# Tag Info

3

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 ...

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

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 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 ...

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 $$A_{xy}(\psi,\rho)\vert x\rangle\langle ... 1 OP's algebraic manipulations are formally correct. But the formal calculation could be wrong for many reasons. For instance: If the summation \sum_n in \Theta is not convergent. If one is not allowed to change order of summation \sum_n and integration \int_{0}^{\infty}\!dt. If the Sokhotsky distribution formula is applied to a singular ... 1 Qmechanic's answer is correct, but you can say more now that you provided more detail in the comment. The function \theta(t) is supposed to be the quantum partition function at imaginary time t, considered as a function of time, and zero for so that it decays according to the gap between the first excited state and the ground state (assuming you add a ... 1 You can compute the trace of an endomorphism using any basis (including non-orthogonal ones). In Dirac notation, you show this by inserting the identity expressed in the new basis and re-arranging:$$\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

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 $$\frac{1}{\gamma^\mu k_\mu - m} = \frac{\gamma^\mu k_\mu + m}{k^2 - m^2}$$ ...

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