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5

Both $L$ and $S$ are quantum angular momenta. Angular momentum is different in quantum mechanics (QM). Almost everything is different in QM. In QM, angular momentum is complete if you give two numbers: the "angular momentum" and its "third component" $M$. So you give the pairs $|l,\ m_l\rangle$ and $|s, \ m_s\rangle$ So, any angular momentum is ...


4

Normally the index at an element tells me "how many atoms of that element there are", but in this case we are discussing values of $x = 0.04$ etc. so this does not make sense. That is still what it means. You just need to beat it with a rubber hammer until it starts making sense. In this particular instance, for example, it means that four atoms out of ...


4

Are $-|0\rangle$, $i|0\rangle$, $\frac{1+i}{\sqrt2}|0\rangle$ valid (that is, physically possible) kets? How are they different from $|0\rangle$? Yes, they are valid, but describe the same physical state. A state can be characterized by its expectation values, and a phase factor $e^{i\varphi}$ cancels with the complex-conjugate factor from the bra. ...


3

No. The right side of your equation is meaningless. You have to contract indices two at a time, not four at a time. Otherwise you don’t get a Lorentz-invariant result.


3

As a third, but equivalent, perspective to the existing answers: When in doubt, you can always put the summation back explicitly. (Just take it back out once you understand the answer and before getting evil looks from your peers for not having used Einstein summation!) In your case, you would have $$ \sum_a \sum_b g^{ab} g_{ab} = ? = \sum_a \sum_a g^{aa} ...


2

As @G.Smith pointed out in his answer, $g^{aa}g_{aa}$ doesn't make sense. You can calculate $g^{ab}g_{ab}$ as a special case of raising and lowering indices. Contracting the $b$ indices first you get $$g^{ab}g_{ab}=\delta^a_a=...$$ I leave out the final step as an exercise for you.


2

If 100% of the medium is composed of the two elements $A$ and $B$, than a easy ways to write the percentages $p_A$ and $$x=p_B=1-p_A$$ is $A_{p_A}B_{p_B}$. Thus, in your case it becomes $A_{1-x}B_x$.


2

Since $a$ is a vector, its entries $a_\mu$ are just numbers and they commute with everything $$ {a\!\!\!/}{a\!\!\!/} = a_\mu\gamma^\mu a_\nu\gamma^\nu = \gamma^\mu \gamma^\nu a_\mu a_\nu,$$ using $\gamma^\mu\gamma^\nu + \gamma^\nu\gamma^\mu = 2 \eta^{\mu\nu}I_4 $ to get $\gamma^\mu\gamma^\nu = 2 \eta^{\mu\nu}I_4 - \gamma^\nu\gamma^\mu $, the expression ...


1

Yes. However, the last term is a bit 'icky' because it looks like $$ \frac{dx_\nu}{d\tau} = \frac{d}{d\tau} x_\nu $$ while what is actually meant is $$ \frac{dx_\nu}{d\tau} = \left(\frac{dx}{d\tau}\right)_\nu $$ It doesn't really make sense to 'lower' the index of a coordinate of the base manifold. However, while I wouldn't recommend it, I think I've seen ...


1

"Space-dependent phasor" makes some sense. Electrical engineers will often just say that's the "complex amplitude of the time-harmonic field" or "time-harmonic phasor".


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