The h Bar

12:23 AM

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Q: How should we handle "multidisciplinary" questions with subparts that are appropriate for different SE networks?

This question was closed as "not suitable for" Physics SE. Based on the comments, the voters seemed to think that it belongs to Biology SE instead. I don't understand this reasoning. Answering this question requires answering two sub-questions: What would be the flux of radiation at the Earth's ...

Silly Goose

2:48 AM

I am struggling to understand why the tensor product of two states is represented, say in position space, as the usual product of the two state’s wave functions

Is this definitional?

Also, what is the meaning in QM of the tensor product of two operators? How is this represented in matrix representations of the operator

Prabhat

7:22 AM

@JohnRennie yeah, I just figure it depends on how much I like my PhD. Anyways, if I am not going for a graduate degree now, I am going to a research job for the next few years. So, scientist or no scientist, my personality, I guess, suits those analytical jobs (also mentioned by my recruiter).

Mr. Feynman

@SillyGoose that's also valid considering countable basis.

It comes from the definition of tensor product

bolbteppa

Mr. Feynman

Good spinor morning

bolbteppa

@DIRAC1930 ^ is pretty clear, at least those $\eta_A = 0$ are arising from it with little effort

You can motivate this approach with creation operators directly from Lie theory and classifying Lie algebras, and the Clifford stuff, and why you'd write a spinor this way, so it's not bad, note isotropy is nowhere until the very last line which is interesting

Mr. Feynman

@SillyGoose The tensor product of $A:\mathcal{H}_1\rightarrow{H}_1$ and $B:\mathcal{H}_2\rightarrow{H}_2$ is unique operator $A\otimes B:\mathcal{H}_1\otimes\mathcal{H}_2\rightarrow{H}_1\otimes\mathcal{H}_2$ such that $(A\otimes B)\lvert\Psi_1\rangle\otimes\lvert\Psi_2\rangle=A\lvert\Psi_1\rangle\otimes B\lvert\Psi_2\rangle$

nickbros123

8:15 AM

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Q: Doubt regarding cancellation of internal torque for a rigid body (Kleppner and Kolenkov)

In chapter "Dynamics of fixed axis rotation", the book says newton's laws are not enough to show that the internal torques of a rigid body cancel out. But section 1.2 in Goldstein says that (using strong law of action and reaction), that the internal torques cancel because they form the exact sa...

newtonian-mechanics rotational-dynamics rigid-body-dynamics

DIRAC1930

@bolbteppa Hmm I'm not too convinced by this. It seems like we just have another rep of the Clifford algebra just written in terms of creation/annihilation operators. The $|0>$ and $|1>$ have always implicitly been there as the basis vectors of the spinor i.e. $\xi = \xi^i \mathbf{e}_i$.

bolbteppa

8:45 AM

If you do it for $D = 5$ you have $|\xi> = \xi_0 |0> + \xi_1 |1> + \xi_2 |2> + \xi_{12} |12>$, have it working up to some signs as of now

It's a general way to derive these $\eta_A = 0$ equations, but it means you have to assume the existence of this $|\xi>$ vector with all those crazy components (which is the whole issue, we're trying to arrive at it from $F = 0$, you can get $|\xi>$ from Lie theory and $X = x^a \gamma_a$ from Dirac's trick and combine the approaches, but still)

bolbteppa

9:12 AM

@DIRAC1930 What about this:

(Where $x_5 = x^5 = x^0$, $x^{1+1} = x^{1'}$ and $x^{2+1} = x^{2'}$ in the old notation)

This looks way better than what he's doing

bolbteppa

9:35 AM

There is a big link to 'lines on a cone' i.e. linear spaces on a quadric

Quadric (algebraic geometry)

In mathematics, a quadric or quadric hypersurface is the subspace of N-dimensional space defined by a polynomial equation of degree 2 over a field. Quadrics are fundamental examples in algebraic geometry. The theory is simplified by working in projective space rather than affine space. An example is the quadric surface x y = z w {\displaystyle xy=zw} in projective space P 3 ...

This is the link, that even talks about spinors...

PM 2Ring

10:36 AM

@bolbteppa Ted Shiffrin in the Math chat is rather fond of the hyperboloid of one sheet, see his avatar. We briefly discussed it the other day. chat.stackexchange.com/transcript/message/62839880#62839880 And I posted this interactive version:

sagecell.sagemath.org/…

ACuriousMind

11:50 AM

@SillyGoose The meaning of $A\otimes B$ is simply that this is the operator that acts on the "left" part of a state as $A$ and on the "right" part of a state as $B$. In matrix form, the tensor product is the Kronecker product

@SillyGoose You can show that functions of the form $f(x)g(y)$ are a basis for $L^2(\mathbb{R}^2)$ (see e.g. this math.SE question). And functions of the form $f(x)\otimes g(y)$ are a basis for the tensor product $L^2(\mathbb{R})\otimes L^2(\mathbb{R})$. Therefore, the map $f(x)\otimes g(y)\mapsto f(x)g(y)$ is an isomorphism between $L^2(\mathbb{R})\otimes L^2(\mathbb{R})$ and $L^2(\mathbb{R}^2)$

Relativisticcucumber

3:47 PM

when i log onto stack exchange and check the notifications, there is a reply to a message sometimes. when i click it, it goes to chat history. from here, i can join the chat room, but i need to scroll up all the way to access and reply to the message in discussion. is there a way around this?

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