Equation:  \documentclass[12pt]{article} \usepackage{mathtools} \begin{document}     \begin{eqnarray}     \langle\pi_n|\hat{H}^{ab}|\pi_m\rangle &amp; =&amp; \Delta_{nm} g\int d^3x|\Psi_p(x-r_m)|^2\left\{|\phi_0(x)|^2+     \sum_{q=1}^{N}\left[\hat{b}_q\left(a\phi_0^*(x)-a\phi_0(x)\right)\right.\right.\\     &amp; &amp; +\left.\left. \phi*a*a*a-\phi^*_0(x)v^*_q(x)\right]\right\}     \end{eqnarray} \end{document}   The third bracket at the end of the equation is of different size than the earlier bracket used in the first line.
5909
I want to align equations (using the align environment from the amsmath package) inside matching \left( and \right) commands. For example:  \documentclass[a4paper,12pt]{article}  \usepackage{amsmath}  \begin{document}  \begin{align}     a \left( b &amp;= c \right)\\     d \left( e &amp;= f \right) \end{align}  \end{document}   But that code gives this error:     Extra }, or forgotten \right.   The error disappears when I remove the delimiters. Is there a workaround for this?
67392
I'm trying to solve this expression when $b \ne 0$:   $$ \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \frac{e^{-{x^2}/{2}}}{1+e^{-(ax+b)}}\,\mathrm dx $$   Using the answers provided to  we can solve $$\frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \frac{e^{-{x^2}/{2}}}{1+e^{-x}}\,\mathrm dx=\frac12$$  However, when we change the power of exponential term in the denominator to $-(ax+b)$ from $x$, the expression can not be solved using the same methods used in the . I tried solving by substituting $z=ax+b$ and it also did not work.
85339
I want to write a three line equation with left curly brace for all three, right curly brace for bottom two ones and some text at right bottom of right curly brace. All with proper alignment. I wrote following code.  \documentclass{article} \usepackage{amsmath,amssymb,amsfonts} \usepackage{flexisym}  \begin{document}  \begin{equation*}     \begin{aligned}         K_{\mathbb{A}_{Ui}} =\begin{cases}             D                =\ g^{ \frac{ (\alpha + r) }{\beta} }\\             \left.\begin {aligned}                 &amp;D_j           &amp;=\ &amp;g^r . H(j)^{r_j} \\                 &amp;D_j\textprime &amp;=\ &amp;g^{r_j}             \end{aligned}\right\} \forall j \in   \mathbb{A}_{Ui}         \end{cases}         \end{aligned}     \end{equation*}  \end{document}   It is giving some problems.   Second and Third lines are aligned but they are not aligned with the first one. I want alignment to occur at start of "D", "=" and "g" for all three. Right now, the text at right of right curly brace is coming at center. I want it to shift down and come at bottom.   Please help. Thanks in advance.
105972
Solve the following equations for $x, (x ∈ N^+)$  $$ (a)\ 2φ(x) = x;$$ $$(b)\ 3φ(x) = x;$$ $$(c)\ 4φ(x) = x.$$  Can anyone help me with this, or give me some hint?
13761
Equation exceeds one column of IEEE pattern in LaTeX how should I correct it? Please help, the code is attached herewith.  \begin{equation}\label{eq11}         p\left( {v,h|u,y,...} \right) = {\rm{RelU}}\left( {\left[ {\hat b + {w^h}\left\{ {\left( {{v^T}{w^v}} \right) \circ \left( {{y^T}{w^y}} \right)} \right\}} \\ \right]*\left[ {\hat a + {w^v}\left\{ {\left( {{h^T}{w^h}} \right) \circ \left( {{y^T}{w^y}} \right)} \right\}} \right]} \right) \end{equation}
102686
I tried this and that, but could not solve this problem. Here is the code that produces this error:  \begin{equation} o=g(h)=\left\{\begin{array}{rcl}     1 &amp; if &amp; h &lt;   0 \\     0 &amp; if &amp; h \le 0  \end{array}\right \end{equation}   The error is given on the line \end{equation} resides. Thanks.
96541
Using Minkowski signature $(+,-,-,-)$, for the Lagrangian density   $${\cal L}=\partial_{\mu}\phi\partial^{\mu}\phi^{\dagger}-m^2\phi \phi^{\dagger}$$  of the complex scalar field, we have the field  $$\phi(x)=\int{\frac{d^3 \vec{p}}{2(2\pi)^3\omega_\vec{p}}}(a(\vec{p})e^{-ipx}+b^{\dagger}(\vec{p})e^{+ipx}).$$  I'm trying to now find an equation for the $a(\vec{p})$ and $b(\vec{p})$ (with the final goal of finding an expression for $[a(\vec{p}),b^{\dagger}(\vec{q})]$ using $[\phi(x),\Pi(y)]$).   What I should note is that we are considering all of this in the Schrodinger picture (t=0) so I suppose the very first thing to do is change all the $x$'s to $\vec{x}$ right?  The strategy I'm struggling to implement and failing at many points along the way:   Find the momentum $\Pi^{\phi}(x)=\frac{\partial L}{\partial \dot{\phi}}=\dot{\phi^{\dagger}}$. Add some combination of $\phi(x)$ and $\Pi(x)$ to get rid of one of the creation/annihilation operators.  Do an inverse Fourier transform to find $a(\vec{p})$ in terms of $\phi(x)$, for example.    None of the major text books seem to actually carry this through, and instead write something like "and it's easy to show…". However I don't find it too easy, especially part 3. as I'm not a Fourier transform expert.   Could anyone either direct me somewhere where the above is computed explicitly (in more than 2/3 lines), or help me understand each of the 3 steps above?  (I realise it is easy to find a reference where this is done for the real scalar field, in which case we have $a(\vec{p})$ and $a^{\dagger}(\vec{p})$. Even still, I find it hard to follow parts.)
33375
I am a bit lost how one has deduced the formula for electric field with electric dipole because of some inconsistency between different sources. The  contains a delta function in the formula, but  gives only the formula without the delta function,  $$\bar{E} = -\nabla \phi = \frac{-p}{4\pi \epsilon_{0} r^{3}} ( \hat{e}_{z}-3\hat{u}_{r}(\hat{u}_{r} \cdot \hat{u}_{z}) ) $$  Is it intentionally left out?  My notes mention that the dipole is with a circular thing $P=Q d$ where $d$ is distance and $Q$ is charge.
38176
I'm having the same problem as it was questioned   I can't get throught the step where I need to show that $\nabla_{E_i}E_j (p)=0$. It only leads to $$ \nabla_{E_i}E_j(p)=\sum_{lk}^n a_{il}(0)\dfrac{\partial b_jk}{\partial x_l}(0)\partial x_k(p) $$ which I can't show that equals zero.  On the linked question, there is also the indication that $\nabla_{E_i}E_j (E_k)=\Gamma^k_{ij}$. This could lead that $\nabla_{E_i}E_j (E_k)(p)=\Gamma^k_{ij}(p)=0$, but $\nabla_{E_i}E_j (E_k)(p)$ is not equal to $\nabla_{E_i}E_j (p)$ which is the one I need.  Can someone help me?
59116
I know I can transform the above product as follows  \begin{eqnarray*}\lim_{n\rightarrow\infty}\left(\frac{1}{2}\frac{3}{4}\frac{5}{6}...\frac{2n-1}{2n}\right)&amp;=&amp;\lim_{n\rightarrow\infty}\prod_{k=1}^n\left(\frac{2k-1}{2k}\right)\\&amp;=&amp;\lim_{n\rightarrow\infty}\prod_{k=1}^n\left(1-\frac{1}{2k}\right)\\&amp;=&amp;\lim_{n\rightarrow\infty}\prod_{k=1}^n\left(1+\frac{-1/2}{k}\right) \end{eqnarray*}  My thoughts are this is going to go to $e^{-x/2}$, but I can't figure out how to go from   $$\lim_{n\rightarrow\infty}\prod_{k=1}^n\left(1+\frac{-1/2}{k}\right)$$  to  $$\lim_{n\rightarrow\infty}\left(1+\frac{-1/2}{n}\right)^n$$
43327
I'm trying to see why this is the case. I have taken  the LHS and added $X\hat{\beta}$ and subtracted $X\hat{\beta}$.  From this I can get both terms on the right but I end up with a cross product term  $2(y-X\hat{\beta})^T(X\hat{\beta} - X\beta)$  that I can't prove equals to zero. I think there is some trick I'm missing, anybody remember it?
65848
$(1+px+qx^2)^8 \equiv  1+8x+52x^2+kx^3...$  Given that expression, calculate the values of p, q, and k.  I'm guessing this is a trinomial that needs to be turned into a binomial, but how would you do it? I have no idea.
28673
Let $\hat{x} = x$ and $\hat{p} = -i \hbar \frac {\partial} {\partial x}$ be the position and momentum operators, respectively, and $|\psi_p\rangle$ be the eigenfunction of $\hat{p}$ and therefore $$\hat{p} |\psi_p\rangle = p |\psi_p\rangle,$$ where $p$ is the eigenvalue of $\hat{p}$. Then, we have $$ [\hat{x},\hat{p}] = \hat{x} \hat{p} - \hat{p} \hat{x} = i \hbar.$$ From the above equation, denoting by $\langle\cdot\rangle$ an expectation value, we get, on the one hand $$\langle i\hbar\rangle = \langle\psi_p| i \hbar  | \psi_p\rangle = i \hbar \langle \psi_p | \psi_p \rangle  = i \hbar$$ and, on the other $$\langle [\hat{x},\hat{p}] \rangle = \langle\psi_p| (\hat{x}\hat{p} - \hat{p}\hat{x}) |\psi_p\rangle =  \langle\psi_p|\hat{x} |\psi_p\rangle p - p\langle\psi_p|\hat{x} |\psi_p\rangle = 0$$ This suggests that $i \hbar = 0$. What went wrong?
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?
85857
I tried the following $$I = \langle X^2,X+1\rangle =\langle X^2,X+1,X^2+2(X+1)\rangle =\langle X^2,X+1,(X+1)^2+1 \rangle$$  Yet no matter how I arrange it, I cannot obtain $1$. Can someone help me out?
111451
By default the style is  \renewcommand{\theequation}{\thechapter.\arabic{equation}}   for example                  x+y =4         (2.1)   How to put \dots in front equation numbering such as              x+y =4      ...(2.1)   while in the text reference remain (2.1).  and how to make it for the array by using \begin{eqnarray} \end{eqnarray}        x = y+ 4                   -Z+ E                  + 4gh     ...(2.1)
30841
So I'm going through my notes and I think I've confused myself. We often imply  $$ |\psi\rangle \to \psi(x)\\ \langle\psi| \to \psi(x)^* $$ for instance when we talk about eigenvalue equations we interpret  $$ \hat{H}|\psi\rangle =E|\psi\rangle $$ as simply  $$ \hat{H}\psi(x)=E\psi(x) $$ but I don't understand why we say $|\psi\rangle \to \psi(x)$ because if that was the case then  $$ \langle\psi|\psi\rangle=|\psi|^2 $$ when clearly $$ \langle\psi|\psi\rangle=\int_{-\infty}^\infty |\psi|^2 dx = 1 $$ I'm obviously missing something simple, could anyone point out where I'm going wrong?