I know what trace is in matrix algebra, but does it have any intuitive physical significance when acted on a quantum mechanical operator? What aspect of an operator $A$ does ${\rm Tr}(A)$ capture? Or is it just a convenient mathematical tool and nothing more?
21815
I was recently thinking what information do we get from a matrix. So if we say the columns (or rows) of a matrix define the basis of a system, say vectors of 3 dimensional space. Then the determinant will tell about the volume of the space enclosed by those vectors. Does trace give any information about the space?
101901
Any traceless $n\times n$ matrix with coefficients in a field of caracteristic $0$ is a commutator (or Lie bracket) of two matrices. What happens when the field has positive caracteristic?  When trying to reproduce the proof I have in the caracteristic $0$ case for the positive caractersitic case, I run into two problems:   multiples of the identity may have trace $=0$. a matrix may have a spectrum equal to the whole field.      Are all traceless matrices commutators? If not, for which $n\in\mathbb{N}\setminus\lbrace 0  \rbrace$ and fields $k$ does it still hold? EDIT given a traceless matrix $M$, I want to know wether there are two matrices $A,B$ with $M=AB-BA$.
71234
I have a question to ask down below, that I have been having some trouble with and would like some help and clarification on.  Suppose A  is an $n \times n$ matrix with (not necessarily distinct) eigenvalues $\lambda_{1}, \lambda_{2}, \ldots, \lambda_{n}$. Can it be shown that:     (a) The sum of the main diagonal entries of A, called the trace of A, equals the sum of the eigenvalues of A.   (b) A $- ~ k$ I has the eigenvalues $\lambda_{1}-k, \lambda_{2}-k, \ldots, \lambda_{n}-k$ and the same eigenvectors as A.  Thank You very much.
64171
Can someone explain  simply?  I know the math, however I don't understand the relevance.   To explain what I'm asking, what is the significance of it when doing quantum mechanical calculations such as getting Planck length or other such equations.
90641
(a) Prove that a square matrix $T$ of rank one has $\text{tr}(T)=0$ if and only if $T^2=0$.       (b) Consider a matrix $A$ of the form $A=aI+T$, where $a\ne0$, $I$ is the identity matrix,   and $T$ has rank one and zero trace. Find the inverse and the determinant of $A$.       (c) Find the inverse of $A$ as above when $T$ has rank one but nonzero trace $\text{tr}(T)=b$.      For which value of $b$ is $A$ not invertible?    I'm still stuck on part (a), but campus buildings are closing soon, so I'll be working from home but would love to get some hints / comments on this question.  I'll have limited access to this site - on my phone.  For part (a), I've been trying to look at the SVD of matrix $A$, since one can read off the rank very easily - by looking at the number of non-zero singular values of $A$.  Then I am trying some block matrix multiplication to see whether $T^2 = 0$, from assuming that $\text{tr}(T) =0$.  So far, no luck.  Do you think I should stick with this SVD approach, or is it better to play around with the definition and properties of nilpotent operators?  Any other hints for the other parts of the question would be greatly appreciated.  Thanks!
21273
Let $G$ be a matrix Lie group. Its Lie algebra $\mathfrak{g}$ comes equipped with a symmetric (2,0) tensor known as its Killing from, denoted $K$ and defined by  $$ K(X,Y) = -\text{Tr}(\text{ad}_X \circ \text{ad}_Y)$$  where $\text{ad}_X \in \text{GL}(\mathfrak{g})$ defined by $\text{ad}_X(Y)= [X,Y]$. Show that $K$ is Ad-invariant, in the sense that  $$  K(\text{Ad}_A(X), \text{Ad}_A(Y)) = K(X,Y) \quad \forall X,Y \in \mathfrak{g}, A \in G $$.  Here $\text{Ad}_A(X) = AXA^{-1} \in \mathfrak{g}$.   I suppose that it suffices to prove it on a basis, but it isn't a big deal to let $X$ and $Y$ be arbitrary. Choosing a basis $E_j$ for $\mathfrak{g}$, I write $$ [E_j,E_k] = {c_{jk}}^l E_l $$ $$ X = a^kE_k, Y = b^kE_k $$ and  $$ AE_kA^{-1} = {d_k}^lE_l. $$ Then $$ [X,[Y,E_m]] = [a^kE_k,[b^lE_l,E_m]] = a^kb^l {c_{kn}}^j {c_{lm}}^n E_j$$ and so the trace of this operator is $a^kb^l{c_{kn}}^j{c_{lj}}^n$. The left hand side is $$  [AXA^{-1},[AYA^{-1},E_m]] = [a^k{d_k}^lE_l,[b^n {d_n}^j E_j,E_m]]= a^kb^n{d_k}^l{d_n}^j {c_{jm}}^r {c_{lr}}^k E_k$$ so the trace of this operator is $$ a^kb^n{d_n}^j{d_k}^l{c_{jk}}^r{c_{lr}}^k $$  assuming the indices are correct. Now these two have to be the same somehow. Using that $\text{Ad}_A{[X,Y]}=[\text{Ad}_A{X}, \text{Ad}_A{Y}]$ may be promising, but I haven't found anything useful with it. A hint would be appreciated.
50221
Suppose we have a particle of mass $m$ that is in an eigenstate $|\psi\rangle$ of the Hamiltonian $\hat{H}=\hat{T}+\hat{V}$, where $\hat{T}$ is the kinetic energy operator and $\hat{V}=V(\hat{r})$ is the potential energy operator. If the potential has lower bound $V_0$, then is it necessary for the energy eigenvalue $E$ of $|\psi\rangle$ to be greater than $V_0$? Classically this is true, since we regard negative kinetic energy as physically unrealizable/meaningless. However, I don't know if I can necessarily say the same in the quantum case. For example, what I'm tempted to do is write  $$\langle\psi | \hat{T}|\psi \rangle = \langle\psi | E-\hat{V}|\psi\rangle$$  and say "If $E&lt;V_0$, then the RHS is necessarily negative, implying the LHS is as well, which we will regard as physically meaningless. If $E=V_0$, then $|\psi\rangle$ is trivial.", but I'm not sure if that's right.   Confused by that, I then wanted to show that, if $E&lt; V_0$, then a non-trivial $|\psi\rangle$ is non-normalizable. However, I'm not entirely sure how to do this.
113328
This relates a little bit to my previous question (), but it's different in a way and more historical.  One of the postulates of quantum mechanics is:     To every observable in classical mechanics there corresponds a linear, Hermitian operator in quantum mechanics ().   However, absent in the list of postulates is what these operators are.  So is the mathematical expression for each individual operator also a postulate that's not listed, or are they derivable from other axioms?  Specifically, where did  expressions come from?  Maybe the question can better be answered historically.  If I had to guess, I'd say it began with the wavefunction.  Some scientists thought perhaps the squared modulus of the wavefunction corresponded to charge distribution or something, but Born was the one who got it right.  He essentially made a guess that the squared modulus corresponded to the probability of finding a particle at a certain location.  What's funny is that he initially wrote down the wrong expression in his paper (), and his paper even got rejected by the first journal he submitted it to, but the right expression is in a corrective footnote.    But again, Born simply guessed what the squared modulus meant.  There was no procedure; he just thought of an idea and then experiment verified it to be correct.  Thus:  $$\langle\psi|\hat{X}|\psi\rangle = \langle\psi|x\rangle\langle x|\hat{X}|x\rangle\langle x|\psi\rangle = \int d\vec{x}\ \psi^*(\vec{x})\ \psi(\vec{x})$$  So basically in the position basis, the position operator is simply $\hat{X} = x$, which seems pretty intuitive.  I believe the momentum operator came about next; and it was "derived" by fiddling around with the de Broglie relation and just substituting classical expressions for momentum into the Schrodinger equation (can't remember where I read that).  The fact that plugging these classical ideas into a quantum equation just happens to match experiment for momentum measurements seems like a pure coincident to me (it's along the same lines as trying to make the Schrodinger equation relativistic; one way of shoehorning relativity into it gives you the Dirac equation and a different way gives you the Klein-Gordon equation).  I imagine the spin operator came about by simple observation.  An electron goes up or electron goes down in a magnetic field.  Not hard to guess what the expression for that operator might be.  So anyway, am I on the right track here?  Have all of the operators come about simply by guessing?  And if I'm wrong, how did they actually come about?  How would I come up with a new operator for a new type of measurement?
17651
Let $A,B$ have the appropriate size. How can we show Von Neumann Trace inequality $ \text{Tr}(AB) \leq \sum_{i=1}^n \sigma_{A,i}\sigma_{B,i} $?  Also, what is the intuition behind this inequality?
114543
I am trying to understand . As far as I could find there are three equivalent definitions and in general many sources start from a different one, still I fail to see their equivalence. I restate the definitions and their equivalences given in the Wikipedia page:   Eigenstate of annihilation operator: $$a|\alpha \rangle =\alpha|\alpha \rangle$$ Displacement operator of the vacuum: $$ |\alpha \rangle =e^{\alpha a^{\dagger}-\alpha^{*} a}|0 \rangle $$ State of minimal Uncertainty: $$ \Delta X= \Delta P = \frac{1}{\sqrt{2}} $$   I fail to see how they are the same! Can someone please explain how to derive these from one another?
35431
I've been studying different scattering processes (from Mandl &amp; Shaw QFT's book, chapter 8) and there's always a purely-mathematical common step I do not understand: the showing-up of the trace. Let me give two specific examples.  $$A_{(l) \alpha \beta}=\sum_{s_1} \sum_{s_2} \Big[ \bar u_{s_2} (\vec p_2') \gamma_{\alpha} v_{s_1} (\vec p_1'))(\bar v_{s_1}(\vec p_1')\gamma_{\beta} u_{s_2} (\vec p_2'))\Big]_{(l)}=Tr\Big[\frac{\not{\!p_2'}+m_l}{2m_l} \gamma_{\alpha} \frac{\not{\!p_1'}-m_l}{2m_l}\gamma_{\beta}\Big] \ \ \ \ (1)$$  $$X= \frac 1 2 \Lambda_{\delta \alpha}^+ (\vec p') \Gamma _{\alpha \beta} \Lambda_{\beta \gamma}^+ (\vec p) \tilde \Gamma _{\gamma \delta}=\frac 1 2 Tr \Big[\Lambda^+ (\vec p') \Gamma \Lambda^+ (\vec p) \tilde \Gamma \Big] \ \ \ \ (2)$$  Where:   $\not{\!A} := \gamma^{\alpha} A_{\alpha}$ $\Gamma$ is a $4 \times 4$ matrix $\tilde \Gamma := \gamma^0 \Gamma^{\dagger} \gamma^0$ $u, v$ are the Dirac Spinors (which are $4 \times 1$ matrices) The Dirac-$\gamma$-matrices are $4 \times 4$ matrices and the adjoint is defined as $\bar w := w^{\dagger}\gamma^{0}$ $\Lambda^+$ is the positive energy projection operator defined as  $$\Lambda_{\alpha \beta}^+ (\vec p) := \Big( \frac{ \not{\!p}+m}{2m} \Big)_{\alpha \beta}$$ $\Lambda^+$ has the following property   $$\Lambda_{\alpha \beta}^+ (\vec p) = \sum_{r=1}^2 u_{r \alpha} (\vec p) \bar u_{r \beta} (\vec p)$$  But what I do not understand is why does the Trace show up in $(1)$, $(2)$   Any help is appreciated.  EDIT  Let's write out the spinor indices explicitly for $(1)$  $$A_l = \Big(\sum_{s_1} u_{s_1 \color{red}{\delta}}(\vec p_2') \bar u_{s_1 \color{blue}{\alpha}}(\vec p_2')\Big) \gamma_{\color{blue}{\alpha}\color{green}{\nu}}\Big(\sum_{s_2}v_{s_2 \color{green}{\nu}}(\vec p_1') \bar v_{s_2 \color{gray}{\beta}}(\vec p_1')\Big)\gamma_{\color{gray}{\beta}\color{red}{\delta}} \ \ \ \ (3)$$  We know the following properties  $$\Lambda^+_{\color{red}{\delta}\color{blue}{\alpha}}(\vec p_2')= \sum_{s_1} u_{s_1 \color{red}{\delta}}(\vec p_2') \bar u_{s_1 \color{blue}{\alpha}}(\vec p_2')=\Big(\frac{\not{\!p_2'}+m_l}{2m_l}\Big)_{\color{red}{\delta}\color{blue}{\alpha}} $$  $$\Lambda^-_{\color{green}{\nu}\color{gray}{\beta}}(\vec p_1')= -\sum_{s_2}v_{s_2 \color{green}{\nu}}(\vec p_1') \bar v_{s_2 \color{gray}{\beta}}(\vec p_1')=-\Big(\frac{\not{\!p_1'}-m_l}{2m_l}\Big)_{\color{green}{\nu}\color{gray}{\beta}}$$  Thus we get  $$A_l = \Lambda^+_{\color{red}{\delta}\color{blue}{\alpha}}(\vec p_2')\gamma_{\color{blue}{\alpha}\color{green}{\nu}}\Lambda^-_{\color{green}{\nu}\color{gray}{\beta}}(\vec p_1')\gamma_{\color{gray}{\beta}\color{red}{\delta}}=-\operatorname{Tr}\Big[\frac{\not{\!p_2'}+m_l}{2m_l}\gamma_{\alpha}\frac{\not{\!p_1'}-m_l}{2m_l}\gamma_{\beta}\Big]$$  Now I have two questions:  1) Why are we allowed to manipulate $\sum_{s_1} \sum_{s_2} \Big[ \bar u_{s_2} (\vec p_2') \gamma_{\alpha} v_{s_1} (\vec p_1'))(\bar v_{s_1}(\vec p_1')\gamma_{\beta} u_{s_2} (\vec p_2'))\Big]_{(l)}$ in such a way to get $(3)$? What I mean is that I do not see what mathematical properties allow us to do so.  2) I get a negative sign. I guess it gets cancelled out due to an antysymmetric swap of certein indices, but what pair specifically?  Thank you :)
4463
Suppose $C$ is an $n\times n$ matrix over complex numbers, with trace $0$.  Are there always $n\times n$ matrices $A,B$ such that $AB - BA = C$?  (Inspired by a recent question which asked for a trace free proof of non-existence of solutions for $C=I$).
115672
This is a variation of the  questioned some time ago.  For a complex Hilbert space $H$ let $T: H \rightarrow H$ be a bounded operator. We call that $T$ is a trace-class operator if the following sum  $$\sum_{i}{\langle |T|e_{i}, e_{i} \rangle} &lt; \infty$$  converges, where $|T| = (T T^{*})^{\frac{1}{2}}$ is the absolute value of the operator.  Assume that  $$\sum_{i}{\langle Te_{i}, e_{i} \rangle}$$  converges for any basis in the space $H$. How to prove that if the aformentioned property holds then the operator is a trace class operator?  The progress on the problem is the following: Given an arbitrary bounded operator $T: H \rightarrow H$, one can use the following decomposition $$T = \big( \frac{T + T^{*}}{2} \big) ^{*} + i \big( \frac{T^{*} - T}{2i} \big) ^{*}$$  The latter line gives the decomposition $$T = A + i B$$ where $A, B$ are normal operators.  For the normal operators we can apply the spectral theorem that proposes that $T$ is unitary equivalent to  $$(UT U^{-1})(f(x)) = g(x) f(x)$$  where  $$U: H \rightarrow L^{2}(X, \mu)$$ $$g \in L^{\infty}(X, \mu)$$  Though this decomposition classify the operator in a broad sence, i see no direct way to conclude the statement. Are the any hints that may extend the previous argument? If not, are there any ways to conclude the statement?
84124
Suppose there're two systems $A$ and $B$, if we're in a pure state $|\psi\rangle\in\mathbb{H}_a\otimes\mathbb{H}_b$. Let $\hat A$ be an operator acting on $\mathbb{H}_a$ and $|\psi\rangle=\sum_{i,j}\psi_{i,j}|a_i\rangle|b_j\rangle$. The expectation value of A with respect to the state $|\psi\rangle$ can be calculated as $$ \langle A\rangle_{\psi}= Tr_{\mathbb{H}_a\otimes\mathbb{H}_b}(A|\psi\rangle\langle\psi|)=\sum_{i,j}\langle a_i|\langle b_j|A\otimes\mathbb{1}_B|\psi\rangle\langle\psi|a_i\rangle|b_j\rangle=... $$ I'm very confused about where these two equalities coming from? In particular, where does the term $|\psi\rangle\langle\psi|$ in the sum come from? Can I understand its function as generating a $\psi$ matrix with different constants $\psi_{i,j}$?
73301
Let $A$ be a $227\times 227$ matrix with entries in $\mathbb{Z}_{227}$ such that all of its eigen values are distinct. What would be its trace? I think it is zero by adding all 227 elements but i am not sure.  Edited: Here I have assumed that eigenvalues are in a base field.
37738
Determine the trace and determinant of the linear operator (on the space $\mathbb{F^{n\times n}}$) that sends the matrix $M\to AMB$ where $A$ and $B$ are $n\times n$ matricies
113256
I am doing my bachelors dissertation based on an article by David Deutsch. He defines the action of a quantum gate as: $$ U = \sum_{x, y \in \mathcal{Z}_{2}} |x \dot{+}y\rangle|y\rangle\langle x|\langle y|  $$ where $\dot{+}$ is the OR operator from boolean algebra. Then, according to me: He then takes the partial trace:  $$ \text{Tr}_{2}[U(|\psi\rangle\langle\psi|\otimes\hat{\rho})U^{\dagger}] $$  and it is equal to: $$ \frac{1}{2}\hat{I} + \text{Re}[(\langle 0|\psi\rangle\langle\psi|1\rangle)(|0\rangle\langle 1|+|1\rangle\langle 0|)] $$  EDIT: where does the $\frac{1}{2}\hat{I}$ comes from?