Can Gerchberg saxton algorithm reconstruct a wave function if we know the squared modulus of its wave function $|\psi(x)|^2$ and the squared modulus of its momentum wave function $|\phi(p)|^2$?
6468
Consider a "wavefunction" $\psi(x)$, which has a Fourier transform $\tilde \psi(p)$  Suppose that we know, for each $x$, $|\psi(x)|^2$, and that we know, for each $p$, $|\tilde \psi(p)|^2$.  Have we enough information to reconstruct the "wavefunction" $\psi(x)$, that is, obtain the phase of the "wavefunction" for all $x$ (up to a global phase, we are only interesting to the relative phases between the $x$)?
98645
Let us consider a (quantum) Harmonic oscillator: $$H=\frac{p^2}{2m}+\frac{1}{2} m \omega^2 x^2$$ Using the annihilation/creation operators defined as: $$a=\sqrt{\frac{\hbar}{2 m \omega}}(x+\frac{i}{m \omega}p)$$ $$a^{\dagger}=\sqrt{\frac{\hbar}{2 m \omega}}(x-\frac{i}{m \omega}p)$$ I end up with $H=\hbar \omega a^{\dagger} a$ Now, let's assume that for any reason I know that the dynamic should be restricted to the two first levels only. One way that I should we could do is to consider: $a \to \sigma_-, a^{\dagger} \to \sigma_+$ Then, we find: $$H=\hbar \omega \sigma_+ \sigma_-=\hbar \omega |1\rangle \langle 1|$$ But what confuses me is that if I do the replacement in $p$ and $x$, I don't find the same. Indeed we have: $$p=i\sqrt{\frac{\hbar m \omega}{2}}(a^{\dagger}-a)$$ $$x=\sqrt{\frac{\hbar}{2 m \omega}}(a^{\dagger}+a)$$ Performing the replacement there, I find: $$p=\sqrt{\frac{\hbar m \omega}{2}}\sigma_y$$ $$x=\sqrt{\frac{\hbar}{2 m \omega}}\sigma_x$$ And as $\sigma_y^2=\sigma_x^2=I$, I would end up with $H \propto I$ from the first Hamiltonian I wrote. Another annoying example. Imagine that my oscillator as in addition an interacting term of the form: $$H_{\text{int}}=p^2*F(t)$$ Where $F(t)$ is a driving field for instance. With the two level approximation and the mapping considered, as $p^2 \propto I$, the interaction would disappear. What is the procedure to cut to its two first level an Harmonic oscillator to get a qubit. I feel like I don't understand anymore. Typically to what the momentum and position should be mapped to ? How to solve the &quot;paradox&quot; I get to ?
43963
In the Schrödinger equation, at any given time $t$ we should jointly add another sub equation, like $$||\psi_t(x)|| = 1$$ where $\psi_t(x) = \Psi(x,t)$, and then try to solve the two equations simultaneously. Why not? I know it does not yield, but I am always baffled, who is doing the normalization of the wave function? Observers, the system, the measuring process, God?
76427
In the book "Quantum Field theory and the Standard Model" by Matthew Schwartz, , the position space wavefunction is defined as   $$\psi(x)=\langle 0|\phi(x)|\psi\rangle, \tag{2.82+2.83}$$  where $|\psi\rangle$ is any state in the Fock space. Then he uses the equations (i) $\partial_t^2\phi_0=(\nabla^2-m^2)\phi_0$ (i.e., the Klein-Gordon equation for the free massive scalar field $\phi_0(\textbf{x},t)$) and (ii) $[H,\phi_0]=-i\partial_t\phi_0$ to derive Eq. 2.85, in following 3 steps:  $$i\langle 0|\partial_t\phi_0(\textbf{x},t)|\psi\rangle=\langle 0|\int \frac{d^3\textbf{p}}{(2\pi)^3}\frac{\sqrt{\textbf{p}^2+m^2}}{\sqrt{2\omega_\textbf{p}}}(a_p e^{-ip\cdot x}-a^\dagger_p e^{ip\cdot x})|\psi\rangle\\ =\langle 0|\sqrt{m^2-\nabla^2}\phi_0(\textbf{x},t)|\psi\rangle.\tag{2.85}$$  This equation is used to successfully derive the Schrodinger equation in quantum mechanics for the state $\psi(\textbf{x},t)\equiv \langle 0|\phi_0(\textbf{x},t)|\psi\rangle$.   The first equality follows from differentiation of $\phi_0(x)$ w.r.t $t$. How does the second equality follow from the first? How are the inputs (i) and (ii) are utilized to derive the second equality from the first? Is it true that the operator $$i\partial_t\sim\sqrt{m^2-\nabla^2}~?$$ If yes, what is the use of doing the middle step?
34486
We know that the Dirac function $$\delta(a)=\lim_{a \rightarrow 0} \delta_{a}(x)$$ can be written as an infinitesimally narrow Gaussian: $$  \delta_{a}(x) := \frac{1}{\sqrt{2\pi a^2}}e^{-x^2/2a^2}$$  Our professor told us that for any value $a&gt;0$, the physical position eigenfunction is $$\psi_{x_0}(x)\cong N_1\delta_a(x-x_0).$$  How can I show that $\psi_{x_0}$ is a physical position eigenfunction?
71314
In all text book and lecture notes that I have found, they write down the general statement \begin{equation} \frac{\delta^n\Gamma[\phi_{\rm cl}]}{\delta\phi_{\rm cl}(x_1)\ldots\delta\phi_{\rm cl}(x_n)}~=~-i\langle \phi(x_1)\ldots\phi(x_n)\rangle_{\rm 1PI} \end{equation} and they show that it is true for a couple of orders.  I heard that Coleman had a simple self contained proof for this statement (not in a recursive way), but I cannot find it. It might have been along the line of comparing to the $\hbar$ expansion but I'm not sure.  Do you know this proof? Is there a good reference for it?  Comment: Weinberg does have a full proof but it is hard and not intuitive.
74985
For a compactly supported distribution $f$, I want to show the non-surjectivity of the map $H:f \mapsto f*f \in \mathcal{E}'$.     My attempt: I have chosen to show that this is not the case by showing that $\text{image } (H)$ does not contain $ \delta_0'$. For instance I restrict myself to $d=1$.   Suppose that $H$ is surjective. Applying the Fourier transform on the convolution, I get $(\hat{f}(z)) ^2 = i z$. The theorem of Paley-Wiener-Schwartz guarantees that $(\hat{f}(z)) $ should be an entire function. However, as it is well-known, no such entire function exists:   If this were true, then $e^{i \pi /4}\hat{f}(z)$ would be the square root function, which cannot be well defined on $\mathbb C$, and even less so entire.    Q1: Is there another way of achieving the result (i.e. with a choice significantly different than $\delta_0'$, and with an approach different than the one using the Fourier transform)?  Q2: Can I generalize this without much effort in the case of $d \ge 1$ (probably using the tensor product)?    Edit: The source is .
82142
We know that every CPTP map $\Phi:\mathcal X\to\mathcal Y$ can be represented via an isometry $U:\mathcal X\otimes\mathcal Z\to\mathcal Y\otimes\mathcal Z$, as $$\Phi(X) = \operatorname{Tr}_{\mathcal Z}[U(X\otimes E_{0,0})U^\dagger],\quad\text{where}\quad E_{a,b}\equiv \lvert a\rangle\!\langle b\rvert.\tag1$$ Showing this is quite easy e.g. from the Kraus representation. If $A_a:\mathcal X\to\mathcal Y$ are Kraus operators for $\Phi$, then $$U_{\alpha a,i0} \equiv \langle \alpha,a\rvert U\lvert i,0\rangle = \langle \alpha\rvert A_a\lvert i \rangle \equiv (A_a)_{\alpha,0}.\tag2$$ We can, of course, replace $E_{0,0}$ with any pure state in (1) without affecting the result. This shows that, given any channel $\Phi$ and any pure state $\lvert\psi\rangle\in\mathcal Z$, we can represent $\Phi$ as in (1) (with $E_{0,0}\to\lvert\psi\rangle$). However, what about the more general case of $E_{0,0}\to\sigma$ with $\sigma$ not pure? To analyse this case, consider a channel written as $$\Phi(X)=\operatorname{Tr}_{\mathcal Z}[U(X\otimes \sigma)U^\dagger]\tag3$$ for some state $\sigma=\sum_k p_k E_{k,k}\in\mathrm{Lin}(\mathcal Z)$ (appropriately choosing the computational basis for $\mathcal Z$). The relation with the Kraus operators reads in this case $$\Phi(X)_{\alpha,\beta} = \sum_{\ell k ij} p_\ell U_{\alpha \ell,i k} X_{ij} U^*_{\beta\ell,jk} = \sum_{\ell,k} (A_{\ell,k}XA_{\ell,k}^\dagger)_{\alpha\beta}\tag4$$ with $$(A_{\ell,k})_{\alpha,i} \equiv  \sqrt{p_\ell} U_{\alpha\ell,ik}, \qquad A_{\ell,k} = \sqrt{p_\ell} (I\otimes \langle \ell\rvert)U(I\otimes \lvert k\rangle).\tag5$$ There is now a difference: the number of Kraus operators must be larger than the rank of $\sigma$ (which determines the number elements spanned by the index $k$ in $A_{\ell,k}$). Indeed, a different way to state this same fact is to notice that the $\Phi$ in (3) is a convex combination of several channels: $$\Phi(X) = \sum_k p_k \Phi_k(X), \qquad \Phi_k(X)\equiv \operatorname{Tr}_{\mathcal Z}[U(X\otimes E_{k,k})U^\dagger]. \tag 6$$  This leads me to the question: can any $\Phi$ be written as in (3) for any $\sigma$? More precisely, given $\Phi$ and $\sigma$, can I always find an isometry $U$ such that (3) holds? The question arises from the fact that, because when $\sigma$ is not pure (3) leads to $\Phi$ be a convex combination of other maps as shown in (6), I would think that there should be maps that are &quot;extremal&quot;, in the sense that they cannot be written as convex combinations of other maps, and that such maps shouldn't be writable as (3) for $\sigma$ not pure.
20995
I am faced with the following question: given a probability density function $f$ over $\mathbb{R}$ with $\int_{\mathbb{R}}f(x)x^2dx=\sigma^2$ given, I am trying to find the "nearest" Gaussian to $\sqrt{f}$ in $L^2$, i.e. compute $$\inf_{a\in \mathbb{R}, b &gt; 0}\|\sqrt{f}-G_{a,b}\|_2^2$$ where $G_{a,b}(x)=a\exp(-bx^2)$ and the $\|\cdot\|_2$ is the classical $L^2$-norm over $\mathbb{R}.$  If one replaces the $L^2$-distance by the relative entropy $H(f|G_{a,b})$, it is easy to check that the infimum is a minimum given by the centered gaussian with variance $\sigma^2.$ For the $L^2$-distance, it seems more involved......
38398
I am a bit confused by the concepts of . In all the courses I am doing at the moment we do transformations of the form: $$ \phi(x) \rightarrow\phi'(x') = \phi(x) $$ and $$ \partial_{\mu}\phi(x)\rightarrow\partial' _{\nu}\phi(x') = \frac{\partial x^{\alpha}}{\partial x' ^{\nu}}\partial_{\alpha}\phi(x) $$ This is all perfectly clear to me. However, I am reading Peskin and Schoder at the moment, and they adapt an "active" point of view (their words), such that the above transformations are: $$ \phi(x) \rightarrow\phi'(x) = \phi(\Lambda^{-1}x) $$ and $$ \partial_{\mu}\phi(x)\rightarrow\partial _{\mu}(\phi(\Lambda^{-1}x)) = (\Lambda^{-1})^{\nu}_{\mu}(\partial_{\nu}\phi)(\Lambda^{-1}x). $$ I don't understand how to interpret this and especially how to derive the second equation.
111159
It is stated that if $\psi(x_0)=\psi'(x_0)=0$ at some $x_0$, then $\psi(x)=0,\forall x$.  Hence, we cannot have the wavefunction and its derivative vanishing at the same point.  My question is, how to prove the above statement?
114900
Suppose $\sigma_{1},\sigma_{2}$ and $\sigma_{3}$ are the Pauli matrices. Given a momentum ${\bf{p}}$, we define the helicity operator: $$ h = \frac{1}{2}\begin{pmatrix} {\bf{\sigma}}\cdot {\bf{\hat{p}}} &amp; 0 \\ 0 &amp; {\bf{\sigma}}\cdot {\bf{\hat{p}}} \end{pmatrix} $$ where $\sigma \cdot {\bf{\hat{p}}} = \sum_{\mu=0}^{3}\sigma^{\mu}\frac{p_{\mu}}{|{\bf{p}}|}$. Let $|\pm \rangle$ be the eigenvectors of $\sigma \cdot {\bf{\hat{p}}}$ with eigenvalues $\pm 1$. Consider the following plane waves: $$\psi_{+} = \frac{1}{\sqrt{2}}e^{i(-Et+{\bf{p}}\cdot {\bf{x}})}\binom{e^{-\frac{\theta}{2}}|+\rangle}{e^{\frac{\theta}{2}}|+\rangle} \quad \mbox{and} \quad \psi_{-} = \frac{1}{\sqrt{2}}e^{i(Et-{\bf{p}}\cdot{\bf{x}})}\binom{e^{\frac{\theta}{2}}|-\rangle}{e^{-\frac{\theta}{2}}|-\rangle}$$ Then $\psi_{+}$ is an eigenstate of $h$ with eigenvalue $1/2$ while $\psi_{-}$ is an eigenstate of $h$ with eigenvalue $-1/2$. Here, $\theta$ is a known parameter. Question: Shouldn't $\psi_{\pm}$ be solutions of the Dirac equation $(i\gamma^{\mu}\partial_{\mu}-m)\psi = 0$? They seem not to since, according to my calculations: $$i\gamma^{0}\partial_{0}\psi_{\pm} = \frac{1}{\sqrt{2}} E ^{i(-Et+{\bf{p}}\cdot{\bf{x}})}\binom{e^{\pm \frac{\theta}{2}}|\pm\rangle}{e^{\mp \frac{\theta}{2}}|\pm\rangle}$$ and, analogously: $$i \gamma^{\mu}\partial_{\mu}\psi_{\pm} = \frac{1}{\sqrt{2}}|{\bf{p}}|e^{i(-Et+{\bf{p}}\cdot{\bf{x}})}\binom{e^{\pm \frac{\theta}{2}}|\pm\rangle}{-e^{\mp \frac{\theta}{2}}|\pm\rangle}$$ while the last term of the equation reads: $$m \psi_{\pm} = m \binom{e^{\mp \frac{\theta}{2}}|\pm\rangle}{e^{\pm \frac{\theta}{2}}|\pm\rangle}$$ and the sum is clearly nonzero. To do such calculations I used the Weyl representation of the Dirac matrices: $$\gamma^{0} = \begin{pmatrix} 0 &amp; I \\ I &amp; 0 \end{pmatrix} \quad \mbox{and}\quad \gamma^{\mu} = \begin{pmatrix} 0 &amp; \sigma^{\mu} \\ -\sigma^{\mu} &amp; 0 \end{pmatrix}$$ Am I doing something wrong? Shouldn't the plane wave solutions above given in terms of eigenstates of $h$ be solutions to the Dirac equation? Moreover, should it depend on the particular representation of the gamma matrices?
27861
Hm, this just occurred to me while answering another question:   If I write the Hamiltonian for a harmonic oscillator as $$H = \frac{p^2}{2m} + \frac{1}{2} m \omega^2 x^2$$ then wouldn't one set of possible basis states be the set of $\delta$-functions  $\psi_x = \delta(x)$, and that indicates that the size of my Hilbert space is that of $\mathbb{R}$.  On the other hand, we all know that we can diagonalize $H$ by going to the occupation number states, so the Hilbert space would be $|n\rangle, n \in \mathbb{N}_0$, so now the size of my Hilbert space is that of $\mathbb{N}$ instead.  Clearly they can't both be right, so where is the flaw in my logic?
78641
In deriving the quantization condition for a bound state in a potential with "no verticle walls" we start with the WKB connection formulas to find the wavefunction in the interior of the well ($x_1&lt;x&lt;x_2$), namely $$\psi(x) \cong \begin{cases}\displaystyle\frac{2C_1}{\sqrt{p(x)}}\sin\left[\frac{1}{\hbar}\int_{x_1}^x{dx\ p(x)}+\frac{\pi}{4}\right]&amp;\quad\text{for}\quad x_1&lt;x\\ \\\displaystyle \frac{2C_2}{\sqrt{p(x)}}\sin\left[\frac{1}{\hbar}\int_x^{x_2}{dx\ p(x)}+\frac{\pi}{4}\right]&amp;\quad\text{for}\quad x&lt;x_2 \end{cases}$$  Our task is to connect the two wave functions in the interior region together. Seeing that the coefficients are easily handled, it is then necessary that the arguments of the sine functions be equivalent. Denoting  $$\theta_1(x) = -\frac{1}{\hbar}\int_{x_1}^x{dx\ p(x)}-\frac{\pi}{4}$$  $$\theta_2(x) = \frac{1}{\hbar}\int_{x}^{x_2}{dx\ p(x)}+\frac{\pi}{4}$$ we have  $$\theta_1(x)+n\pi = \theta_2(x)\quad\text{for}\quad n = 0,1,2,3..$$  since  the relative minus can be absorbed into either coefficient. Thus $$-\frac{1}{\hbar}\int_{x_1}^x{dx\ p(x)}-\frac{\pi}{4}+n\pi= \frac{1}{\hbar}\int_{x}^{x_2}{dx\ p(x)}+\frac{\pi}{4}$$ \begin{equation}\int_{x_1}^{x_2}{dx\ p(x)} = \left(n-\frac{1}{2}\right)\pi\hbar\quad\text{for}\quad n=0,1,2,3..\end{equation} This is our quantization condition for the allowed energies.   However, following the derivation in Griffiths or Sakurai, $n=0$ is not included. Why?   My ideas: Well this would imply  $$\theta_1(x) = \theta_2(x)$$ \begin{equation}\int_{x_1}^{x_2}{dx\ p(x)} = -\frac{\pi}{2}\hbar\end{equation} \begin{equation}\int_{x_1}^{x_2}{dx\ \sqrt{2m(E-V(x))}} = -\frac{\pi}{2}\hbar\end{equation} I am trying to imagine why a negative value of the action integral might be prohibited. Am I correct in assuming that if we were to restrict all possible values of the energy to be real, it would that imply $$p(x)&gt;0\ $$ Can someone provide a proof as to why  $$p(x)&gt;0\rightarrow E\in\mathbb{R}. $$It would then follow the the integral of a strictly positive integrand must itself be positive, which means we cannot have $-\frac{\pi}{2}\hbar$.
4632
The total wavefunction of an electron $\psi(\vec{r},s)$ can always be written as $$\psi(\vec{r},s)=\phi(\vec{r})\zeta_{s,m_s}$$ where $\phi(\vec{r})$ is the space part and $\zeta_{s,m_s}$ is the spin part of the total wavefunction $\psi(\vec{r},s)$. In my notation, $s=1/2, m_s=\pm 1/2$.   Question 1 Is the above statement true? I am asking about any wavefunction here. Not only about energy eigenfunctions.   Now imagine a system of two electrons. Even without any knowledge about the Hamiltonian of the system, the overall wavefunction $\psi(\vec{r}_1,\vec{r}_2;s_1,s_2)$ is antisymmetric. I think (I have this impression) under this general conditions, it is not possible to decompose $\psi(\vec{r}_1,\vec{r}_2;s_1,s_2)$ into a product of a space part and spin part. However, if the Hamiltonian is spin-independent, only then can we do such a decomposition into space part and spin part.   Question 2  Can someone properly argue that how this is so? Please mention about any wavefunction of the system and about energy eigenfunctions.
104773
The probability amplitude for a free particle with momentum $\ p$ and energy $E$ is the complex wave function: $$\psi_{(\vec x , t)}=e^{i(\vec k\cdot \vec x -\omega t)}$$  is there any uncertainty on the free particle with a definite momentum $\vec p$!?
91861
For one dimensional quantum mechanics $$[\hat{x},\hat{p}]=i\hbar. $$  Does this fix univocally the form of the $\hat{p}$ operator? My bet is no because $\hat{p}$ actually depends if we are on coordinate or momentum representation, but I don't know if that statement constitutes a proof. Moreover if we choose $\hat{x}\psi=x\psi$ is the answer of the following question yes?  For the second one   $$(\hat{x}\hat{p}-\hat{p}\hat{x})\psi=x\hat{p}\psi-\hat{p}x\psi=i\hbar\psi, $$  but I don't see how can I say that $\hat{p}$ must be proportional to $\frac{\partial}{\partial x}$. I don't know if trying to see that $\hat{p}$ must satisfy the Leibniz rule and thus it should be proportional to the $x$ derivative could help. Or using the fact that $\hat{x}$ and $\hat{p}$ must be hermitian  Any hint will be appreciated.