Consider the following:  $$[ \hat H, \hat x]=\left[-\frac{\hbar^2 \hat p^2}{2m}+V,\hat x\right]\ne0 \text{ in general}$$ But  $$[ \hat H, \hat x]=\left[i\hbar \frac{\partial }{\partial t},\hat x\right]=0 $$ Why are these not in contradiction?
51101
In non-relativistic QM, on one hand we have the following relations:  $$\langle x | P | \psi \rangle ~=~ -i \hbar \frac{\partial}{\partial x} \psi(x),$$  $$\langle p | X | \psi \rangle ~=~ i \hbar \frac{\partial}{\partial p} \psi(p).$$  On the other hand, despite the similarities, the relations cannot be directly applied to energy and time:  $$\langle t | H | \psi \rangle ~=~ i \hbar \frac{\partial}{\partial t} \psi(t),$$  $$\langle E | T | \psi \rangle ~=~ -i \hbar \frac{\partial}{\partial E} \psi(E).$$  Just wondering, how can one mathematically prove that the "classical time" (which means no QFT or relativity involved), unlike its close relative "position", is not a Hermitian operator?  I ask your pardon if you feel the question clumsy or scattered. But to be honest, if I can clearly point out where the core issue of the problem is, I may have already answered it by myself :/
22053
The formula we need to prove is $$ \displaystyle \iint_{x^{2}+y^{2} \leq 1} e^{x^{2}+y^{2}} d x d y \leq\left[\int_{-\frac{\sqrt{\pi}}{2}}^{\frac{\sqrt{\pi}}{2}} e^{x^{2}} d x\right]^{2}  $$ As I already done the left side $$ \iint_{x^{2}+y^{2} \leq 1} e^{x^{2}+y^{2}} d x d y  = \int_{-\pi}^{\pi} \mathrm{d}\varphi \int_{0}^{1} e^{r^2}r\mathrm{d}r  = 2\pi \int_{0}^{1} e^{r^2}r\mathrm{d}r = \left.\pi e^{r^2}\right|_{0}^{1} = \pi(e-1) $$ I still have no idea about the right side.
63304
Consider the time-dependent  (or some equation in Schrödinger form) written down as $$ \tag 1 i \partial_{0} \Psi ~=~ \hat{ H}~ \Psi . $$ Usually, one likes to write that it has a formal solution of the form $$ \tag 2 \Psi (t) ~=~ \exp\left[-i \int \limits_{0}^{t} \hat{ H}(t^{\prime}) ~\mathrm dt^{\prime}\right]\Psi (0). $$ However, this form for the solution of $(1)$ is actually built by the method of successive approximations which actually returns a solution of the form $$ \tag 3 \Psi (t) ~=~ \hat{\mathrm T} \exp\left[-i \int \limits_{0}^{t} \hat{H}(t^{\prime})~\mathrm  dt^{\prime}\right]\Psi (0), \qquad t&gt;0, $$ where $\hat{\mathrm T}$ is the time-ordering operator.  It seems that $(3)$ doesn't coincide with $(2)$, but formally $(2)$ seems to be perfectly fine: it satisfies $(1)$ and the initial conditions. So where is the mistake?
27764
In lectures we're told:  $$\dfrac {\partial y} {\partial x} = \dfrac 1 {\dfrac {\partial x} {\partial y}}$$  as long as the same variables are being held constant in each partial derivative.  The course is 'applied maths', i.e non-rigorous, so don't confuse me.  But anyway, if we have:  $$\xi = x - y \qquad \eta = x$$  Then $\dfrac {\partial x} {\partial \xi} = 0$ and $\dfrac {\partial \xi} {\partial x} = 1$.  The rule presumably fails because one of the partial derivatives is $0$.  But then isn't this rule useless? Since I cannot use it without checking that the partial derivative isn't in fact $0$, but then I've worked out the partial derivative manually anyway.
33369
The answer given is that it's not because it's a parabola and hence, would fail the horizontal line test, i.e., two values of $x$ will have the same value of $f(x)$.  However, how can we prove the same algebraically? My solution:  Let $f(x_1) = f(x_2)$  $\implies x_1^2 - 2x_1 = x_2^2 - 2x_2$  $\implies (x_1 - 1)^2 -1 = (x_2 - 1)^2 - 1$  $\implies (x_1 - 1)^2 = (x_2 - 1)^2$  $\implies x_1 = x_2$ (I believe that I've made the mistake here because we can't really take square root of both the sides.)  My question is: how do we prove that the function is not one-to-one algebraically?
60815
Assuming the eigenvalue of position operator $\hat x$ equal to $k$, can I not write:  $$\begin{align} \langle\psi_n|x|\psi_m\rangle &amp;= \langle x\psi_n|\psi_m\rangle \\ &amp;=\langle k\psi_n|\psi_m\rangle \\ &amp;=k\langle\psi_n|\psi_m\rangle \\ &amp;=k\delta_{nm}  \end{align}$$  But I know that $\langle x \rangle =0$ in case of even potentials (I don't know how that happens) and what I have written above is wrong, at least in case of even potentials.  Taking the example of infinite 1D square well, the states are : $$ \psi_{n} \left(x\right)=A\sin\left(\frac{nx\pi}{L}\right)dx $$ then, $$ \langle\psi_n|x|\psi_m\rangle =A\int_{-\infty}^{\infty}\sin\left(\frac{mx\pi}{L}\right)x\sin\left(\frac{nx\pi}{L}\right)dx $$ If m=n=1, $$ &lt;x&gt;=A\int_{-\infty}^{\infty}x\sin^{2}\left(\frac{x\pi}{L}\right)dx $$ if we apply an even potential then the equation gets reduced to$$ &lt;x&gt;=\frac{1}{L}\int_{-L}^{L}x\sin^{2}\left(\frac{x\pi}{L}\right)dx=0 $$ while in case of a potential(neither even nor odd), the equation leads to $$&lt;x&gt;=\frac{2}{L}\int_{0}^{L}x\sin^{2}\left(\frac{x\pi}{L}\right)dx=L/2 ~? $$  Here $n=m=1$ but $ &lt;x&gt;=0 $ for even potentials, which is confusing me! It should be $k$ right?
3441
This is what I have so far, but I don't feel like its a full explanation   The even function gets rid of the negation   Since $ h(x) $ is even then $ h(x)=h(-x) for \ all \ x \ $  $ f(x) = g(h(x)) $  $ f(-x) = g(h(-x)) = g(h(x)) \ or \  g(-(-h(x))) = g(h(x)) $  therefore $  f(x) $ is even  Thanks in advance!!
58203
I stumbled on this question while answering the following question:  Given a general wavefunction $\psi(x)$ and an operator $\hat Q$, with eigenvalue spectrum $\{q_r\}$ and corresponding eigenfunctions $\{u_r\left(x\right)\}$, show that the expectation value of $\hat Q$ is given by $$\langle\hat Q\rangle = \int_{-\infty}^\infty{\psi^*\hat Q \psi \,dx}$$  I started of with the ususal definition of the expected value and inserted the identity operator in position representation: $$\langle\hat Q\rangle = \langle \psi|\hat Q|\psi\rangle = \langle \psi|\left(\int dx|x\rangle\langle x|\right) \hat Q\left(\int  {dx}'|x'\rangle\langle x'|\right)|\psi\rangle$$  Taking $\langle\psi|$ and $|\psi\rangle$ into the integral gives: $$\left(\int dx\langle \psi|x\rangle\langle x|\right) \hat Q\left(\int  {dx}'|x'\rangle\langle x'|\psi\rangle\right)$$  Pulling the 2nd integral into the first:$$\int dx\int  {dx}'\langle \psi|x\rangle\langle x'|\psi\rangle\langle x| \hat Q|x'\rangle$$  This is the point where I'm stuck. If $\langle x|\hat Q|x'\rangle = \hat Q \delta(x-x')$, then the last expression would evaluate to the answer. However I  don't know why this should be true and even if it is true I wouldn't know how to prove this equality.
39724
Suppose that $\hat{x}$ is the least squares solution of $Ax = b$.  That is, $\min{|r|}, r = Ax - b$ with respect to $x$.  Is $\hat{x}$ also the least squares solution of $JAx = Jb$?  I suspect so, but I'm not sure how to prove it.  If $J$ is invertible, it's easy enough to prove as you can just cancel it out from both sides.  But what if it's not invertible?
59035
$$L~=~ \frac{1}{2} \dot{q}  \sin^2{q} $$  Is it zero or not defined?
101749
If $f \in \operatorname{Hol}(D),f(\frac{1}{2}) + f(-\frac{1}{2}) = 0$, prove that $|f(0)| \leq \frac{1}{4}$  $D = \{ z \in \mathbb{C} : |z| &lt; 1 \} $  My thoughts so far: Let's say $f(0) = a$. Define $g = \frac{z-a}{1-\bar{a}z}$ and $h(z) = (g(f(z))$ Now all the conditions for the Shwarz lemma are met, and I can conclude that $|h(\frac{1}{2})| \leq  \frac{1}{2}$ and $|h(-\frac{1}{2})| \leq  \frac{1}{2}$. The idea would then be to multiply the two inequalities together and try to somehow separate $a$, but the algebra gets really messy and I feel like I'm doing something wrong. Any help would be appreciated!
1802
I`m trying to prove this claim and I need some advice how to continue, $$(A \setminus B) \cup (A \setminus C) = B \Leftrightarrow A=B , C\cap B=\varnothing$$ what I did is: $$(A \setminus B) \cup (A \setminus C) = (A \cap B') \cup (A \cap C') = A \cup (B' \cap C')$$ thanks!
101244
Testing with my calculator in degree mode, I have found the following to be true:  $$\tan \left(90 - \frac{1}{10^n}\right) \approx \frac{180}{\pi} \times 10^n, n \in \mathbb{N}$$  Why is this? What is the proof or explanation?
87089
I have tried and tried but cannot for the life of me see how one equation follows onto the other... can anybody help??   $$\Omega(\theta)=-b \coth(\operatorname{arsinh}(e^{a\theta}   \sinh c_0 ))$$  $$\implies \Omega(\theta)=\sqrt{c_1 e^{-2a\theta} + b^2}$$  note that: $c_0=-\operatorname{arcoth}\left(\frac{\Omega_0}{b}\right)$  Eichberger quotes (at the bottom of page 5) "By inserting (11) into (8), using identities of the hyperbolic transcendental functions and carefully observing $±$ signs, we obtain $Ω$ as a function of $θ$."
58379
I just got some question about the hat matrix in linear models.  My first question is: Why in a balanced one-way layout $(n_1=...=n_c=n_0)$, all leverages $h_{ii}$ have the same value $\frac{1}{n_0}$? I know that $h_{ii}$ is the $(i,i)$ entry in the hat matrix $H=X(X^TX)^{-1}X^T$.  But I couldn't see any relationship between this expression and the result.   My second question is:  when discussing the leverage points in a general linear model, we know that  $$0\leq h_{ii}\leq1$$ $$\sum_{i=1}^n h_{ii}=p$$ where $p$ is the number of parameters to be estimated, and $$Var(\hat {Y_i})=\sigma^2h_{ii}=\frac{\sigma^2}{\frac{1}{h_{ii}}}$$ Then it says $\frac{1}{h_{ii}}$ is roughly the number of observations needed to estimate $\hat{Y_i}$. Why is that?   My third question is:   Continuing with my second question, it then says, if $h_{ii}$ is very close to 1, then variance of the $i$th residual very lose to zero (already know that $Var(E_i)=\sigma^2(1-h_{ii})$)  so $Y_i-\sum_{j=1}^p\hat{\beta_j}x_{ij}\simeq0$.  Isn't this expression for the $i$th residual? Why we can conclude this from 0 variance of residual? I mean, can't $E_i$ be constant so that the variance is also zero?   My last question is:  Continuing with the above, it then conclude that $\hat{Y_i}\simeq Y_i$ and hence almost one degree of freedom needs to be used to just fit this one observation.  Could anyone explain this result to me? Isn't $\hat{Y_i}=Y_i$ exactly what we want?  Since I think it means we estimate the $i$th observation perfectly.
73833
I'm trying to show the product of two Legendre symbols. When I compile the following  \documentclass{article} \begin{document} $$\left(\frac{x}{p}\right) \left(\frac{x}{p}\right)$$ \end{document}   it yields     as expected. However, when I switch the variables, i.e.  \documentclass{article} \begin{document} $$\left(\frac{x}{p}\right)\left(\frac{p}{x}\right)$$ \end{document}   it yields     How do I resolve this? Why do the values within the fractions matter?
80900
I'm working through an econometrics textbook and it's proving that   $$ \sigma^2 = E(\hat{\sigma}^2) = \frac{SSR}{n-2} $$  I followed the proof (an example of which is shown on ) until it reaches this step:  \begin{align} E\left[\displaystyle\sum\limits_{i=1}^n \hat{u}_{i}^{2}\right] &amp;= E\left[\displaystyle\sum\limits_{i=1}^n (u_{i}-\bar{u})^{2}\right] + (\hat{\beta}_{1}-\beta_{1})^{2} E\left[\displaystyle\sum\limits_{i=1}^n (x_{i} - \bar{x})^2\right] \\ &amp;+ 2 (\hat{\beta}_{1}-\beta _{1}) E\left[\displaystyle\sum\limits_{i=1}^n (u_{i}-\bar{u})(x_{i}-\bar{x})\right] \\ &amp;= \dots \end{align}  The book says that:   The expected value of the first term should be $(n-1)\sigma^2$ The expected value of the second term should be $\sigma^2$. The expected value of the third term should be $2 \sigma^2$ because the term itself can be written as $2(\hat{\beta}_1-\beta_1)^2 \hat{\sigma}_x^2$.   For the first term, I can work backwards and get that:  \begin{align} \hat{\sigma}^2 &amp;= \frac{1}{n-1} \displaystyle\sum\limits_{i=1}^n (u_{i}-\bar{u})^{2} \\ &amp;= \sigma^2 \end{align}  but it seems like this only works if we assume that $\hat{\sigma}^2$ is an unbiased estimator.  For the second term:  \begin{align} (\hat{\beta}_{1}-\beta_{1})^{2} E\left[\displaystyle\sum\limits_{i=1}^n (x_{i} - \bar{x})^2\right] &amp;= (\hat{\beta}_{1}-\beta_{1})^{2} E\left[(n-1) \hat{\sigma}_x^2\right] \\  &amp;= (\hat{\beta}_{1}-\beta_{1})^{2} (n - 1) E\left[\hat{\sigma}_x^2\right] \\ &amp;= (\hat{\beta}_{1}-\beta_{1})^{2} (n - 1) \hat{\sigma}_x^2 \end{align}  but this doesn't match what the text shows, so I'm a bit confused as to the final step for this term.   For the third term: \begin{align} 2 (\hat{\beta}_{1}-\beta _{1}) E\left[\displaystyle\sum\limits_{i=1}^n (u_{i}-\bar{u})(x_{i}-\bar{x})\right] &amp;= 2 (\hat{\beta}_{1}-\beta _{1}) E\left[\displaystyle\sum\limits_{i=1}^n u_{i}(x_{i}-\bar{x})\right] \\ &amp;= 2 (\hat{\beta}_{1}-\beta_{1}) \displaystyle\sum\limits_{i=1}^n E\left[u_{i}(x_{i}-\bar{x})\right] \\ &amp;= 2 (\hat{\beta}_{1}-\beta_{1}) \displaystyle\sum\limits_{i=1}^n E\left[u_{i}\right]E\left[(x_{i}-\bar{x})\right] \\ \end{align}  I would have thought that because $E[u_i] = 0$, this term would cancel out, so I'm not sure how to derive $2\sigma^2$ from it.  Can someone give me a hint in the right direction for these steps? I'm firmly trying to work through all of the proofs myself with only a little guidance from the book, but I'm not quite there yet.  I noticed , which links to , so maybe with more reading I'll see where to apply that. (If that isn't it, hints are welcome, but I'll still work through those questions to see if it helps)