# Tag Info

4

In terms of your ordinary matrix multiplication, you have, for the case of a 4x4 matrix $M = g_{ab}$: $M\cdot M^{-1} = I$, which is the same thing as $g_{ab} g^{bc} = \delta_{a}{}^{c}$ and $Tr\left(M\cdot M^{-1}\right) = 4$, which is the same thing as $g_{ab}g^{ab} = \delta_{a}{}^{a} = 4$

4

I quite like your characterization of the partial trace! I think you perceive a conflict with the Wikipedia definition because you are only taking part of the latter: given an operator $T\in L(V\otimes W)$, the requirement that its partial trace obey $$\text{Tr}_W(T)\in L(V)$$ simply says that the partial trace over $W$ be an operator on $V$, but that ...

4

The trace defined as you did in the initial equation in your question is well defined, i.e. independent from the basis when the basis is orthonormal. Otherwise that formula gives rise to a number which depends on the basis (if non-orthonormal) and does not has much interest in physics. If you want to use non-orthonormal bases, you should adopt a different ...

4

Every Hermitian traceless matrix $H$ is in $\mathfrak{su}(N)$ since $\mathrm{Tr}(H) = 0$ and so $$\exp(\mathrm{tr}(\mathrm{i}H)) = \det(\exp(\mathrm{i}H)) = 1$$ so $\exp(\mathrm{i}H)$ is unitary with determinant $1$, hence in $\mathrm{SU}(N)$. The gauge field is always in the Lie algebra of the gauge group since it is introduced to cancel terms that are ...

3

Consider a one-parameter family of operators $X + \epsilon Y$, and let $f$ be an analytic function. Then we formally use linearity of the trace to obtain \begin{align} \mathrm{tr}[f(X + \epsilon Y)] = \mathrm{tr}\left[\sum_{n=0}^\infty c_n(X+\epsilon Y)^n\right] = \sum_{n=0}^\infty c_n\mathrm{tr}[(X+\epsilon Y)^n] \end{align} But notice that \begin{align} ...

3

The mistake you made is this: $\eta^{\mu}_{\nu} \neq \eta_{\mu\nu}$. When you raise index $\mu$ from downstairs to upstairs, the matrix elements change. $\eta^{0}_{0} = 1$, $\eta_{00} = -1$. That is why if you take the trace of $\eta_{\mu\nu}$, you get 2, but if you take the trace of $\eta^{\mu}_{\nu}$ you get 4.

3

Linear algebra result states that if $A$ is an $m\times n$ matrix and B is an $n\times m$ matrix, then $\text{Tr}(AB)=\text{Tr}(BA)$. The proof is elementary, using the definition of product of matrices. So in the case of vectors, let $A=|\psi⟩$ and $B=|\chi⟩$. Note that $A,B$ need not be square. It therefore follows that ...

3

Good luck. To check the cancellation for particular groups like $E_8\times E_8$ and $SO(32)$, you will indeed have to get through similar group-theoretical tasks. Similar trace formula for the traces of $E_8$ transformations are especially yummy, including the factor of $1/30$. The orthogonal case is easier even if one is not an intimate friend of all ...

2

The oscillatory part is nothing but Thomas-Fermi approximation or more riguresly, this is a version (someone should correct me if I am wrong) Weyl's formula Regrading on how to obtain the WKB from the trace formula: You can read the 2 papers by Berry and Tabor on how they derived a trace formula (like that of Gutzwiller) but to the case of integrable ...

2

You should write the indices on the gamma matrices. So your expression is actually \begin{align} \text{tr}[\gamma^\mu(\gamma^\alpha k_\alpha + \gamma^\beta p_\beta + \gamma^\delta q_\delta + m) \gamma_\mu(\gamma^\rho k_\rho + \gamma^\sigma p_\sigma + m) ]. \end{align} Then you use the trace technology to evaluate the traces. For example, the $m^2$ term has ...

2

The problem with the reasoning above is that it considers only the effect the measurement has on the covariance matrix and does not consider the displacement vector. While the covariance matrix of the resulting state does not depend on the particular measurement result, this is not true for the displacement vector. As a result, when doing the partial trace, ...

2

You go from $$\text{Tr}\big[\gamma^{\alpha} \gamma^0 (-\gamma^0 \gamma^{\beta} + 2g^{0 \beta})\big]$$ to $$-\text{Tr}\big[\gamma^{\alpha} \gamma^{\beta}\big] + 2 \text{Tr}\big[\gamma^{\alpha} \gamma^{\beta}\big]$$ but this is wrong. The correct result is $$\text{Tr}\big[\gamma^{\alpha} \gamma^0 (-\gamma^0 \gamma^{\beta} + 2g^{0 ... 2 The trick is the following. Write$$\sum_{a',b,b'} \langle a'|b'\rangle \langle b'|X|b''\rangle\langle b''|a'\rangle = \sum_{b,b'} \langle b'|X|b''\rangle \sum_{a'} \langle b''|a'\rangle\langle a'|b'\rangle .$$But$$\sum_{a'} |a'\rangle \langle a'|$$is the identity operator, so the latter sum evaluates to$$\langle b''|b'\rangle.2 f(\hat{X}) usually "means" \sum a_n \hat{X}^n. So, big hint: Let y be your infinitesimal and let \hat{Y} be some operator. \begin{align*} &\mathrm{Tr}f(\hat{X}+y\hat{Y})-\mathrm{Tr}f(\hat{X})=\\ &\int \langle q| \sum a_n (\hat{X}+y \hat{Y})^n|q\rangle \mathrm{d}q-\int \langle q| \sum a_n \hat{X}|q\rangle \mathrm{d}q \end{align*} expand ... 2 In the rest frame \not p \rightarrow - m \gamma^0 and (\not p - m)u(\mathbf{p},s) = 0 \;\; \rightarrow \;\; (I + \gamma^0) u(\mathbf{0},s) = 0 \\ {\bar u}(\mathbf{p},s)(\not p - m) = 0 \;\; \rightarrow \;\; {\bar u}(\mathbf{0},s) (I + \gamma^0) = 0 \\ (\not p + m)v(\mathbf{p},s) = 0 \;\; \rightarrow \;\; (I - \gamma^0) v(\mathbf{0},s) = 0 \\ {\bar ...

2

Substitute two values for $\mu$ and $\nu$. If $\mu=\nu$, then using $(\gamma^\mu)^2 =\pm1$ and $tr(\gamma ^5) =0$ you have finished. If $\mu \neq\nu$, then use $tr(\gamma ^\mu\gamma ^\nu) =0$ with the two remaining $\gamma$ .

2

Start noticing that ${(\gamma^{\alpha})}^2 =1\cdot g^{\alpha \alpha}$ and that $$\textrm{tr} (\gamma^{\mu}\gamma^{\nu}\gamma^5)= \textrm{tr}\left(\frac{1}{g^{\alpha \alpha}}{(\gamma^{\alpha})}^2\gamma^{\mu}\gamma^{\nu}\gamma^5\right)=\frac{1}{g^{\alpha \alpha}}\textrm{tr} (\gamma^{\alpha}\gamma^{\alpha}\gamma^{\mu}\gamma^{\nu}\gamma^5).$$ Now choose ...

2

Let $H_A \otimes H_B$ be your Hilbert space, and $O$ be an operator acting on this composite space. Then $O$ can be written has $$O = \sum_{i,j} c_{ij} M_i \otimes N_j$$ where the $M_i$'s and $N_j$'s act on $H_A$ and $H_B$ respectively. Then the partial trace over $H_A$ defined as $$tr_{H_A}(O) = \sum_{i,j} c_{ij} tr(M_i) N_j ,$$ and similarly for $H_B$.

2

To take the partial trace you need to build the sum over the matrix elements w.r.t. the same input and output basis, as you probably already used to calculate the partial traces you gave. In Dirac notation this is often written as: $$tr_A(L_{AB}) =\sum_i \langle i|_A L_{AB} |i\rangle_A=\langle0|0\rangle\langle 0|0\rangle ... 1 \newcommand{\tr}{\mathrm{tr}}For \mathrm{SO}(N), the adjoint representation can be seen as the antisymmetric part of the tensor product of the fundamental representation with itself. In general, we have the following property for such an antisymmetric representation: Given a representation (V,\rho), the representation V\otimes V decomposes as ... 1 You can represent \rho_{AB} by tr_B(\rho_{AB})\otimes tr_A(\rho_{AB}) only if the two subsystems are not entangled, i.e. \rho_{AB}=\rho_A \otimes \rho_B. 1 this paper might help.^1 It's written pedagogically and hence is easier to read. It goes on to discuss a lot more than just color decomposition too. ^1 Scattering Amplitudes, Henriette Elvang, Yu-tin Huang. 1 The first question asks why in the computation of the divergent part of MS 51.49, we can set k_i=0. To see this, we do an intuitive counting of the power of l in the integrals. Notice each fermion propagator effectively contributes \frac{1}{l^4} and the integral measure d^4l\sim l^3dl, so it is \frac{1}{l} divergent at large l. Thus if one ... 1 You can compute the trace of an endomorphism using any basis (including non-orthogonal ones). In Dirac notation, you show this by inserting the identity expressed in the new basis and re-arranging:$$\begin{align*} \sum\limits_{|s\rangle \in B} \langle s^*| \rho |s\rangle &= \sum\limits_{|s\rangle \in B} \langle s^*| \left( \sum\limits_{|t\rangle \in ...

1

Hint: Start by representing $\psi$ and $\rho$ in the basis $\{ \vert x\rangle,\vert y\rangle\}$. Shouldn't be too difficult to calculate the action of $U$ once you've done that. If you don't know how to take the partial trace, post the intermediate result and ask back. These steps will produce an equation like  A_{xy}(\psi,\rho)\vert x\rangle\langle ...

1

I would consider using water with a dye like a deep blue to be nearly opaque. Illuminate with an array of LEDs. Each bright spot reflected from the droplet is an LED. You can add a few strategic red LEDs among all white as reference. Working out the most convenient geometry will tell if you need the LEDs on some curved surface in space or if a flat panel ...

1

There may be some problems with Teflon particles. Prior to droplet separation from the syringe, the Teflon particle can roll down along the surface of the droplet, so you'll only have your tracer at the bottom of the droplet. This movement of the particle along the surface will not take place when the droplet is in free fall. This movement will be less ...

1

This is classic raytracing problem with well known pitfalls. If you trace from the camera to the scene (backward raytracing), you can get most of the things right. Having a physical camera with real lenses is not a problem, just consider the lens system as part of the scene. For a simple pinhole sensor, this works fairy well, but requires special treatment ...

1

Here we want to craft a counterexample. Let $A$ be a $2\times2$ diagonal matrix $\text{diag}(\lambda_1,\lambda_2)$. Then its exponential is $e^A = \text{diag}(e^{\lambda_1},e^{\lambda_2})$, whereas its trace is $e^{\lambda_1}+e^{\lambda_2}$. If $B$ is another diagonal matrix then it commutes with $A$ and therefore $e^Ae^B = e^{A+B}$. If ...

1

I believe that an example will help clarify your confusion about notation (as examples usually do). Consider a system of two qubits, $A$ and $B$, with Hilbert spaces $V_A$ and $V_B$ spanned by two orthonormal eigenbasis of $\sigma_z$, $|0\rangle_A$ and $|1\rangle_A$; and $|0\rangle_B$ and $|1\rangle_B$. Now suppose that we have a Bell state, ...

Only top voted, non community-wiki answers of a minimum length are eligible