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Both $$\sum_i |i\rangle \langle i | $$ and $$\sum_j |j\rangle \langle j | $$ are summations over basis vectors. The indices $i,j$ run over the same values – values of indices that identify the basis vectors in the same basis (set of vectors) – but the particular values of the indices $i,j$ are independent. Can you calculate how much is the expression below?...


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It is generally a good idea to keep units out of your expressions for as long as possible. If you do that you are unlikely to run into trouble. Still, if you do have issues, you can just come up with you own typographic conventions if the notes are just for yourself. For instance, you could always keep your units to the far right and put curly backets or ...


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It's an axiom which leads to a formalism that has great agreement with experiment. Asking why an axiom is what it is isn't really useful.


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This combination represents the following. $f$ in your reference is the number of particles per cubic space volume per cubic velocity volume. When you multiply it by some volume in velocity space, you obtain the regular density — number of particles per volume. But, as $f$ itself depends on velocity, to obtain the total density of all particles with all ...


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In the 'strict' sense, you should only apply the summation convention to a pair of indices if one is raised and another is lowered. For example, consider a vector $v$ and a dual vector $f$ (i.e. a map from vectors to numbers). Then one can compute $f(v)$, the number that results from $f$ acting on $v$. In components, this would be written as $f_i v^i$, ...


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There is nothing wrong by summing up indices when both indices are either up or down. It is just a matter of convention. However the meanings can be different if you are in a Relativistic theory. When you sum one up and one down indices in Relativity it means you have a Lorentz invariant quantity because you are combining covariant and contravariant ...


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I believe it should be: $$h^\prime_{\mu\nu} = h_{\mu\nu} + 2\epsilon\,\partial_{(\mu}\xi_{\nu)}$$ where the parentheses denote the symmetric part of the tensor in the $\mu$ and $\nu$ indices: $$h^\prime_{\mu\nu} = h_{\mu\nu} + \epsilon\,\left(\partial_{\mu}\xi_{\nu} + \partial_{\nu}\xi_{\mu} \right)$$ It is a generalization of the usual definition: $$ ...


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Notation-wise, you can use either of \begin{align} A\rho & = \sum_{ijk} p_k \Big(|i \rangle \langle j | \otimes |i \rangle \langle j |\Big) \Big(| k \rangle \langle k | \otimes |k\rangle \langle k |\Big) \\ & = \sum_{ijk} p_k |i \rangle \langle j |k \rangle \langle k | \otimes |i \rangle \langle j |k \rangle \langle k | . \end{align} As you well ...


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Normally the notation $(n_b|n_f)$ denotes the dimension of a super vector space of Grassmann-even dimension $n_b$ and Grassmann-odd dimension $n_f$. When writing a super vector as a column vector, it is standard to order the Grassmann-even sector before the Grassmann-odd sector. However, the authors introduce a non-standard ordering $(n_{b_1}|n_f|n_{b_2})$ ...


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Physicists do you the rules of calculus but can be sloppy in their notation. The symbols $\delta, \Delta$ and $d$ often seem to be used interchangeably to mean a (small or infinitesimal) change in something or better still a final value minus an initial value. So in your equation $dU$ is the change in internal energy of a system or final internal energy ...


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Bra-ket notation is just a useful short hand for some well-defined objects in functional analysis (or linear algebra if you work in finite dimensions). To understand what is allowed and what isn't, you would better know what those concepts are, so let's quickly recap: A ket $|\psi\rangle$ is just a vector in some Hilbert space $\mathcal{H}$. A bra $\...


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Others have pointed out that time is not an operator in the Schrodinger equation (I'll link to posts when I find them), but this isn't the end of the story. For example, how should one denote the monentum operator acting on a bra? $\partial_x \psi(x,t)$ could be denoted as $\langle \psi | \hat p^\dagger$ (preferred) but I've also seen variants like $\...



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