I have been doing some practice problems from "Gentle introduction to Quantum Computing". I am a little bit lost with this one:  The generalized Pauli group $\mathcal G_n$ is defined by all elements of $\mathcal G_n$ being of the form $\mu A_1\otimes A_2 \otimes \ldots\otimes A_n$ where $A_j\in\left\lbrace I, X, Y, Z\right\rbrace$ and $\mu\in\left\lbrace 1, i, -1, -i\right\rbrace$.     Show that generalized set of Pauli elements for a basis for the linear transformations on the vector space associated with an n-qubit system.   Is there any formal proof for this problem? And how do I approach it?
109175
Is it possible to decompose a hermitian and unitrary matrix $A$ into the sum of the Pauli matrix Kronecker products?   For example, I have a matrix 16x16 and want it to be decomposed into something like $$A =\sum_{i,j,k,l}h_{ijkl}\sigma_i\otimes\sigma_j\otimes\sigma_k\otimes\sigma_l$$ I can't find any clear explanation how it can be done. I found only this for two-dimentional case:
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1) Is $\mathbb{Z}/\langle n\rangle$ cyclic and of order $n$? Why or why not? ($\mathbb{Z}/\langle n\rangle$ is defined to be the factor group of $\mathbb{Z}$ determined by $\langle n\rangle$.)  I guess I'm having trouble understanding factor groups (quotient groups make much more sense) and cyclic groups. My textbook isn't the best at explaining what's going on. I just chose some past homework problems that looked related to my problems.  This is my first abstract algebra course and I'm floundering a lot.
77135
There's a particular property of the elements of $\mathrm{GL}(2,q)$, the general linear group of $n\times n$ matrices over a finite field of order $q$, that I don't understand.  I know that the order of $\mathrm{GL}(2,q)$ is $(q^2-1)(q^2-q)=q(q+1)(q-1)^2$, since there are $q^2-1$ possible vectors for the first column, excluding the $0$ vector, and $q^2-q$ possible vector for the second column, excluding all multiples of the first.  So the order of any element must divide $q(q+1)(q-1)^2$ by Langrange. However, there is a further detail that      any element of $\mathrm{GL}(2,q)$ must have order dividing $q(q-1)$ or $(q-1)(q+1)$.    Is there a reason why one can narrow down the order to divide one of those smaller factors of $q(q+1)(q-1)^2$? Thanks!
43097
I need to see an example of how Hamiltonian, i.e. any Hermitian matrix, can be decomposed into a linear combination of Pauli matrices. I would prefer an option to do this in larger than 2 dimensions, if that is possible.
88756
I am struggling with the end of my proof.   So far I have used the fact that $S_n=\langle T\rangle$ where $T=\{(i\text{ }j)|1\leq i&lt;j\leq n\}$ to show that $S_n$ is generated by the set $\{(1\text{ }2),(1\text{ }3),\cdots,(1\text{ }n)\}$ of $n-1$ transpositions.   I am struggling with proving that $S_n$ is also generated by the set $\{(1\text{ }2),(2\text{ }3),\cdots,(n-1\text{ }n)\}$ of $n-1$ transpositions. Essentially I want to show that each $(1\text{ }i)$ is a product of these transpositions to reach this conclusion.   Here is what I have so far...    Consider any $(1\text{ }i)$ transposition in $S_n$. Notice by the composition of a permutation that \begin{eqnarray*} (1\text{ }i-1)(i-1\text{ }i)(1\text{ }i-1)&amp;=&amp;\begin{bmatrix}     1       &amp; i-1 &amp; i   \\     i-1       &amp; 1 &amp; i    \end{bmatrix} \begin{bmatrix}     1       &amp; i-1 &amp; i   \\    1       &amp; i &amp; i-1  \end{bmatrix} \begin{bmatrix}    1       &amp; i-1 &amp; i   \\     i-1       &amp; 1 &amp; i \end{bmatrix}\\ &amp;=&amp; \begin{bmatrix}     1       &amp; i-1 &amp; i   \\     i-1       &amp; 1 &amp; i    \end{bmatrix} \begin{bmatrix}    1       &amp; i-1 &amp; i   \\     i       &amp; 1 &amp; i-1 \end{bmatrix}\\ &amp;=&amp; \begin{bmatrix}    1       &amp; i-1 &amp; i   \\     i       &amp; i-1 &amp; 1 \end{bmatrix}\\ &amp;=&amp; (1\text{ }i) \end{eqnarray*}  Now we want to use this property, call it STAR, to show that $S_n$ is also generated by the set  $$Y=\{(1\text{ }2),(2\text{ }3),\cdots,(n-1\text{ }n)\}$$ of $n-1$ transpositions. Notice for each $(i\text{ }j)$ in $Y$ that $j-i=1$. Suppose $n=2$. Then $S_2=\{(1),(1\text{ }2)\}$, and we done because $2-1=1$. Suppose $n=3$. Then $S_3$ is generated by the set $\{(1\text{ }2),(1\text{ }3)\}$. By STAR,  $$ (1\text{ }3)=(1\text{ }2)(2\text{ }3)(1\text{ }2) $$ $$ (1\text{ }2)=(1\text{ }1)(1\text{ }2)(1\text{ }1) = (1\text{ }2) $$ Therefore $S_3$ is generated by $\{(1\text{ }2),(2\text{ }3)\}$ and for each $(i\text{ }j)$ we have $j-i=1$.    And here is where I am struggling. I do not know how to apply induction to this. I want to show that each side part will continue breaking down using the property I stated until all of $S_n$ is created. Here is why I think this: $$ (1\text{ }1)(1\text{ }2)(1\text{ }1)=(1\text{ }2) $$ $$ (1\text{ }2)(2\text{ }3)(1\text{ }2)=(1\text{ }3) $$ $$ \cdots $$ $$ (1\text{ }n-1)(n-1\text{ } n)(1\text{ } n-1)=(1\text{ }n) $$ Any input would be much appreciated. Best.
104129
This is a problem from Introduction to Representation Theory by Etingof et al. I have no idea how to approach it. My friend wrote a program that uses Gaussian elimination to find a basis, which starts with x, y, [x, y], [y, [x, y]], ... but freezes once it gets to elements with 9 or more occurrences of x, y.  I expect that there is some way to characterize the nonzero elements of this Lie algebra (i.e., elements that are "immune" to being shown to be zero through repeated application of the Jacobi identity) which could be turned into a way to enumerate them. I don't see what it is, though.  It's also totally unclear to me how to show that some set we know spans the Lie algebra (proven using some criterion as described in the previous paragraph) is actually linearly independent, unless we find a faithful representation... but I have no idea how to find a faithful representation for an infinite-dimensional Lie algebra.  Hints would be greatly appreciated.  Edit: After looking at the partial output, we can form the "straightforward" conjecture that:   If $n \geq 4$ is even, then there is one basis element of grade $n$, namely $\mathrm{ad}([x, y])^{(n-4)/2}[y, [y, [y, x]]]$. If $n \geq 5$ is odd, then there are two basis elements of grade $n$, namely $[x, z]$ and $[y, z]$ where $z$ is the single basis element of grade $n-1$.   I also have a different conjecture, which is mutually exclusive with the above: that if we flatten out all nonzero elements into the form ${\rm ad}(e_1) {\rm ad}(e_2) \ldots {\rm ad}(e_{n-1}) e_n$ using the Jacobi identity, then in the sequences $e_1, \ldots, e_{n-1}, e_n$ and $e_1, \ldots, e_n, e_{n-1}$, any span containing two x's must contain at least two y's and any span containing five y's must contain at least two x's, otherwise we can use the Jacobi identity to rearrange the element somehow to show that it vanishes. If this is true, then the long-range behaviour must be:   For $n \geq 8$ there is exactly one basis element of grade $n$, namely $[... [y, [x, [y, [y, [x, y]]]]] ...]$ (the pattern $y x y$ repeats).   I still don't see an approach to proving that none of the elements in this sequence actually vanishes, though.  Edit 2: Here is the hint given by Etingof in an email:     No, there are no published solutions, since these problems are used as   a homework for classes. This is indeed a nontrivial problem.      This is the positive part of the affine Kac-Moody algebra of type   A_2^2, which is the twisted loop algebra for sl(2). So what you need   to do is to relate this Lie algebra to loops into sl(2). I see from   your post that you are on the right track.
114252
In quantum mechanics any two level system can be mapped onto effective spin variables. If the system is defined by two energy levels, $|E_1\rangle$ and $|E_2\rangle$, the Hamiltonian is  $$ H = \left(\begin{array}{cc} E_1 &amp; 0 \\ 0 &amp; E_2 \end{array}\right) \, .$$  This can be recast as  $$ H = \frac{E_1+E_2}{2} 1\!\!1 + \frac{E_1-E_2}{2} \sigma_3\, .$$  $1\!\!1$ is the identity operator and $\sigma_3$ is the third pauli matrix. This is very nice because we can now apply our knowledge and intuition of spin dynamics to any two level system.  My question is the following: for a real spin I can measure expectation values of $\sigma_{1,2}$ by physically rotating my system and align its $x$ or $y$ axis with my measuring apparatus. How do I do this (in the lab!) with a pseudo-spin system? Imagine I am looking at a cold atom and I want to know if the system is in an symmetric or anti-symmetric superposition of $|E_1\rangle$ and $|E_2\rangle$, how do I do it?
77555
Let $\langle g\rangle$ be a cyclic group of order $n$. Suppose $1\leq q \leq n-1$, I want to show that $g^q$ generates $\langle g\rangle$ if and only if $\gcd(n,q)=1$.  Suppose $g^q$ generates $\langle g\rangle$, then   $$ 1,g^q,g^{2q},...,g^{(n-1)q} $$  are all distinct and $g^{nq}=1$, since $n$ is the order of the group. Note that this is the lowest multiple of $q$ for which this is the case. Therefore $\operatorname{lcm}(q,n)=qn$ and it follows that $q$ and $n$ are coprime.  On the other hand, if $\gcd(q,n)=1$, then $\operatorname{lcm}(q,n)=qn$, we have $g^{nq}=1$ and hence all of  $$ 1,g^q,g^{2q},...,g^{(n-1)q} $$  are distinct again and $g^q$ is a generator.  Is my proof correct at all and is there maybe a more elegant argument?
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I was reading the paper . The author defines a generic two qubit density matrix as $$ \rho=\frac{1}{4}\left( I \otimes I + \vec{r_{\rho}} \cdot \vec{\sigma}\otimes I + I \otimes \vec{s_{\rho}} \cdot \vec{\sigma} + \sum_{i,j=x,y,z}T^{ij}_{\rho} \sigma_i \otimes \sigma_j \right) \, . \tag{1} $$  How is it obtained and what are the constraints over $T^{ij}_{\rho}$ ? Also, seeing that it has some symmetry can a general 3 qubit density matrix be written as  \begin{align} \rho = \frac{1}{8} &amp;\left( I \otimes I \otimes I + \vec{r_{\rho}} \cdot \vec{\sigma} \otimes I \otimes I + I \otimes \vec{s_{\rho}}\cdot \vec{\sigma} \otimes I + I \otimes I \otimes \vec{t_{\rho}}.\vec{\sigma} \right. \\ &amp;+ \sum_{i,j=x,y,z}T^{ij}_{\rho} \sigma_i \otimes \sigma_j \otimes I + \sum_{i,j=x,y,z}U^{ij}_{\rho} \sigma_i \otimes I \otimes \sigma_j \\ &amp;\left. + \sum_{i,j=x,y,z}W^{ij}_{\rho} I \otimes \sigma_i \otimes \sigma_j +\sum_{i,j,k=x,y,z}X^{ij}_{\rho} \sigma_i \otimes \sigma_j \otimes \sigma_k \right) ? \end{align} Here $\vec{r_{\rho}}, \vec{s_{\rho}}, and \vec{t_{\rho}}$ are 3 dimensional vectors with real components and each having magnitude $\le 1$.  EDIT:  I get the fact that tensor of pauli matrices acts as a basis but can't get the condition on $T_{ij}$. I was able to work backwards to see that $T_{\rho}^tT_{\rho}$ has to be such that its maximum eigen value is $\le 1$ so that CHSH inequality is only violated at maximum upto $2\sqrt{2}$. So if this condition is not followed then $(1)$ should not be a valid density matrix. The form given in $(1)$ is already hermitian and has trace 1. So for $T_{\rho}^tT_{\rho}$ having maximum eigen value $\ge1$ $(1)$ might not be a positive operator but I am unable to prove that.
103191
Does anyone know, where I can find a reference (preferably a book) which says that the general linear group $\text{GL}_{2}(\mathbb{Z})$ is generated by the set  $$\left\{\begin{bmatrix} 1&amp;0\\0&amp;-1\end{bmatrix}, \begin{bmatrix} 0&amp;1\\1&amp;0\end{bmatrix}, \begin{bmatrix} 1&amp;1\\0&amp;1\end{bmatrix}\right\}$$  Thank you in advance.
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Proving this can be done as follows: consider a finite group G and elements $g_i \in G$ for some integer $i$. Now consider $\langle g_i \rangle = \{g_i^n: n\geq 0\}$, a generator. It can be proved that $\langle g_i \rangle \leq G$ and that the order of $g_i$ is equal to the order of $\langle g_i \rangle$, so $|\langle g_i \rangle| \leq |G|$.  We can now use Lagrange's theorem which states that if $H \leq G$ then $|H|$ divides $|G|$ and we're done. But...  Is there a simpler way to prove this fact?
61432
We have a general linear group $GL_{2}(\mathbb{C})$ that acts on the set of $M_2(\mathbb{C})$ set of $2\times 2$ matrices by conjugation.   I want to classify the orbits of its action.  What I know:  $$ Orb_{G}(A) = \left\{ BAB^{-1} : \text{where } B \in GL_2(\mathbb{C})\right\}.$$
27388
I have a homework question that asks     Find necessary and sufficient conditions on $1 \leq i &lt; j \leq n$ so that $(i \, j)$ and $(1 \, 2 \, \dotsc \, n)$ generate $S_n$.   Here is what I have done so far.  Call $c = (1 \, 2 \, \dotsc \, n)$.  I made the observation that $$ \overbrace{c c \dotsb c}^{n - j + 1} (i \, j) \overbrace{c^{-1} c^{-1} \dotsb c^{-1}}^{n - j + 1} = (1 + i - j + n \, 1). $$ Thus, $\langle (i \, j), c \rangle = \langle (1 \, i - j + n + 1), c \rangle$, so it suffices to look at $\langle (1 \, k), c \rangle$ for positive integers $2 \leq k \leq n$.  I've checked that, for the value $k=2$, $\langle (1 \, 2), c \rangle = S_n$.  This immediately implies $\langle (1 \, n), c \rangle = S_n$.  I suspect that no other values of $k$ will work, but am not sure how to prove it.
77409
The following is a problem from the Berkeley Problems book.      Let $G$ be a group with three normal subgroups $N_1$ , $N_2$, and   $N_3$. Suppose $N_i \cap N_j = \{ e\}$ and $N_iN_j  = G$ for all   $(i,j )$ with $i \ne j$ . Show that $G$ is abelian and $N_i$, is   isomorphic to $N_j$ for all $i, j$ .   Attempt:   I have not got very far. The only thing that comes to my mind is that as $|HK|=\frac{|H||K|}{|H \cap K|}$, $|N_1|=|N_2|=|N_3|=m$, and order of $|G|=m^2$. I have made other futile attempts of trying to show m prime, finding an isomorphism between the $N's$. I think the fact that the equation $N_i N_j=G$, could be seen as, left coset of $N_j$ with the multiplying elements restricted to only $N_i$ could provide me an answer, but I haven't got anywhere with this.   Thanks in advance.
34229
Let $GL_2(R)$ be the general linear group of $2\times2$ matrices and $SL_2(R)$ the special linear group of $2\times2$ matrices. Show that the commutator subgroup of $GL_2(R)$ is $SL_2(R)$.  I can show that the commutator subgroup is contained in $SL_2(R)$ as if $A,B \in GL_2(R)$ then $$\operatorname{det}(ABA^{-1}B^{-1})=\operatorname{det}(A)\operatorname{det}(B)\operatorname{det}(A^{-1})\operatorname{det}(B^{-1})=1.$$   But how can I show the reverse inclusion? That is, that $SL_2(R)$ is contained in the commutator subgroup of $GL_2(R)$.  Any help will be appreciated.
113461
I know that $S_n$ is generated by a number of things, like all transpositions, all transpositions of the form $(1\,a)$, the transpositions $(1\,2),(2\,3),\dots,(n-1\, n)$, and  just the two elements $(123\dots n),(1\,2)$.      Suppose $n$ is prime. If you just have $(123\dots n)$ and some arbitrary transposition $(a\, b)$, how does this also generate $S_n$?    Can you somehow get to $(1\,2)$ or reduce it to some other previous case?
77052
I'm solving this question in Hungerford's book:     8​. Let $G$ be the multiplicative group generated by the real matrices $a=\begin{pmatrix}2&amp;0\\0&amp;1\end{pmatrix}$ and $b=\begin{pmatrix}1&amp;1\\0&amp;1\end{pmatrix}$. If $H$ is the set of all matrices in $G$ whose (main) diagonal entries are $1$, then $H$ is a subgroup that is not finitely generated.   I've already proved that $H$ is a subgroup of $G$ and every element of $H$ is of the form:   $$         \begin{pmatrix}          1 &amp; x \\         0 &amp; 1 \\         \end{pmatrix} $$  I thought this question really weird, $H$ is a subgroup of a group generated by $a$ and $b$ but $H$ itself can't be generated by $a$ or $b$, how can be possible?  My question is how to show $H$ is infinitely generated? I didn't find any good candidates to generate $H$.  Thanks in advance