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1

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 ...


0

The expression you obtained is almost correct (check the signs in your residue calculations more carefully). It is still true that $|G(r,t;r,t_0)|^2=G(r,t;r,t_0)^*G(r,t;r,t_0)=G_A(r,t_0;r,t)G(r,t;r,t_0)$ is the product of the advanced and retarded Green functions (it's just another perspective on the same math). In Fourier space, the product is transformed ...


2

The delta function is not really a function, it is a distribution, In the strict sense both $\delta (x)$ and $e^{ikx}$ are not normalizable when $n=m$ One way to prove your equations is to use fourier transforms Using Placherels theorem the fourier transform $F([f(x)]k)$ for the function $f(x)$ is given by ...


0

Generally put, Fourier Transform is a decomposition of your signal to its harmonic components. Each point in Fourier space correspond to a certain frequency (be it temporal, or spatial, at the most common cases), and the higher the value of the transform there, that frequency becomes more dominant in the actual signal. You can look at it like a 'histogram ...


3

I'd like to ask what is the meaning of the value obtained from X(jω) with certain frequency ω Consider for a moment, the synthesis equation where we 'construct' $x(t)$ out of a weighted 'sum' (integral) of the orthonormal basis functions of time: $e^{j\omega t}$ $$x(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty}\mathrm{d}\omega\:X(\omega)\:e^{j\omega ...


0

Your Fourier transform eq. is incorrect. it should be $e^{-jwt}$ not $e^{jwt}$ Physical meaning: Split your integral into real and imaginary parts, the Fourier transform equation is split into integral with sine and cosines instead of $e^{jwt}$, but with same frequency. $F = \int \Re{x(t)} cos(wt) dt + j \int \Im {x(t)} sin(-wt) dt$. You could think of ...


2

Since the Fourier transform is a way to express the time evolution of your quantity such as the voltage in frequency space, you can interpret it as the value as the density distribution of your quantity in frequency space. Note that it is a continuous distribution, thus there is a density. The reason for the unit is simple: it's Volt/Hertz = Volt * Second. I ...


2

As a complement to another answer, I'll demonstrate moving from the coordinate representation (wave function) to the momentum representation. Recalling that $$\Psi(x,t) = \int dp \frac{e^{i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}}\Phi(p,t)$$ and putting this expression into the (coordinate representation of the) TDSE, we have $$i\hbar\frac{\partial}{\partial ...


4

Why is the vector |S⟩ represented as Ψ for both bases when working out the components for the quantum mechanics case above? The first of the final two equations is simply an expression for the sifting property of the delta 'function'. $$f(x) = \int dx' f(x')\delta(x - x') $$ Let's back up just a bit and write the state (ket) as a weighted 'sum' of ...


1

Take the inverse Fourier transform of your last equation, $\psi(x,t)=\frac{1}{\sqrt{2\pi\hbar}}\int e^{i p x/\hbar}\phi(p,t) dp$, to see that the "coefficients" of $\psi(x,t)$ are not the same in the two representations: in the $x$ representation, the coefficients are $\psi$ and in the $p$ representation, the coefficients are $\phi$.


7

Let me rephrase those precise equations in the language of finite-dimensional linear algebra. You have a vector $A$ and two bases $\beta=\{e_i\}_i$ and $\beta'=\{e_i'\}_i$. This means you can write the components of $A$ with respect to $\beta$ as $$ A_i=e_i·A=\sum_j\delta_{ij}e_j·A $$ and the components with respect to $\beta'$ as $$ ...


1

The safest way to start with is the representation-free Schrodinger equation, $$ i\hbar\frac{\partial}{\partial t}|\Psi(t)⟩= \hat{H}|\Psi(t)⟩. $$ Referring to your case, we take the separable Hamiltonian: $H=\frac{p^2}{2m}+V$ so that $$ i\hbar\frac{\partial}{\partial t}|\Psi(t)⟩= \left(\frac{\hat{p}^2}{2m}+V\right)|\Psi(t)⟩. $$ Now is the time the ...


1

After revising the (very informative) responses to my previous queries, my revised understanding (which addresses my queries) is: If we measure the position then $\Psi(x,t)$ collapses to $\Psi(x,t) = \delta(x-z)$ (where $z$ is the measured position), using the Fourier transform we get: $$\Phi(p,t) = \frac{1}{\sqrt{2 \pi \hbar}} ...


0

The act of measuring cannot be represented by an operation in Linear Algebra, because postulate of QM says that upon measurement of observable $A$ with eigenvectors and eigenvalues given by $|a_i>$ and $a_i$ respectively on $|\psi>$ . $$|\psi>=\sum_i c_i |a_i>$$ Gives $|\psi_{after}>$. where $|\psi_{after}>$ can be any of $|a_i>$ with ...


2

I think you are getting something wrong in the 2nd and 3rd point, but i am going to try and give you an explanation. If i got something wrong in your questions, please do point it out. 1) If you measure the position, then Ψ(x)=δ(x-z) where z is the position that you measured(not a variable). If you plug this into the Fourier transform of Ψ(the first ...



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