это происходило в разных ситуациях, и я не мог точно определить источник проблемы.
Итог, я создал блокнот без кода, только уценка ячеек с латексным кодом. попытался создать PDF, используя верхнее меню-> file-> download as-> pdf через latex.
позже я попытался создать PDF непосредственно из терминала, но возникла та же проблема.
PDF был создан без последних 15 строк или около того.
это происходило раньше, но не последовательно и не всегда одни и те же 15 строк отсутствовали, в прошлый раз это была целая страница.
у меня есть макпро.
имя файла не содержит пробелов.
Я думаю, что причиной этой ошибки является некоторая часть латексного кода, которая не вызывает ошибку в записной книжке, но в то время как преобразование становится проблематичным. я поместил здесь весь код на случай, если кто-нибудь найдет его передо мной.
код:
# 1. {-}
given a potential of the type : $V(x\to-\infty)=V(x\to\infty)=0$ , we can assume the same wave function on both sides of the block .
$\longrightarrow k_1=k_2=\frac{\sqrt{2mE}}{\hbar}\longrightarrow\hspace{20pt}S = \begin{bmatrix} t & -\frac{t}{t^{\star}}r^{\star} \\ r & t \end{bmatrix}$
a matrix is unitary if it's conjugate is also inverse . meaning , $S\bullet S^{\star}=I$
in our case :
$\begin{bmatrix} t & -\frac{t}{t^{\star}}r^{\star} \\ r & t \end{bmatrix}\bullet\begin{bmatrix}t^{\star} & r^{\star} \\ -\frac{t^{\star}}{t}r & t^{\star}\end{bmatrix}=\begin{bmatrix}|r|^2+|t|^2&0\\0&|r|^2+|t|^{2}\end{bmatrix}=I$
hence , S is unitary .
# 2.A and 2.B {-}
given a particle with a delta potential function , the protocle of solution is known :
$\psi(x)=\left\{\begin{array}{l}A_{+}e^{ikx}+A_{-}e^{-ikx}\hspace{60pt}x<0\\B_{+}e^{ikx}+B_{-}e^{-ikx} \hspace{60pt}x>0\end{array}\right.\hspace{30pt}(k=\frac{\sqrt{2mE}}{\hbar})$
stiching the bounderies , considering the discontinuity of the derivative at point of diverging potential :
- $\psi^{'}(0_{+})-\psi^{'}(0_{-})=-\frac{2m\lambda}{\hbar^2}\psi(0)\hspace{10pt}\longrightarrow\hspace{10pt}B_{+}-B_{-}-A_{+}+A_{-}=-2\frac{\lambda}{i\hbar}\sqrt{\frac{m}{2E}}(A_{+}+A_{-})$
- $\psi(0_{+})=\psi(0_{-})\hspace{10pt}\longrightarrow\hspace{10pt}A_{+}+A_{-}=B_{+}+B_{-}$
under the restriction of generality, we can force the constants to match a scenario of a particle starting solely at $x=-\infty$ and obtain the coefficents of transmition and reflection :
$\begin{array}{l}A_{+}=1\\A_{-}=r_1 \\B_{-}=0\\ B_{+}=t_1 \end{array}\hspace{10pt}\longrightarrow\hspace{10pt}\begin{array}{l}r_1=-\frac{q}{1+q} \\ t_1 = \frac{1}{q+1}\end{array}\hspace{10pt}\text{denoting}\hspace{10pt}q=-i\frac{\lambda}{\hbar}\sqrt{\frac{m}{2E}}$
finally , using the formulas mentioned in class we obtain the scatter and transfer matrices :
$T=\begin{bmatrix} 1-q & -q\\ q & 1+q \end{bmatrix}\hspace{30pt}S=\begin{bmatrix} \frac{1}{q+1} & -\frac{q}{q+1}\\ -\frac{q}{q+1} & \frac{q}{1-q} \end{bmatrix}$
# 2.C {-}
we can obtain a solution in a striaght-forward manner by applying the same rules with two boundry points and double the equations.
rather , we can twich the former solution to fit a general delta function $\delta(x-a)$ and then multiply matrices in order to obtain the desired coefficent .
for the general case $\delta(x-a)$ we get the following wave functions :
$\psi(x)=\left\{\begin{array}{l}A_{+}e^{ikx}+A_{-}e^{-ikx}\hspace{60pt}x<a\\B_{+}e^{ikx}+B_{-}e^{-ikx} \hspace{60pt}x>a\end{array}\right.\hspace{30pt}(k=\frac{\sqrt{2mE}}{\hbar})$
- $\psi^{'}(a_{+})-\psi^{'}(a_{-})=-\frac{2m\lambda}{\hbar}\psi(a)\hspace{10pt}\longrightarrow\hspace{10pt}B_{+}e^{ika}-B_{-}e^{-ika}-A_{+}e^{ika}+A_{-}e^{-ika}=-2q(A_{+}e^{ika}+A_{-}e^{-ika})$
- $\psi(a_{+})=\psi(a_{-})\hspace{10pt}\longrightarrow\hspace{10pt}A_{+}e^{ika}+A_{-}e^{-ika}=B_{+}e^{ika}+B_{-}e^{-ika}$
and again forcing the scenario of a particle arriving solely from $x=-\infty$ we obtain the following general matrices:
$T=\begin{bmatrix} 1-q & -qe^{-2ika}\\ qe^{2ika} & 1+q \end{bmatrix}\hspace{30pt}S=\begin{bmatrix} \frac{1}{q+1} & -\frac{q}{q+1}e^{-2ika}\\ -\frac{q}{q+1}e^{2ika} & \frac{q}{1-q} \end{bmatrix}$
finally , the transmission coefficent of a particle arriving solely at $-\infty$ and passing two delta potentials will be the product of two T matrices for a and -a :
$\underset{\delta(x-a)}{\underbrace{\begin{bmatrix} 1-q & -qe^{-2ika}\\ qe^{2ika} & 1+q \end{bmatrix}}}\underset{\delta(x+a)}{\underbrace{\begin{bmatrix} 1-q & -qe^{2ika} \\ qe^{-2ika} & 1+q \end{bmatrix}}}=\underset{T_2}{\underbrace{\begin{bmatrix} ...& = \frac{1}{t_2} \\ ... & \overbrace{(1+q)^2-q^2e^{4ika}} \end{bmatrix}}}\hspace{10pt}\longrightarrow\hspace{10pt}t_2=\frac{1}{(1+q)^2-q^2e^{4ika}}$
# 3.A {-}
given a potential composed of a N identical delta potentials train , and given that the corrseponding transfer matrix obeys the following rule :
$T(k)\cdot\vec{V_{\pm}}=\lambda_{\pm}\vec{V_{\pm}}\hspace{10pt}\longrightarrow\hspace{10pt}T(k)\cdot\underset{=P}{\underbrace{\begin{pmatrix}|&|\\\vec{V_+}&\vec{V_-}\\|&|\end{pmatrix}}}=\begin{pmatrix}|&|\\\vec{V_+}&\vec{V_-}\\|&|\end{pmatrix}\cdot\underset{=\Lambda}{\underbrace{\begin{pmatrix} \lambda_+ &0\\ 0 & \lambda_- \end{pmatrix}}}\hspace{10pt}\longrightarrow\hspace{10pt}T(k)=P\cdot\Lambda\cdot P^{-1}$
as shown in class , we can treat the matrices as equal and ignore the phase accumalated since we are only interested in the transmission coefficent which is equal for all x . therefore :
$T_{tot}=T^N=\left(P\cdot\Lambda\cdot P^{-1}\right)^N=P\cdot\Lambda^N\cdot P^{-1}=P\cdot\Lambda^N\cdot P^T$
where the last equality was due to the orthogonality of the bases.
the vector of amplitudes $\vec{A}=\begin{pmatrix}A_{+}\\A_{-}\end{pmatrix}$ can be projected using the eigenvectors : $\vec{A}=C_{+}\vec{V_{+}}+C_{-}\vec{V_{-}}$
and finally we can write in matrix notation the equation describing a partice incident at N delta potentials :
$\vec{B}=T_{tot}\cdot\vec{A}=P\cdot\Lambda^N\cdot P^T\cdot\left(C_{+}\vec{V_{+}}+C_{-}\vec{V_{-}}\right)=\begin{pmatrix}|&|\\\vec{V_+}&\vec{V_-}\\|&|\end{pmatrix}\cdot\begin{pmatrix} \lambda^N_+ &0\\ 0 & \lambda^N_- \end{pmatrix}\cdot\begin{pmatrix}\leftarrow&\vec{V_+}&\rightarrow\\\leftarrow&\vec{V_-}&\rightarrow\end{pmatrix}\cdot\left(C_{+}\vec{V_{+}}+C_{-}\vec{V_{-}}\right)=C_{+}\lambda^N_+\vec{V_+}+C_{-}\lambda^N_-\vec{V_-}$
the transmission coefficent will just be the ratio of the refracted ampliture by the incident :
$t_{tot}=\frac{B_+}{A_+}\hspace{5pt}\text{,while forcing}\hspace{10pt}B_-=0\hspace{10pt}\longrightarrow\hspace{10pt}t_{tot}=\cfrac{C_-\lambda^N_-(v_-^{(1)}v_+^{(2)}-v_-^{(2)}v_+^{(1)})}{v_{+}^{(2)}(C_+v_+^{(1)}+C_-v^{(1)}_-)}\underset{\text{after simplification}}{=}\cfrac{C_-\lambda^N_-v_-^{(1)}}{v_{+}^{(2)^2}(C_+v_+^{(1)}+C_-v^{(1)}_-)}$
# 3.B {-}
manually performing the multiplication , we obtain the matrix T(k):
$T(k)=\begin{pmatrix}\lambda_+v_+^{(1)^2}+\lambda_-v_-^{(1)^2}&\lambda_+v_+^{(2)}+\lambda_-v_-^{(1)}\\\lambda_+v_+^{(1)}v_+^{(2)}+\lambda_-v_-^{(1)}v_-^{(2)}&\lambda_+v_+^{(2)^2}+\lambda_-v_-^{(2)^2}\end{pmatrix}\hspace{10pt}\longrightarrow\hspace{10pt}tr(T)=\lambda_++\lambda_-\hspace{15pt} \text{and}\hspace{15pt}det(T)=\lambda_+\lambda_-$
- let's assume $tr(T)\le 2$. we also know that det(T)=1 . for these two equations to hold , $\lambda_-$ must equal 1.
in that case it is clear that $t_{tot}>0$.
- let's assume $tr(T)>2$ . applying the same logic as before we conclude that $\lambda_-$ can take any value but 1 . especially $\lambda_-$ can be smaller then 1 , but in that case we obtain:
$\underset{N>>1}{\lim}t_{tot}=\cfrac{C_-v_-^{(1)}}{v_{+}^{(2)^2}(C_+v_+^{(1)}+C_-v^{(1)}_-)}\cdot\underset{\text{smaller then 1}}{\underbrace{\lambda^N_-}}\to 0$
we can ultimately conclude that t>0 if and only if tr(T)$\le$2.