How does one prove that the Dirac Delta distribution is the eigenfunction of the position operator $\hat{x}$? In math, why does $\langle x’|x\rangle = \delta(x’-x)$?
36068
I have an infinite basis which associates with each point, $x$, on the $x$-axis, a basis vector $|x\rangle$ such that the matrix of $|x\rangle$ is full of zeroes and a one by the $x^{\mathrm{th}}$ element. The book on Quantum Mechanics by Shankar says that the inner product between a basis vector and itself is not one, why not? Why can't these basis vectors be normalized to one, only to the ?
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?
41029
I've seen this identity in my electrodynamics book: $$\delta (f(x))=\sum_i{ \frac{1}{|{df\over dx}(x_i)|}\delta (x-x_i)}$$  Where $x_i$ shows the $i$th zero of $f(x)$. How can I prove it? I've tried the integral definition of delta function, but doesn't work.
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.
23262
Let $E$ be  a complex Hilbert space. Let $T\in \mathcal{L}(E)$.  I have two questions:     Why it is not true that for an arbitrary operator $T\in \mathcal{L}(E)$, we have $\langle Tu\;|\;u\rangle=0,\;\forall u\in E \Longrightarrow T=0$? And is  this property true for normal operators?   I think it is true for self adjoint operators because the norm of a self adjoint operators is given by  $$\left\|T\right\|= \sup\big\{\;\left|\langle Tu\;|\;u\rangle \right|,\;\;u \in E\;, \left\| u \right\| = 1\;\big\}$$  Thank you.
31063
In my quantum mechanics textbook it says that the relation between the basis $|x\rangle$ and $|p\rangle$ is given by:  $\langle p | x \rangle = \Large \frac{e^{-ip x/ \hbar}}{\sqrt{2\pi \hbar}} \, .$  However, i'm not sure how to go about proving this relation. My idea was that the eigenstates of momentum can be written as below:  $\phi(x,p) = \Large \frac{1}{\sqrt{2\pi \hbar}} e^{-ipx/\hbar},$  which form an orthogonal basis.  But, we can expand any state, $|\text{state}\rangle$, in terms of the othorgonal basis right? So is my notation below correct for the expansion of the state $| x \rangle$ ?  $|x\rangle = \int_{-\infty}^{+\infty} \phi(p', x) |p'\rangle dp'$  If it were correct then the following would also be correct?  Using the fact that $\langle p | p' \rangle = \delta(p - p')$, $\langle p | x \rangle$ can be computed:  \begin{align} \langle p | x \rangle =&amp; \int_{-\infty}^{+\infty} \phi(p', x) \langle p | p' \rangle dp' \\ =&amp; \int_{-\infty}^{+\infty} \phi(p', x) \delta(p - p') dp' \\ =&amp; \phi(p, x) \\ =&amp; \Large \frac{e^{-ip x/ \hbar}}{\sqrt{2\pi \hbar}} \end{align} as required.  Does the logic work here?
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.
17610
Some websites and textbooks refer to $$\Delta x \Delta p \geq \frac{\hbar}{2}$$ as the correct formula for the uncertainty principle whereas other sources use the formula $$\Delta x \Delta p \geq \hbar.$$ Question: Which one is correct and why?  The latter is used in the textbook "Physics II for Dummies" (German edition) for several examples and the author also derives that formula so I assume that this is not a typing error.  This is the mentioned derivation:     $\sin \theta = \frac{\lambda}{\Delta y}$      assuming $\theta$ is small:      $\tan \theta = \frac{\lambda}{\Delta y}$      de Broglie equation:      $\lambda = \frac{h}{p_x}$      $\Rightarrow \tan \theta \approx \frac{h}{p_x \cdot \Delta y}$      but also:      $\tan \theta = \frac{\Delta p_y}{p_x}$      equalize $\tan \theta$:      $\frac{h}{p_x \cdot \Delta y} \approx \frac{\Delta p_y}{p_x}$      $\Rightarrow \frac{h}{\Delta y} \approx \Delta p_y \Rightarrow \Delta p_y \Delta y \approx h$      $\Rightarrow \Delta p_y \Delta y \geq \frac{h}{2 \pi}$      $\Rightarrow \Delta p \Delta x \geq \frac{h}{2 \pi}$   So: Which one is correct and why?
95226
If a particle is totally localized at $x=0$, its wave function $\Psi(x,t)$ should be a Dirac delta function $\delta(x)$. Accordingly, its Fourier transform $\Phi(p,t)$ would be a constant for all $p$, thus the particle's momentum is totally uncertain. I guess this is what the Uncertainty Principle told us.  But in another hand, since the particle is totally localized at $x=0$, it is not moving and therefore static. Its velocity should be 0, so is its momentum, which contradicts the conclusion above.  So, what's wrong with my thinking?  Is it that there shouldn't be a TOTALLY STATIC particle, or it's in a non-normalizable state, so the uncertainty principle is not applicable?
103756
Statement: Let $M_x$ denote the multiplicative operator acting on $L^2([0,1], \, dx)$ by $M_x(f) = xf$. Show that $M_x$ has no eigenvalues  Attempt: Let $M_x(f) = xf$ then we should have $M_x(f) = \lambda f$  then we get $M_x(f) = \lambda f \Rightarrow xf = \lambda f \Rightarrow xf-\lambda f \Rightarrow (x-\lambda)f = 0$  Now $f \neq 0$ so that implies that $(x-\lambda) = 0$ but this is achieved when $x = \lambda$, so I don't see how it doesn't have eigenvalue, when we achieve the desired result.
74327
What is the square root of the Dirac Delta Function? Is it defined for functional integrals? Can it be used to describe quantum wave functions?  \begin{align} \int_{-\infty}^{\infty} f(x)\sqrt{\delta(x-a)}dx \end{align}
25952
How to prove that the radial basis function $k(x, y) = \exp(-\frac{||x-y||^2)}{2\sigma^2})$ is a kernel? As far as I understand, in order to prove this we have to prove either of the following:   For any set of vectors $x_1, x_2, ..., x_n$ matrix $K(x_1, x_2, ..., x_n)$ = $(k(x_i, x_j))_{n \times n}$ is positive semidefinite. A mapping $\Phi$ can be presented such as $k(x, y)$ = $\langle\Phi(x), \Phi(y)\rangle$.   Any help?
107693
Taken from "An Introduction to Quantum Physics" by Stefanas Trachanas (Question 2.4).  I need to prove that for an arbitrary statistical distribution:  $$ \Delta A ≤ \frac{|a_{max}-a_{min}|}{2} $$  Where $a_{min}$ &amp; $a_{max}$ are the minimum &amp; maximum possible values of the distribution,respectively.  I have proved this for 2 special cases: The case in which the mean is at the centre of the range, &amp; the case with only two possible outcomes. How do I prove the general case?
70896
Firstly, apologies in advance for using an abuse of notation by placing the Dirac-delta inside an integral. But for my level of understanding, I have no choice.  This question involves one of the Dirac-delta properties: $$\delta(bt)=\frac{1}{\mid b \mid}\delta(t)$$ where $b \gt 0$ and $b \in \mathbb{R^+}$; which was proven in this  asked by myself.  This is essentially a word by word copy of a  given on page 470:      Given $$\delta(bt)=\frac{1}{\mid b \mid}\delta(t)$$ and by    considering an integral of the form  $$\int  f(t)\delta(h(t))\mathrm{d}t$$ then making a change of variables to $z =  h(t)$, we may show that $$\displaystyle\delta(h(t)) =  \operatorname*{\Large\sum}\limits_{i}\cfrac{\delta(t − t_i)}{\mid  h^{\prime}(t_i)\mid}$$ where the $t_i$ are those values of $t$ for    which $h(t) = 0$ and $h^{\prime}(t)$ stands for   $\cfrac{\mathrm{d}h}{\mathrm{d}t}$.   Could someone please show the steps in this derivation as I have no idea how they reached $$\color{#180}{\displaystyle\delta(h(t)) = \operatorname*{\Large\sum}\limits_{i}\cfrac{\delta(t − t_i)}{\mid h^{\prime}(t_i)\mid}}$$   Many thanks.
122753
Okay, I know that in quantum mechanics the quantum observable is obtained from the classical observable by the prescription   $$ X \rightarrow x,\quad P \rightarrow -i\hbar\frac{\partial}{\partial x} $$   in the position basis. Now my questions are:    What if $x$ or $p$ appears in the denominator in a classical expression?  How to promote this to a quantum expression? What would be the meaning of division by an operator?   My expression likely contains a mixture of $x$ and $p$. For e.g., it could contain terms like $$\frac{p}{x^2}$$ or $$\frac{xp}{(x^2 + a^2)^{3/2}}.$$    How to resolve products of non-commuting operators like $x$, $p$ in a satisfactory way?
16985
We know that for a position variable $x$ and momentum $p$, the uncertainties of the two quantities are bounded by  $$\Delta x \Delta p \gtrsim \hbar$$  Now, this is usually first explained with $x$ being a simple linearly measured position and $p$ being linear momentum.  But it should apply to any good coordinate and its conjugate momentum.  It should, for instance, apply to angle $\phi$ about the $z$ axis, and angular momentum $L_z$:  $$\Delta \phi \Delta L_z \gtrsim \hbar$$  The thing is, $\Delta \phi$ can never be greater than $2\pi$.  I mean, you have to have some value of $\phi$ and $\phi$ only runs from 0 to $2\pi$.  Therefore  $$\Delta L_z \gtrsim \hbar/\Delta \phi \geq \hbar/2\pi$$  But, uh-oh!  This means it is impossible for $\Delta L_z$ to be zero, and we should never be able to have angular momentum states with definite $L_z$ values.  Of course, it doesn't mean that.  But I have never figured out how this is not in contradiction with the Schroedinger eqn. calculations that give us states with definite values of $L_z$.  Can anyone help me out?  One answer I anticipate is that $\phi$ is sort of "abstract" in that if you chose your origin at some other point you will get completely different values of $\phi$ and $L_z$, and ipso facto, usual considerations don't apply.  I don't think this will work, though.  Consider a "quantum bead" sliding around on a rigid circular ring and you get the exact same problem with no ambiguity in $\phi$ or $L_z$.  (Well, there will be some limited ambiguity in $\phi$, but still, there won't be in $L_z$.)
11306
I have been going through some problems in Sakurai's Modern QM and at one point have to calculate $\langle \alpha|\hat{p}|\alpha\rangle$ where all we know about the state $|\alpha\rangle$ is that $$\langle x|\alpha\rangle=f(x)$$ for some known function $f$. ($|\alpha\rangle$ is a Gaussian wave packet.) Sakurai says that this is given by:  $$\langle p\rangle = \int\limits_{-\infty}^{+\infty}\langle\alpha|x\rangle\left(-i\hbar\frac{\partial}{\partial x}\right)\langle x|\alpha\rangle dx.$$  I am wondering how we get to this expression. I know that we can express   $$|\alpha\rangle =\int dx|x\rangle\langle x|\alpha\rangle$$  and  $$\langle\alpha|=\int dx\langle\alpha|x\rangle\langle x|,$$   so my thinking is that we have:  $$\langle\alpha|\hat{p}|\alpha\rangle =\iint dx dx'\langle\alpha|x\rangle\langle x|\hat{p}|x'\rangle \langle x'|\alpha\rangle, $$  and if we can 'commute' $|x\rangle$ and $\hat{p}$ this would become: $$\iint dxdx'\langle\alpha |x\rangle\hat{p}\langle x|x'\rangle \langle x'|\alpha\rangle,$$  which is the desired result as $$\langle x|x'\rangle=\delta(x-x').$$ Is this approach valid?  I think my question boils down to: Does the operator $\hat{p}$ act on the basis kets $|x\rangle$ or on their coefficients? In the latter case, if we had some state $|\psi\rangle = |x_0\rangle$ for some position $x_0$, then would we say that for this state $$\langle p\rangle =\langle x_0|\left(-i\hbar\frac{\partial}{\partial x}\right)|x_0\rangle = 0$$ as the single coefficient is $1$ and the derivative of $1$ is $0$?