# 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 ...

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 ...

3

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.

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

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

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, ...

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 \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 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 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 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

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 $H_B$.

1

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: $$tr_A(L_{AB}) =\sum_i \langle i|_A L_{AB} |i\rangle_A=\langle0|0\rangle\langle 0|0\rangle ... 1 I believe that an example will help clarify your confusion about notation (as examples usually do). Consider a system of two qubits, A and B, with Hilbert spaces V_A and V_B spanned by two orthonormal eigenbasis of \sigma_z, |0\rangle_A and |1\rangle_A; and |0\rangle_B and |1\rangle_B. Now suppose that we have a Bell state, ... 1 A basis for the Hilbert space A \otimes B could be written :$$|e_{i_1 i_2}\rangle = |e_{i_1}\rangle \otimes |e_{i_2}\rangle \tag{1}$$We may user the notation I representing a composite index :  I = (i_1 i_2), so that |e_{I}\rangle = |e_{i_1}\rangle \otimes |e_{i_2}\rangle  The density matrix could be written :$$\rho_{I'I} = \rho_{ \large ...

1

The source you linked to looks for the divergent part of the integral in a high energy limit ($m \to 0$). If you want to compute the whole thing, here's how you would do it. Focus on just one of the two traces: 1) "Rationalize" all the propagators by doing something like $$\frac{1}{\gamma^\mu k_\mu - m} = \frac{\gamma^\mu k_\mu + m}{k^2 - m^2}$$ ...

1

It's useful to know how matrix multiplication is defined: For $n \times n$ matrices, $A$ and $B$, denote the entry in the $i$th row and $j$th column by $A^i_j$ and $B^i_j$ respectively. Then for $C = AB$, the entries are given by $$C^i_j = A^i_kB^k_j$$ (summation convention of course), which you can check by working out a few examples. Now when we have ...

1

OP's algebraic manipulations are formally correct. But the formal calculation could be wrong for many reasons. For instance: If the summation $\sum_n$ in $\Theta$ is not convergent. If one is not allowed to change order of summation $\sum_n$ and integration $\int_{0}^{\infty}\!dt$. If the Sokhotsky distribution formula is applied to a singular ...

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