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olgerm
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what is SR transformarion for bispinor? I have heard that it is different than 4-vectors transfomation.
Yes.king vitamin said:The Wikipedia article gives a pretty nice overview of how Lorentz transformations act on bispinors: https://en.wikipedia.org/wiki/Bispinor
olgerm said:Yes.
##\chi=v/c##?
Good catch, it seems that Wikipedia is using notation from these lecture notes.olgerm said:Yes.
χ=v/c?
Yes, one simply uses the following formula, which you should attempt to prove (it is a good exercise for somebody working on this stuff):olgerm said:It is possible to write that as one (4,4) matrix. So that it does not include matrix exponentation.
Can you send me a link or explain what you meant by vector representation?king vitamin said:and is not true for the vector representation which is also four-dimensional)
olgerm said:Can you send me a link or explain what you meant by vector representation?
According to this, bispinor is a direct sum of two spinors. I thought that bispinor should be a direct product of two spinors, in which case it would be equivalent to a vector.king vitamin said:bispinors: https://en.wikipedia.org/wiki/Bispinor.
I'm also used to that last interpretation in the context of supergravity, e.g. in the susy-transfo of vielbeine.Demystifier said:According to this, bispinor is a direct sum of two spinors. I thought that bispinor should be a direct product of two spinors, in which case it would be equivalent to a vector.
A Dirac spinor is the direct sum of two different Weyl spinors, one transforming according to the (1/2,0) the other according to the (0,1/2) representation of the proper orthochronous Lorentz group. You need both spinors to also enable the space-reflection symmetry (parity conservation) of electrodynamics and QCD, i.e., the Dirac spinor refers to the representation ##(1/2,0) \oplus (0,1/2)## (which is reducible for the proper orthochronous Lorentz group but irreducible as a representation of the orthchronous Lorentz group, which additionally includes space reflections).Demystifier said:According to this, bispinor is a direct sum of two spinors. I thought that bispinor should be a direct product of two spinors, in which case it would be equivalent to a vector.
There is no difference in this case.Demystifier said:According to this, bispinor is a direct sum of two spinors. I thought that bispinor should be a direct product of two spinors, in which case it would be equivalent to a vector.
There is no difference between the direct sum and direct product of two vector spaces or any finite number of vector spaces. You need an infinite number of them to see a difference (not to be confused with the dimensions of the vector spaces, that makes no difference). Check the definitions.vanhees71 said:There's a difference between ##(1/2,0) \oplus (0,1/2)## and ##(1/2,1/2)=(1/2,0) \otimes (0,1/2)##.
Demystifier said:According to this, bispinor is a direct sum of two spinors. I thought that bispinor should be a direct product of two spinors, in which case it would be equivalent to a vector.
haushofer said:I'm also used to that last interpretation in the context of supergravity, e.g. in the susy-transfo of vielbeine.
If I have two finite-dimensional vector spaces with dimension ##d_1## and ##d_2##, then the direct sum of the two vector spaces has dimension ##d_1 + d_2##, while the direct product has dimension ##d_1 d_2##. Therefore, they cannot be equivalent in general. (This particular argument isn't so great for the example in the above post where ##d_1 = d_2 = 2##, but it clearly shows that the quoted statement is incorrect.)martinbn said:There is no difference between the direct sum and direct product of two vector spaces or any finite number of vector spaces.
I think you are just unfamiliar with the notation being used. The dimension of the ##(s,s')## irrep of the Lorentz group has dimension ##(2s+1)(2s'+1)##, so this irrep is four-dimensional. (As @vanhees71 claimed, it is the four-vector irrep.)martinbn said:Also (1/2,1/2) is very strange, to say the least, it suggests two dimensions, while the space is four dimensional!
No, the direct product has dimension ##d_1 + d_2##. Look up the definition.king vitamin said:If I have two finite-dimensional vector spaces with dimension ##d_1## and ##d_2##, then the direct sum of the two vector spaces has dimension ##d_1 + d_2##, while the direct product has dimension ##d_1 d_2##. Therefore, they cannot be equivalent in general. (This particular argument isn't so great for the example in the above post where ##d_1 = d_2 = 2##, but it clearly shows that the quoted statement is incorrect.)
I see, the notations are for the representation spaces, not vectors in them. Then he definitely means tensor product, not direct product.I think you are just unfamiliar with the notation being used. The dimension of the ##(s,s')## irrep of the Lorentz group has dimension ##(2s+1)(2s'+1)##, so this irrep is four-dimensional. (As @vanhees71 claimed, it is the four-vector irrep.)
Yes, a second case of unfortunate terminology in this thread. Certain physics textbooks use the terms direct product and tensor product interchangeably when referring to representations (which I imagine drives mathematicians nuts). I think that everyone above meant tensor product (I did), but they'd have to clarify for themselves.martinbn said:I see, the notations are for the representation spaces, not vectors in them. Then he definitely means tensor product, not direct product.
king vitamin said:I think that everyone above meant tensor product (I did), but they'd have to clarify for themselves.
vanhees71 said:A Dirac spinor is the direct sum of two different Weyl spinors, one transforming according to the (1/2,0) the other according to the (0,1/2) representation of the proper orthochronous Lorentz group. You need both spinors to also enable the space-reflection symmetry (parity conservation) of electrodynamics and QCD, i.e., the Dirac spinor refers to the representation ##(1/2,0) \oplus (0,1/2)## (which is reducible for the proper orthochronous Lorentz group but irreducible as a representation of the orthchronous Lorentz group, which additionally includes space reflections).
Right, hence why my original reply referred to the representation being block diagonal. What I meant in your quote (which I hope is clear from the context) is that I was later using the word "direct product" interchangeably with "tensor product" in my post #14. This is probably bad practice from a mathematical POV, but in my defense, many textbooks (Merzbacher, Weinberg, Zee...) do the same, so I also do so sometimes.George Jones said:The original post refers to a direct sum (or direct product) of (inequivalent) representations, not to a tensor product of representations.
It seems that here we have a semantical confusion, because what mathematicians call tensor product, physicists often call direct product. See e.g. https://www.quora.com/What-is-the-m...-the-context-of-physics-and-quantum-mechanicsmartinbn said:Then he definitely means tensor product, not direct product.
In your dictionary, what would be the difference between direct product and tensor product?George Jones said:The original post refers to a direct sum (or direct product) of (inequivalent) representations, not to a tensor product of representations.
king vitamin said:Right, hence why my original reply referred to the representation being block diagonal. What I meant in your quote (which I hope is clear from the context) is that I was later using the word "direct product" interchangeably with "tensor product" in my post #14.
king vitamin said:This is probably bad practice from a mathematical POV, but in my defense, many textbooks (Merzbacher, Weinberg, Zee...) do the same, so I also do so sometimes.
Demystifier said:In your dictionary, what would be the difference between direct product and tensor product?
Even in the category of vector spaces the direct product and direct sum of an infinite family are different. For example the direct product of ##\{V_i\}_{i=1}^\infty## is the vector space of all sequences ##(v_1,v_2,\dots)## with ##v_i\in V_i## and componentwise operations. The direct sum is the vector space of sequences with only finitely many nonzero terms.George Jones said:In other categories, direct sums and direct product can look much different.
We are talking about the representations of the Lorentz group not simply about direct sums/products of vector spaces. The Dirac-spinor representation of the covering group of the orthochronous Lorentz group (related to particles with spin 1/2) is definitely different and non-equivalent from the "fundamental representation" of ##\mathrm{SO}(1,3)## (related to particles with spin 1).martinbn said:There is no difference between the direct sum and direct product of two vector spaces or any finite number of vector spaces. You need an infinite number of them to see a difference (not to be confused with the dimensions of the vector spaces, that makes no difference). Check the definitions.
What you wrote is unclear to me. You use the tensor product notation when you talk about the direct product. Also ##(1/2,1/2)## is very strange, to say the least, it suggests two dimensions, while the space is four dimensional!
it should beking vitamin said:Yes, one simply uses the following formula, which you should attempt to prove (it is a good exercise for somebody working on this stuff)
In this instance and historically, bi-spinor refers to the 4-component spinor of Dirac [tex]\Psi_{D} = \begin{pmatrix} \psi_{\alpha} \\ \bar{\chi}^{\dot{\alpha}} \end{pmatrix} ,[/tex] or that of Majorana [tex]\Psi_{M} = \begin{pmatrix} \psi_{\alpha} \\ \bar{\psi}^{\dot{\alpha}} \end{pmatrix} , \ \ \alpha , \dot{\alpha} = 1,2 \ ,[/tex] where [itex]\psi_{\alpha} \in (\frac{1}{2} , 0)[/itex] is the 2-component (left-handed) Weyl spinor, and [itex]\bar{\chi}_{\dot{\alpha}} \in (0 , \frac{1}{2})[/itex] is the 2-component (right-handed) Weyl spinor, i.e., the two inequivalent fundamental representations of [itex]\mbox{SL}(2 , \mathbb{C}) \cong \mbox{Spin}(1,3)[/itex], the universal covering group of the Lorentz group [itex]\mbox{SO}(1,3)^{\uparrow}[/itex].Demystifier said:According to this, bispinor is a direct sum of two spinors.
If I replace your “direct product” by tensor product, and your “vector” by Lorentz vector, your statement will still be only partially correct. This is obvious because you obtain spin-0 and spin-1 from the tensor product of two spin-1/2, and geometrically, spin-0 corresponds to a Lorentz scalar, and spin-1 corresponds either to a Lorentz vector, or to an anti-symmetric Lorentz tensor.I thought that bispinor should be a direct product of two spinors, in which case it would be equivalent to a vector.
samalkhaiat said:spin-1 corresponds either to a Lorentz vector, or to an anti-symmetric Lorentz tensor.
Yes, think of the vector potential and the field tensor of Maxwell. Also, the spin of a Lorentz vector (1/2,1/2) is 1/2+1/2 = 1and the number of components is equal to the dimention of (1/2,1/2) which is [2(1/2) +1][2(1/2) +1]=4. Now, the spin of the anti-symmetric tensor (1,0) is equal to 0+1=1 and the number of complex components is equal to [2(1) + 1][2(0) + 1] = 3.PeterDonis said:Is this correct? A 4-vector has 4 components, but an antisymmetric 4-tensor has 6 independent components.
samalkhaiat said:think of the vector potential and the field tensor of Maxwell
samalkhaiat said:the number of complex components is equal to [2(1) + 1][2(0) + 1] = 3.
Why do you need them to be 4?I gave you the rule to calculate the spin and dimention of a representation. The vector representation (1/2,1/2) is real, while the tensor (1,0) is complex. For Maxwell field tensor, we associate the real representation (1,0) + (0,1).PeterDonis said:Which is 6 real components, not 4.
samalkhaiat said:Why do you need them to be 4?
samalkhaiat said:I gave you the rule to calculate the spin and dimention of a representation. The vector representation (1/2,1/2) is real, while the tensor (1,0) is complex. For Maxwell field tensor, we associate the real representation (1,0) + (0,1).
An irreducible representation [itex](j,k)[/itex] has 2, in general, unrelated numbers. For non-zero mass, these are the spin and dimension: [tex]\mbox{Spin}(j,k) = j + k ,[/tex][tex]\mbox{dim}(j,k) = (2j +1)(2k +1).[/tex] The rule of identification with objects on Minkowski space-time is as follows: The irreducible representation [itex](j,k)[/itex] corresponds to a massive spin-[itex](j + k)[/itex] Lorentz tensor with [itex]\mbox{dim}(j,k)[/itex] independent components. For example the irrep [itex](1,1)[/itex] corresponds to a (massive) spin-2, traceless symmetric Lorentz tensor: [itex]G_{ab} = G_{ba}, \ \eta^{ab}G_{ab} = 0[/itex]. Check the rule: [itex]\mbox{dim}(1,1) = 9[/itex] and this is exactly the number of independent components in a traceless symmetric tensor in 4 dimensions.PeterDonis said:I don't. I'm trying to understand how spin-1 can "correspond to an antisymmetric Lorentz tensor", as you said, when "spin-1", to me, means 4 independent components, but "antisymmetric Lorentz tensor" means 6. Basically what I understand you to be saying is that my interpretation of "spin-1" is wrong--"spin-1" is a more general term that includes multiple representations, only one of which (the Lorentz vector) has 4 independent real components.
Would (1, 0) + (0, 1) still be considered spin 1?