p,t)$ in terms of its Fourier components, i.e.: $\phi(p,t) = \int dx \phi(x,t) e^{i p x}$ Step 2: Notice that the $\nabla$ operator acts now only on the exponential factor inside the integral. Doing the derivative results in: $\int dx (-x^2) \phi(x,t) e^{i p x}$ Step 3: Multiply the equation by by a factor of $e^{-i p r}$ ...



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