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These relations are found in every book on QM, but the usual notation is $$ X|x\rangle=x|x\rangle $$ and $$ P|p\rangle=p|p\rangle $$ To go from these equations to the ones you've written, you just have to project them into the position basis $|x'\rangle$ (and use $\langle x'|x\rangle=\delta(x-x')$ and $\langle x'|p\rangle\sim\exp[ipx]$). Edit Important: ...


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Quick answer My question is how does the presence of nonzero $J(x)$ results in a non-trivial spacetime dependent value of $\langle 0|\phi(x)|0\rangle$? The equation $\phi(x)=\mathrm e^{-iPx}\phi(0)\mathrm e^{iPx}$ works both for $J=0$ and $J\neq 0$. Therefore, $$ \langle \phi(x)\rangle_J= {}_J\langle 0|\mathrm e^{-iPx}\phi(0)\mathrm e^{iPx}|0\rangle_J ...


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The work done onto the spring is $dE/dt=F(t) \dot x(t)$. You should not look at the direction of $F$ alone, but at the the direction of the motion as well.


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If the Fourier transform of $f(x)$ is $\tilde{f}(k)$, then the Fourier transform of $df/dx$ is $ik\tilde{f}(k)$. Proof: $$\frac{df}{dx} = \frac{d}{dx} \int \frac{dk}{2\pi} \tilde{f}(k) e^{ikx} = \int \frac{dk}{2\pi} \left[ik\tilde{f}(k) \right] e^{ikx}.$$ This explains the momentum factors, so we've reduced the task to showing $$\Gamma(x, y, z) \sim ...


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It depends upon the shape of your laser pulse. The time bandwidth product is defined as: $$\Delta v \Delta t \geqslant K $$ Where both $\Delta v$ and $\Delta t$ are measured as FWHM (full width half max). The factor $K$ depends upon the shape of your laser pulse. $K = 0.441$ for a Gaussian shaped pulse, $0.315$ for a sech pulse, $0.142$ for a Lorentzian ...


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A qualitative understanding can be gained from the animation provided by 'TheGhostOfPerdition', each small peak on the dark blue wave (envelope of the wave packet) has an associated $k_{0}$ for which $\phi(k)$ peaks at $k = k_0$. So you can work out the velocity $v_g : = \frac{d \omega}{d k}|_{k = k_0}$ of the dark blue (envelope of wave packet) using the ...


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I believe that the question is asking you to perform the Fourier transform on the given function and instead of plotting the resulting complex function in 3D, to convert the values from the Fourier transform into the magnitude and phase. That is, take the complex numbers you get from the transform $\mathbf{F}(y)=a(\omega)+ib(\omega)$, convert to ...



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