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February 16

Quadratic roots

Is is possible to have two equal complex roots of a quadratic equation, where a,b,c are real numbers is a quadratic equation? — Preceding unsigned comment added by 117.227.197.26 (talk) 13:06, 16 February 2013 (UTC)[reply]

No - only complex conjugate, if they are complex. Using the standard form of the quadratic equation:

ax^{2}+bx+c=0

where as usual a, b, c are real numbers, think about the quadratic formula solution:

x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}

so for negative discriminant,

b^{2}-4ac<0

the roots are complex conjugate. M∧Ŝc²ħε И_τlk 13:25, 16 February 2013 (UTC)[reply]

Altenative method - if the roots are equal and both equal to k then

ax^{2}+bx+c=a(x-k)^{2}

\Rightarrow ax^{2}+bx+c=ax^{2}-2akx+ak^{2}

\Rightarrow b=-2ak

\Rightarrow k=-{\frac {b}{2a}}

and so k is real since a and b are real. Gandalf61 (talk) 14:10, 16 February 2013 (UTC)[reply]

Notation for representation theory of the Lorentz group

The notations D(Λ) = "representation of the Lorentz group" and (m, n) = "finite dimensional irreducible representations" seem clear enough, however the notation (say) "D^{(1/2, 0)}" (i.e. including the superscript) is confusing...

My burning questions are...

is D^(2m) = (m, 0) for half-integer m, or equivalently D^(m) = (m/2, 0) for integer m? Or...
is D^(m) = (m, 0) for integer or half-integer m?

Just alternative notations/conventions?...

In this paper by Gábor Zsolt Tóth (see appendix A4); letting m and n be integers:

D^(m) = (m/2, 0) ⊕ (0, m/2),

D^(m) = (m/2, m/2),

D^{(m, n)} = (m/2, n/2) ⊕ (n/2, m/2),

what does the last expression have for equivalent notation:

D^{(m, n)} = D^(?) ⊕ D^(?) ?

while...

The WP article takes m, n to be half-integers in (m, n), so does that translate to

"D^{(2m, 2n)} = (m, n) ⊕ (n, m)" ?

and this has what notation:

"D^{(m, n)} = D^(?) ⊕ D^(?) ?

In all... is the statement

D^{(m, n)} = (m/2, n/2) ⊕ (n/2, m/2) = "(2m + 1)(2n + 1)-dimensional irreducible representations of D(Λ)"

true?

Are any of the differing conventions, where to put the 1/2 factor, or just the choice of what is integer and half-integer, is "the" standard?

Thanks in advance for any and all replies. Best, M∧Ŝc²ħε И_τlk 13:23, 16 February 2013 (UTC)[reply]

Don't get lost in the notation. The finite dimensional real irreducible representations of the spin group of the Lorentz group all have one of the following forms (for n and m integers):

(n/2,m/2)\oplus (m/2,n/2),(n/2,n/2).

These are irreducible as real representations, which means you cannot decompose them further without breaking invariance under complex conjugation. However those of the first kind do decompose as a sum of two complex conjugate represnetations. The source you cite calls the first kind of representation

D^{(n,m)}

and the second kind

{\widetilde {D}}^{(n)}

. The fact that these are irreducible implies that there is no decomposition of

D^{(n,m)}

into a direct sum of two real representations D's. Sławomir Biały (talk) 14:40, 16 February 2013 (UTC)[reply]

It might be added that the above exhaustive list applies to O⁺(1;3) which includes parity inversions (the orthochronous Lorentz group). In physics SO⁺(1;3) (the proper orthochronous Lorentz group) is often of interest because parity is not a symmetry in every theory (see e.g. weak interactions). For SO⁺(1;3) the irreducible representations are of the form

(m/2,n/2)

for n and m integers.YohanN7 (talk) 16:18, 16 February 2013 (UTC)[reply]

Excellent explanations of the facts - if that's all there is to it (the notation and it's meaning), that's fine. Thanks (again)! M∧Ŝc²ħε И_τlk 23:43, 16 February 2013 (UTC)[reply]