I have been given the following complete systems of eigenvectors  $$\mathbf{Q}\mid\mathbf{q} \rangle=\mathbf{q}\mid\mathbf{q} \rangle, \quad \mathbf{P}\mid\mathbf{p} \rangle=\mathbf{p}\mid\mathbf{p} \rangle, \quad  \langle \mathbf{q} \mid\mathbf{p} \rangle=\frac{1}{(2\pi\hbar)^{n/2}}e^{i\mathbf{q}\cdot\mathbf{p}/\hbar},$$  but I cannot see where the 3rd expression comes from, I cannot understand why it isnt just 1? I think it may be somehow related to the delta function of which I have the integral representation   $$\frac{1}{(2\pi\hbar)^n}\int_{\mathbb{R}^n}e^{i\mathbf{p}\cdot\mathbf{q}/\hbar}d^np=\delta^n(\mathbf{q}).$$
19797
I'm curious whether I can find the overlap $\langle q | p \rangle$ knowing only the following:   $|q\rangle$ is an eigenvector of an operator $Q$ with eigenvalue $q$. $|p\rangle$ is an eigenvector of an operator $P$ with eigenvalue $p$. $Q$ and $P$ are Hermitian. $[Q,P] = i \hbar$.   I'm asking because books and references I've looked in tend to assume that $Q$ is a differential operator when viewed in the $P$-basis, then show that it satisfies the commutation relation. (I haven't read one yet that proves that the form given for $Q$ is the only possible one.) I've heard, though, that we can work purely from the hypothesis of the commutation relation and prove the properties of $Q$ and $P$ from it.
31677
If $B_1 = \{v_1,\ldots,v_n\}$ and $B_2 = \{v_1',\ldots,v_n'\}$ are basis for a vector space V, and $D_1= \{\delta v_1,\ldots, \delta v_n\}$ and $D_2 = \{\delta v_1',\ldots, \delta v_n'\}$ are the corresponding dual bases of $V^*$ prove that, if $P$ is the change-of-basis matrix from $B_1$ to $B_2$, then $(P^{-1})^T$ is the change-of-basis from $D_1$ to $D_2$  (this is a theorem from Schaum's outline of theory and problems of linear algebra)  My knowledge of linear algebra is not great, but so far I have that   \begin{align} \text{For any }v \in V, [v]_{B_1}&amp;=P[v]_{B_2}\\ \text{so } (\alpha_1,\ldots,\alpha_n)(v_1,\ldots,v_n)^T&amp;=P(\beta_1,\ldots,\beta_n)(v_1',\ldots,v_n')^T \end{align} for some $\alpha_1,\ldots,\alpha_n,\beta_1,\ldots,\beta_n \in F$ (where $F$ is a field)  How do I use the fact that $D_1$, $D_2$ are dual basis to further this proof? And also, where does the transpose of $P^{-1}$ come in? Can I use that $D_1 = \delta B_1$ and $D_2 = \delta B_2$ (where I am guessing $\delta$ is the kronecker delta?)  Any hints or suggestions are greatly appreciated, I know that my knowledge is quite limited with regard to this, links to text or sources that could explain a problem like this would be great.
25250
In the following question we regard vectors in $\Bbb R^n$ as columns and define the dot product in the usual way which means that $x · y$ = $x^{t}y$  a) If A is an n × n matrix show that $A_{ij}$ = $e_i \cdot A e_j$ where $e_i$, $i =1, \dots , n$ are the standard basis vectors in $\Bbb R^n$  b) Show that a matrix $P$ is orthogonal if and only if ($Px$) · ($Py$) =$x · y$ for all $x, y$ ∈ $\Bbb R^n$  Working -   I got part (a) alright but have no idea of where to start with part (b) except to use the information that an n x n matrix $P$ is orthogonal if $P^{-1}$= $P^{t}$ and that $P^{t}$$P$=$I$, $P$$P^{t}$=$I$, where $I$ is the identity matrix. Any suggests are gracefully appreciated. Feel free to edit my question for clarity. Thank you
65684
I'm having trouble with part of homework exercise. Let $\mathcal{P}_3(\mathbb{C})$ be the vector space of complex polynomials of degree 2 or lower and $\alpha,\beta\in\mathbb{C},\alpha\ne \beta$.  Let $L:\mathcal{P}_3(\mathbb{C})\to\mathbb{C}^2$ given by $L(p)=(p(\alpha), p(\beta))^T$ and I want to find the matrix representation of $L$ with respect to the bases $\{1,X,X^2\}$ and $\{\mathbf{e_1},\mathbf{e_2}\}$.  I usually apply $L$ to the elements of the basis for $\mathcal{P}_3$, but with this definition of $L$ I don't know how to do it. Can someone give me a pointer?
64715
Preliminaries: In their QFT text, Peskin and Schroeder give the KG propagator (eq. 2.50) $$  D(x-y)\equiv\langle0|\phi(x)\phi(y)|0\rangle = \int\frac{d^3p}{(2\pi)^3}\frac{1}{2\omega_\vec{p}}e^{-ip\cdot(x-y)},  $$ where $\omega_\vec{p}\equiv\sqrt{|\vec{p}|^2+m^2}$.  For lightlike separations, we can choose a frame where $x-y$ is purely in the time-direction and the propagator can be put into the form (2.51) $$ D(x-y)=\frac{1}{4\pi^2}\int^\infty_m d\omega\sqrt{\omega^2-m^2}e^{-i\omega (y^0-x^0)} \tag{1}\label{timelike_prop}, $$ where I use the $\text{diag }\eta=(-,+,+,+)$ convention. Now, one has the following integral representation of the modified Bessel function () \begin{align} K_1(z) &amp;= z\int^\infty_1 dt \sqrt{t^2-1} e^{-zt} \\        &amp;= \frac{z}{m^2} \int^\infty_m dt \sqrt{t^2-m^2} e^{-zt/m}, \tag{2}\label{int_rep} \end{align} where we go to the second line by rescaling the integration variable $t \to t/m$.  Comparing \eqref{timelike_prop} with \eqref{int_rep} suggests $$ D(x-y)=\frac{m}{(2\pi)^2|y-x|}K_1(m|y-x|), $$ where we have written the time separation in terms of the Lorentz invariant $i (y^0-x^0)=|y-x|$.  (Note: there is an issue in what I've written here in that the integral representation \eqref{int_rep} is only valid for $|arg z|&lt;\pi/2$ and $|y-x|$ is on the imaginary axis ($|arg z|=\pi$), but I think one could infinitesimally displace $z$ off of the imaginary axis to get a convergent integral.  Check me on that.) Anyway, for spacelike separations, we can choose a frame where $y-x=\vec{y}-\vec{x}\equiv\vec{r}$.  Performing the polar integrations yield $$ D(x-y)=\frac{-i}{2(2\pi)^2 r}\int^\infty_{-\infty}dp\frac{p e^{ipr}}{\sqrt{p^2+m^2}}. $$ Finally, PS claim that taking a contour integral in the upper half plane (making sure to avoid the branch cut at +im) will give $$ D(x-y)=  \frac{1}{(2\pi)^2r}\int^\infty_m d\rho \frac{\rho e^{-\rho r}}{\sqrt{\rho^2-m^2}}, \tag{3}\label{spacelike_prop} $$ where $\rho\equiv-ip$. Question: I know from plugging into Mathematica that the spacelike propagator \eqref{spacelike_prop} can also be expressed as a modified Bessel function $K_1$.  Moreover, the integration bounds of \eqref{spacelike_prop} and \eqref{int_rep} are even the same.  However, I don't see how to transform the spacelike propagator integral \eqref{spacelike_prop} into the form of \eqref{int_rep}.  Any ideas? (I'd prefer, if at all possible, to use the integral representation that I've quoted \eqref{int_rep} and used for the timelike case rather than some other representation of the modified Bessel function.)
124076
Let $V$ be the vector space over $\mathbb{R}$ consisting of all polynomials of degree $\leqslant n$. If $f,g\in V$, let $$\langle f,g \rangle = \int_\limits{0}^{1}f(t)g(t)dt$$ Find the matrix of this scalar product with respect to the basis $\{1,t,\dots,t^n\}$.  What does it mean to find a matrix of the scalar product? Thanks in advance!
19516
For a real (or complex) inner product space $V$, the inner product can be expressed in terms of the norm as either $$ \langle x,y\rangle=\frac{1}{4}(\|x+y\|^2-\|x-y\|^2) $$ or $$ \langle x,y\rangle=\frac{1}{4}(\|x+y\|^2-\|x-y\|^2+i\|x+iy\|^2-i\|x-iy\|^2) $$ respectively.  These identities are not hard to verify directly once they are given. I would like to know, is there a systematic way one would derive these from scratch without knowing them beforehand, or are both polarization identities just clever observations?
101271
I was going through my notes on the unitary irreducible representations of the Poincare group and the subsequent construction of one particle states and I stumbled across the following steps in the method of induced representations: ($\mathscr{H}$ is the Hilbert space and $\mathscr{H}_p$ is the subspace of one-particle states with momenta p. $\mathcal{S}_k$ is just the little group of the standard momentum $k$. $L(p)$ is the standard Lorentz boost; $L(p)k=p$)   (1) Let $|k,\sigma\rangle$ be a basis for $\mathscr{H}_k$ and let $\mathcal{D}$ be the (finite) unitary irrep of $\mathcal{S}_k$ induced by the unitary irrep $\mathcal{U}$. Then $\mathcal{D}$ acts on $\mathscr{H}_k$ by mixing the spin polarization states,  \begin{align} \mathcal{U}(h)|k,\sigma\rangle=\sum_{\sigma'}\mathcal{D}_{\sigma'\sigma}(h)|k,\sigma'\rangle,~~~ \forall h\in\mathcal{S}_k,~~~\mathcal{U}(h)\equiv\mathcal{U}(h,0).  \end{align} (2) Define the basis for $\mathscr{H}_p$ by \begin{align} |p,\sigma\rangle=\mathcal{U}(L(p))|k,\sigma\rangle. \end{align} (3) The action of Lorentz transformations on these states is then \begin{align} \mathcal{U}(\Lambda)|p,\sigma\rangle=\sum_{\sigma'}\mathcal{D}_{\sigma'\sigma}\big(h(\Lambda,p)\big)|\Lambda p,\sigma'\rangle  \end{align} where $h(\Lambda,p)=L^{-1}(\Lambda p)\Lambda L(p)\in \mathcal{S}_k$ is known as the Wigner rotation.  (4) Finally, we extend this action to an arbitrary state in the Hilbert space $|\Psi\rangle=\sum_{p,\sigma}\Psi_{\sigma}(p)|p,\sigma\rangle\in\mathscr{H}$ for some coefficients $\Psi_{\sigma}(p)$ by, \begin{align} \mathcal{U}(\Lambda)|\Psi\rangle&amp;=\sum_{p,\sigma}\big[\mathcal{U}(\Lambda)\cdot\Psi\big]_{\sigma}\big( p\big)| p,\sigma\rangle,\notag\\ \text{ where } \big[\mathcal{U}(\Lambda)\cdot\Psi\big]_{\sigma}\big(p\big)&amp;=\sum_{\sigma'}\mathcal{D}_{\sigma\sigma'}\big(h(\Lambda,\Lambda^{-1}p)\big)\Psi_{\sigma'}(\Lambda^{-1}p). \end{align}  These are my own notes which are largely influenced by Weinbergs construction, so they may be incorrect. I understand all of the steps except for one part of step 4; The part where I say $|\Psi\rangle=\sum_{p,\sigma}\Psi_{\sigma}(p)|p,\sigma\rangle\in\mathscr{H}$ for some coefficients $\Psi_{\sigma}(p)$. What are these coefficients? Are they the wavefunctions? Do they depend continuously on $p$, if so should the sum over $p$ be replaced by an integral? Should these coefficients be square-integrable; $\Psi_{\sigma}(p)\in L^2(\Sigma_m,\mu,\mathbb{C})$ where $\Sigma_m$ is the mass shell and $\mu$ a Lorentz invariant measure? Is this expansion of an arbitrary state even correct?
71857
Help me find $\hat{B^\dagger}$, when we know that $$\hat{B}=i\frac{d}{dr}$$ with the condition that $\hat{B}$ is defined in spherical coordinates. My approach: $$ \langle\psi|\hat{B}\psi\rangle=\int_{0}^{\infty} \psi^* i\frac{d}{dr} \psi dr=\psi^*\psi|_{0}^{\infty}-\int_{0}^{\infty}\psi i\frac{d}{dr}\psi^*dr=\langle\hat{B}\psi|\psi^*\rangle$$ And so I get $ \hat{B^\dagger} = -i\frac{d}{dr} $. Could someone confirm if this is correct?
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.
7310
I am having difficulty solve this problem in my homework:  (In my notation, $[x;y]$ represents a matrix of 2 rows, 1 column)     Let $\mathbf{x}=[x_1;x_2]$, $v_1$=[−3;5] and $v_2=[7;−2]$ and let $T\colon\mathbb{R}^2\to\mathbb{R}^2$  be a linear transformation that maps $\mathbf{x}$ into $x_1v_1+x_2v_2$. Find a matrix $A$ such that $T(\mathbf{x})$ is $A\mathbf{x}$ for each $\mathbf{x}$.   I am pretty clueless. So I assume that I start off with the following:  $x_1v_1 + x_2v_2 = x_1[−3;5] + x_2[7;−2]$  But I do not know what to do from here, or if this is even the correct start!
79770
How is the momentum operator derived in Dirac formalism? I am reading Quantum Mechanics by Sakurai and he gives the following derivation. But I don't understand how he goes from the third equation to the last equation in (1.7.15). What I don't understand is where the partial derivative with respect to $x^{\prime}$ comes from. Here is the derivation from the book.     Momentum Operator in the Position Basis      We now examine how the momentum operator may look in the $x$-basis - that is, in the representation where the position eigenkets are used as base kets. Our starting point is the definition of momentum as the generator of infinitesimal translations:      $$\begin{align} \biggl(1 - \frac{ip\Delta x'}{\hbar}\biggr)\lvert\alpha\rangle &amp;= \int dx' \mathcal{J}(\Delta x')\lvert x'\rangle\langle x'\lvert\alpha\rangle \\ &amp;= \int dx' \lvert x' + \Delta x'\rangle\langle x'\lvert\alpha\rangle \\ &amp;= \int dx' \lvert x'\rangle\langle x' - \Delta x'\lvert\alpha\rangle \\ &amp;= \int dx' \lvert x'\rangle\biggl(\langle x'\lvert\alpha\rangle - \Delta x'\frac{\partial}{\partial x'}\langle x'\lvert\alpha\rangle\biggr).\tag{1.7.15} \end{align}$$      Comparison of both sides yields      $$p\lvert\alpha\rangle = \int dx'\lvert x'\rangle\biggl(-i\hbar\frac{\partial}{\partial x'}\langle x'\lvert\alpha\rangle\biggr)\tag{1.7.16}$$      or      $$\langle x'\rvert p\lvert\alpha\rangle = -i\hbar\frac{\partial}{\partial x'}\langle x'\lvert\alpha\rangle,\tag{1.7.17}$$
54646
I have seen the scattering matrix defined using initial ("in") and final ("out") eigenstates of the free hamiltonian, with  $$\left| \vec{p}_1 \cdots \vec{p}_n \; \text{out} \right\rangle = S^{-1} \left| \vec{p}_1 \cdots \vec{p}_n \; \text{in} \right\rangle$$  so that  $$\left\langle \vec{p}_1 \cdots \vec{p}_n \; \text{out} \mid \vec{q}_1 \cdots \vec{q}_m \; \text{in} \right\rangle = \left\langle \vec{p}_1 \cdots \vec{p}_n \; \text{in} \mid S \mid \vec{q}_1 \cdots \vec{q}_m \; \text{in} \right\rangle.$$  1) What are "in" and "out" states?   2) Are they Fock states?   3) In Schrödinger or Heisenberg or interaction representation?   4) How are they related? (I believe that I see what they handwavily represent physically, but not formally.)  My main issue is that, if "in" and "out" states are one-particle eigenstates of the free hamiltonian, i.e. if $\left| \vec{p}_1 \text{out} \right\rangle$ describes a free particle with momentum $\vec{p}_1$, and $\left| \vec{p}_1 \text{in} \right\rangle$ also describes a free particle with momentum $\vec{p}_1$, then $\left| \vec{p}_1 \text{out} \right\rangle = \left| \vec{p}_1 \text{in} \right\rangle$ ... which is false. Still, books (some at least) describe these "in" and "out" states like that.  Moreover, I have seen (e.g. in Wikipedia, but also ) that the scattering matrix is a map between two different Fock states, and I don't understand that.   5) Do states of the interacting system live in the same Fock space that asymptotic free states?   6) And if not, where do they live?  Understandable references would be appreciated.
10343
I'm currently working at the start of Alexei Tsvelik's book Quantum Field Theory in Condensed Matter Physics. I'm kinda stumped on a few essential steps.  Starting with the action:  $$S = \int dt \int d^3 x \mathcal{L}_{\mathrm{EM}} = \int dt \int d^3 x \left( -\frac{1}{4} F_{\mu \nu} F^{\mu \nu} + j_\mu A^{\mu}\right).$$  In the absence of any sources $j_\mu = 0$ and $c=1$, which after pulling in the negative sign implies  $$S = \frac{1}{2}\int dt \int d^3 x  \,\,\left(E^2 - B^2\right)$$ The next step is to take $\vec{E} = -\vec{\nabla} \phi + \partial_t \vec{A}$ and $\vec{B} = \vec{\nabla} \times \vec{A}$. Letting $\phi = 0$ and directly inserting these into the above yields:  $$S = \frac{1}{2} \int dt \int d^3 x  \,\left( (\partial_t \vec{A})^2 - (\vec{\nabla} \times \vec{A})^2 \right) $$  Now considering $\vec{A} = \vec{A}_0 + \delta \vec{A}$ as some minimal change in our vector potential, with $\vec{A}_0$ gives a minimum for the action and putting this into the above expression we have: $$\delta S = \int dt \int d^3 x \left(\partial_0 \vec{A}_0 \partial_0 \delta \vec{A} + \left(\vec{\nabla} \times \vec{A}_0\right)\cdot\left(\vec{\nabla} \times \delta \vec{A} \right) \right)+ \mathcal{O}(A^2)\,\,\,\,\,\,\, (1)$$ where I have the term  $$\mathcal{O}(A^2) = \int dt \int d^3 x \left( (\partial_0 \vec{A}_0)^2 + (\partial_0 \delta \vec{A})^2 - (\vec{\nabla}\times\vec{A}_0)^2-(\vec{\nabla}\times \delta\vec{A})^2\right)$$  The next step is where I lose our author and go from "Oh ok I get this" to "completely clueless". He rewrites $\delta S$ as:  $$\delta S =  \int dt \int d^3 (\delta x) \vec{A} \mathcal{F(\vec{A}_0)}+ \mathcal{O}(A^2)$$  where $\mathcal{F} = \frac{\delta S}{\delta A}$. What is the purpose of $x\to \delta x$ in the above expression? Is the functional derivative $\mathcal{F}$ a vector quantity? He has it bolded with a bolded $\vec{A}$ in the above expression. One last question takes the above expression further. We assume $\delta \vec{A}$ vanishes at infinity, as it should, and integration by parts yields:  $$\delta S =  -\int dt \int d^3 x \left( \partial^2_0 \vec{A}_0 - (\vec{\nabla} \times \vec{\nabla})\times \vec{A}_0  \right)\delta \vec{A}\,\,\,\,\,\,\,\, (2)$$ How exactly did we go from (1) to (2)? Thanks I appreciate your time!
75890
This question can be formulated in two ways. Let there be two $d$-dimensional orthonormal bases $B_{1}$ and $B_{2}$. I refer to the elements of $B_{1}$ by $\lvert\nu_{i}\rangle$ and to the elements of $B_{2}$ by $\lvert\omega_{j}\rangle$. These two bases are mutually unbiased namely: $\lvert\langle\omega_{j}\lvert\nu_{i}\rangle\lvert^{2}=1/d$ for all $\lvert\omega_{j}\rangle$ and $\lvert\nu_{i}\rangle$. We define some vector $\lvert\tau\rangle$ such that $\lvert\tau\rangle$ is orthonormal to one element of $B_{2}$, say $\lvert\omega_{m}\rangle$ and it is mutually unbiased with respect to all the elements in $B_{1}$: $\lvert\langle\tau\lvert\nu_{i}\rangle\lvert^{2}=1/d$ for all $\lvert\nu_{i}\rangle$. Can we conclude that $\lvert\tau\rangle$ is equal to a member of $B_{2}$ say $\lvert\omega_{p}\rangle$ for some $p\neq m$ up to a phase difference?  Another way to make the same conclusion is to ask this: Let there be three orthonormal bases $B_{1}$, $B_{2}$ and $B_{3}$. $B_{1}$ and $B_{2}$ are mutually unbiased, so are $B_{2}$ and $B_{3}$. Can we conclude that $B_{1}$ and $B_{3}$ are either mutually unbiased or have equivalent Hadamard matrices? This question for me is particularly important since in the literature it is always talked about sets of pairwise mutually unbiased bases. But there's no proof given for the fact that three mutually unbiased bases should necessarily be pairwise MUBs.
23257
I am trying to derive Eq. (7.25) (p. 117) of Polyakov's book:  $$ \delta \Psi (C) ~=~ \int_{0}^{2\pi} {\rm P} \left(F_{\mu\nu}(x(s)) \exp \oint_C A_\mu dx^\mu \right)\dot{x}_\nu \delta x_\mu(x) \, {\rm d} s, \tag{7.25} $$  where the non-abelian phase factor around a closed loop $C$ is defined as  $$ \Psi(C) ~=~ {\rm P}\exp \left(\oint A_\mu dx^\mu \right) = {\rm P}\exp \left(\int_{0}^{2\pi} A_\mu \dot{x}_\mu\, {\rm d}s \right). \tag{7.1} $$  It seems that he is using the relation given on p. 116:  $$ \delta \, {\rm P} \exp \int_{0}^{2\pi} M(\tau) {\rm d}\tau ~=~ \int_{0}^{2\pi}{\rm d}t\,{\rm P} \left(\delta M(t) \exp \int_{0}^{2\pi}M(\tau){\rm d}\tau\right). \tag{7.24b} $$  Matching with (7.25) I find $\delta A_\nu = F_{\mu\nu} \delta x_\mu$. This relation seems to be saying that if I change the position of the loop at the parameter $s$ by $\delta x_\mu(s)$ then the vector potential changes by $\delta A_\nu(x(s)) = F_{\mu\nu}(x(s)) \delta x_\mu(s)$.    I don't know how to derive this relation. Is it legitimate?
66641
Why can we write an arbitrary object  $v_{a \dot{b} }$ our transformations in this basis act on as  $$ v_{a \dot{b} } = v_{\nu} \sigma^{ \nu}_{a \dot{b} } = v^0 \begin{pmatrix}    1&amp;0 \\ 0&amp;1 \end{pmatrix}  + v^1 \begin{pmatrix}    0&amp;1 \\ 1&amp;0 \end{pmatrix}  +v^2 \begin{pmatrix}    0&amp;-i \\ i&amp;0 \end{pmatrix}  + v^3 \begin{pmatrix}   1&amp;0\\0&amp;-1 \end{pmatrix} $$  Formulated differently: How do we know that the vector space for the $(\frac{1}{2},\frac{1}{2})= (\frac{1}{2},0) \otimes (0,\frac{1}{2})$ representation of the Lorentz group is the space of hermitian $2\times2$ matrices? The vector space for the $(\frac{1}{2},0)$ representation is $\mathbb{C}^2$ and I guess the same is true for the $(0,\frac{1}{2})$  representation, but I can't put it together to end up with hermitian matrices.   EDIT: I found in the book  by Matthew Robinson the following explanation.      Recall that just as any real matrix can be written as the sum of a symmetric matrix and an antisymmetric matrix, any complex matrix can be written as the sum of a Hermitian matrix and an anti-Hermitian matrix. However, the two indices on our matrix $v^{a \dot b}$ transform under representations of $SU(2)$. Notice that in the generators of these copies of $SU(2)$, both sets of generators $N^-$ and $N^+$ are Hermitian (cf. (3.229)). So, we’ll limit our discussion to the case where $v^{a \dot b}$ is a Hermitian $2 \times  2$ matrix.   If anyone could help me understand this line of thought my problem would be solved.  Why does this allow us to " limit our discussion to the case where $v^{a \dot b}$ is a Hermitian $2 \times  2$ matrix"?  I understand that our representation here acts on complex $2\times2$ matrices. But I don't understand why we can restrict to hermitian matrices.