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## Hot answers tagged trace

5

The Einstein field equations $$R_{\mu\nu}~-~\frac{1}{2}Rg_{\mu\nu}~=~8\pi GT_{\mu\nu}$$ for zero stress energy means that the Ricci Curvature $R_{\mu\nu}$ is proportional to the metric with $R_{\mu\nu}~=~\frac{1}{2}Rg_{\mu\nu}$. This is called an Einstein spacetime, and for a constant Ricci scalar $R~=~R_{\mu\nu}g^{\mu\nu}$ this is a spacetime of constant ...

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 doesn'...

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 ...

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

You can easily see this isn't the case by considering the special case of the stress-energy tensor equal to zero i.e. the vacuum solutions. These include the Minkowski metric, which is flat, but also the Schwarzschild and Kerr metrics and of course gravitational waves.

4

This follows from the cyclicity of the trace, i.e. the property that $$\mathrm{Tr}(AB)=\mathrm{Tr}(BA),$$ which extends to the cyclic permutation $\mathrm{Tr}(ABC\cdots XYZ)=\mathrm{Tr}(BC\cdots XYZA)$ for larger products. Thus, if you expand $f$ in its Taylor series, you get \begin{align} \mathrm{Tr}\left(f(G^\dagger G)\right) & = \mathrm{Tr}\left(\sum_{...

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

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

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: $$tr_A(L_{AB}) =\sum_i \langle i|_A L_{AB} |i\rangle_A=\langle0|0\rangle\langle 0|0\rangle (|1\rangle\langle0|)_B+\langle1|0\... 3 Linear algebra result states that if A is an m\times n matrix and B is an n\times m matrix, then \text{Tr}(AB)=\text{Tr}(BA). The proof is elementary, using the definition of product of matrices. So in the case of vectors, let A=|\psi⟩ and B=|\chi⟩. Note that A,B need not be square. It therefore follows that \text{Tr}(|\psi⟩⟨\chi|)=\text{Tr}(⟨\... 3 An A_{ab}B_{bc} yields a C_{ac}. Contracting all indices, but the outer ones of your expression yields a [(\not{p}_2-m)\gamma^\mu(\not{p}_1+m)\gamma^\nu]_{dd}. Now executing the dd contraction is just the trace. 2 The problem with the reasoning above is that it considers only the effect the measurement has on the covariance matrix and does not consider the displacement vector. While the covariance matrix of the resulting state does not depend on the particular measurement result, this is not true for the displacement vector. As a result, when doing the partial trace, ... 2 A basis for the Hilbert space A \otimes B could be written :$$|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 (...

2

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, |\Psi\rangle_{... 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 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 ,$$and similarly for H_B. 2 Start noticing that {(\gamma^{\alpha})}^2 =1\cdot g^{\alpha \alpha} and that$$ \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 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 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 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 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.$$2 You go from$$ \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 \beta})\big]=-\text{Tr}\...

2

In the rest frame $\not p \rightarrow - m \gamma^0$ and $$(\not p - m)u(\mathbf{p},s) = 0 \;\; \rightarrow \;\; (I + \gamma^0) u(\mathbf{0},s) = 0 \\ {\bar u}(\mathbf{p},s)(\not p - m) = 0 \;\; \rightarrow \;\; {\bar u}(\mathbf{0},s) (I + \gamma^0) = 0 \\ (\not p + m)v(\mathbf{p},s) = 0 \;\; \rightarrow \;\; (I - \gamma^0) v(\mathbf{0},s) = 0 \\ {\bar v}(\... 2 You are halfway correct. The trace of a combination of \gamma-matrices does not depend on the representation in which they are expressed. Sakurai primarily uses the Dirac-Pauli representation, while Peskin and Schroeder use the Weyl chiral representation. This difference in representation should no affect the traces of matrix combinations; the traces ... 2 G^\dagger G and G G^\dagger are unitary equivalent in the finite dimensional case : There exists the polar decomposition G = R U where R \geq 0 and U unitary. Then G^\dagger G = U^\dagger R^2 U = U^\dagger (G G^\dagger) U. But from this it follows that G^\dagger G and G G^\dagger have the same eigenvalues. In the infinite dimensional case ... 1 You can represent \rho_{AB} by tr_B(\rho_{AB})\otimes tr_A(\rho_{AB}) only if the two subsystems are not entangled, i.e. \rho_{AB}=\rho_A \otimes \rho_B. 1 You can compute the trace of an endomorphism using any basis (including non-orthogonal ones). In Dirac notation, you show this by inserting the identity expressed in the new basis and re-arranging:$$\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 C}...

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 \$B=\text{diag}(\mu_1,\...

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