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| bio | website | en.wikipedia.org/wiki/… |
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| visits | member for | 11 months |
| seen | 35 mins ago | |
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A Sufi prayer of love
O love, O pure deep love, be here, be now, be all.
Dissolve me into your stainless endless radiance,
Make me your servant, your breath, your core.
:- by sufi mystic Rumi
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1d |
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Quantum mechanics and everyday nature Sorry, I read only first line :) |
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1d |
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Quantum mechanics and everyday nature I think QM is not really needed to explain their experiment. Wave nature of light too explains it well. In order for two waves to form constructive or destructive interference their polarization should either be random or (anti/)parallel. |
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2d |
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How does relativity explain gravity, without assuming gravity "They distort it because they are pulled down by .. what?" :- They are not "pulled down" by anything. Its just that energy interacts with spacetime as per Einstein's equation and makes it curved. |
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2d |
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If air contains oxygen and nitrogen, why won't these two elements form water? For chemistry stack exchange it would make a good question I think. Why we don't have oceans filled with N2O instead of water? |
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2d |
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If air contains oxygen and nitrogen, why won't these two elements form water? @anna May be what he wants to ask is why not nitrogen and oxygen make water like liquid. |
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2d |
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Why is this not a realisable operation on a quantum system? @wemblem Can you please edit your question to make clear what you are trying to ask? In particular what is an "affine map" ? As per your definition a "physical operation" should be an affine map; then why don't you simply check if the map is "affine" or not? (whatever that means) |
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May 20 |
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Criticism of String Theory by other string theorists D. Friedan (who has done quite a lot of work in string theory) has criticized string theory in this paper. |
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May 20 |
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Why is the temperature zero in the ground state? @joshphysics. If (according to your argument) a system at zero temperature is necessarily in its ground state then it means that system is isolated. But for an isolated system we should use the other definition of temperature which is in terms of rate of change of entropy with energy. |
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May 19 |
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Why is the temperature zero in the ground state? @joshphysics Will your argument work for continuous spectrum too? |
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May 12 |
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Some Dirac notation explanations $\dagger$ is transpose + complex conjugate. |
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May 9 |
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Irreducible Representations Of Lorentz Group This situation should be analyzed more carefully. But since $Q$'s (in supersymmetric extension of Poincare algebra) don't commute with general Lorentz transformations so I guess you are right. Choice of such an L(p) will not be possible in this case. But you can choose L(p) which takes |k,..> to |p,..> where |p,..> is some linear combination of eigenstates of Q and J's. |
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May 9 |
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Irreducible Representations Of Lorentz Group But even in case of Poincare algebra $J_3$ can be diagonalized in $V_p$ for those $p$'s which are related to $k$ by a boost along z direction. |
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May 9 |
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Irreducible Representations Of Lorentz Group $J$ and $P$ do not commute. But yes, if algebra $G_k$ were such that some element $Q$ of it commuted with $P$ then you could find some linear combination $|p,n>$ of $|p,\sigma>$'s such that $|p,n>$ be eigenstate of $Q$. This situation can not be realized in case of Poincare algebra. However, little group method also applies to some other algebras (namely supersymmetric extensions of Poincare algebra) where a situation similar to this holds. There we have elements in $G_k$ which commute with $P$. |
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May 9 |
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Will adding heat to a material increase or decrease entropy? added 836 characters in body |
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May 8 |
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Irreducible Representations Of Lorentz Group Physical reality of a Hilbert space doesn't depend upon the basis you choose to work with in it. Anyway, in our case for every $p$ there is a group $G_p$ which keeps $V_p$ fixed. You will find e.g. a "boosted $J_3$" in $G_p$ which will get diagonalized in $V_p$ with eigenbasis $|p,\sigma>$, if $|k,\sigma>$ was eigenbasis of $J_3$ in $V_k$. Thus $\sigma$ appearing in $|p,\sigma>$ can be thought of as spin measured in that boosted frame which is obtained from the inertial frame in which you found spin eigenvalue of $|k,\sigma>$ to be $\sigma$, by applying the boost that takes you from $k$ to $p$ |
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May 7 |
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Irreducible Representations Of Lorentz Group As I said $V_p$ has nothing in common with $V_k$ except for the dimension. And since its dimension is same as that of $V_k$ so I can choose any arbitrary basis of $V_p$ and label it with $\sigma$; But this doesn't mean that $|p,\sigma>$ is eigenvalue of (say) $J_3$. In fact $G_k$ was defined such that its elements fix momentum $k$; so in particular there is no guarantee that elements of $G_k$ will also fix some $p$ different from $k$. In general they will not. So elements of $G_k$ (in general) may not have any eigenvectors in $V_p$ for $p$ different from $k$. |
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May 7 |
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Irreducible Representations Of Lorentz Group No. In basis $|k,\sigma>$, label $\sigma $ can be taken as eigenvalue of (say) $J_3$; but for $p$ different from $k$, $\sigma$ appearing in basis $|p,\sigma>$ is not in general spin eigenvalue or anything like that. Initially action of $G_k$ is only defined on $V_k$. $V_p$'s are completely different vector spaces which have nothing in common with $V_k$ except for the dimension. To know how $G_k$ acts on $V_p$ (and hence what are eigenvectors of (say) spin in $V_p$) one should follow the definition of action of Lorentz operators that we have given later. |
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May 7 |
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Irreducible Representations Of Lorentz Group This answer is actually reverse of Weinberg's approach. Weinberg begins with a Hilbert space H that is assumed to be irreducible representations of Poincare algebra; and then finds a basis of it using little group method. Here we begin from a finite dimensional representation of little group and construct a Hilbert space representation of Poincare group using it. In particular, while in Weinberg (1) is a definition of basis element $|p,\sigma>$ in our case it defines action of $U(L_p)$. These two approaches are completely equivalent. |
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May 7 |
revised |
Irreducible Representations Of Lorentz Group added 52 characters in body |
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May 7 |
answered | Irreducible Representations Of Lorentz Group |
