# Tag Info

2

One gets there by noting that $\langle x | p \rangle = e^{i p x/\hbar}$ is a plane wave, and you have to throw on the test wave function to talk about the derivative operation. So, you are worried about $$\langle x| P | \Psi \rangle = \int dp ~p~ e^{i p x/\hbar} ~\langle p | \Psi \rangle$$ From there you note that you can get the ...

2

I believe in the last line, the plane-wave functions $u_k(x)$ should carry different coordinates and momenta, e.g $$[a(k)^\dagger,a(k')]u_k(x)u_{k'}(x')$$ You may note that the commutator $[\phi(x),\pi(x')]=i\hbar\delta(x-x')$ holds if one choses $[a_k,a_{k'}^\dagger]=\delta_{kk'}$. However, this indirect reasoning is no proof that this choice is unique. ...

2

In Mathematica: Refine[-(c^2/(4 \[Pi] c r)) Integrate[ DiracDelta[r - c t] Exp[ I \[Omega] t], {t, -\[Infinity], \[Infinity]}], Assumptions -> {r \[Element] Reals, c > 0}] The output is $$-\frac{e^{\frac{i r \omega }{c}}}{4 \pi r}=-\frac{e^{irk_0}}{4 \pi r}$$ The extra factor of $c$ is eliminated since the $\delta$ function has an argument of ...

0

I assume that by zero frequency, you mean zero momentum transfer. Zero momentum transfer corresponds to the $k=0$ value of the Fourier transform. The value of this part of the fourier transform is the integral of the scattering strength over all space. So you can think of this value as being the total amount of stuff that is there. Another thing to keep in ...

1

First, a somewhat minor point is that $x = 0,0.01a,0.02a,...a,1.01a,....2a....100a$ actually gives a list of 10001 points, not 10000 points. I will assume that you actually meant to say $x = 0,0.01a,...a,1.01a,....2a....99.99a$. Second, you say that $$V(x)=\sum_{K}e^{iKx}V_{K}$$ where $K =\frac{2\pi n}a$ and $n=0,1,2,3$, but this gives a non-Hermitian ...

4

The reason you can get rid of the integral and the exponential is due to the uniqueness of the Fourier transform. Explicitly we have, \begin{align} \int \frac{ \,d^3p }{ (2\pi)^3 } e ^{ i {\mathbf{p}} \cdot {\mathbf{x}} } \left( \partial _t ^2 + {\mathbf{p}} ^2 + m ^2 \right) \phi ( {\mathbf{p}} , t ) & = 0 \\ \int d ^3 x \frac{ \,d^3p }{ (2\pi)^3 ...

2

The idea is that the equivalence must hold for $\textit{all}$ values of $\vec{x}$. Another way to note it, you can see $\left(\frac{\partial^2}{\partial t^2} +|\mathbf{p}|^2+m^2\right)\phi(\mathbf{p},t)=0$ as the Fourier transform of your function. And which is your function? 0, and which is 0-s Fourier transform, well zero.

6

The functions $e^{i \bf p \cdot \bf x}$ as functions of $\bf x$ are linearly independent for different $\bf p$'s, hence every coefficient in the linear superposition (that is, in the integral) must be zero.

2

(a) Your answers are correct. (b) Yes,it is simpler to write out the wave packet in the momentum basis. (This is effectively equivalent to working out the three-dimensional Fourier transform of the given Gaussian wave packet in position basis.)

2

As very often in these sorts of proof, we just need to use a completeness relation: We have, \begin{align} \langle p | \psi _0 \rangle & = \langle p | \int dq |q \rangle \langle q | \psi _0 \rangle \\ & = \int dq e ^{ - i p q } \langle q | \psi _0 \rangle \\ & = \frac{1}{ \sqrt{ a \sqrt{ \pi }}} \int dq e ^{ - i p q } e ^{ - q^2 / 4 a ^2 ...

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