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Suppose we have a general unitary $2\times 2$ matrix. The condition $\hat{U} \hat{U}^\dagger = I$ imposes four constraints; therefore, we can express it in terms of four real parameters. Equivalently, we may consider four linearly independent $2\times 2$ matrices which represent the generators of the transformation $$\hat{U} = \exp (i \alpha_i \hat{G}_i).$$ One of the generators can be identified as $$\hat{U} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}e^{i \phi}$$ The remaining three unitary matrices have the property $\det U =1.$ The three matrices representing the Hermitian generators of the $SU(2)$ group are linearly independent from the identity and are therefore traceless. One suitable choice of three Hermitian traceless generators are the Pauli spin-matrices.

My question is:

My textbook [Modern Particle Physics by Mark Thomson page 212] states that

"The remaining three unitary matrices form a special unitary $SU(2)$ group with the property $\det U =1.$"

But how can these form a group if they do not include the identity? I am assuming the group operation is matrix multiplication.

Jbag1212
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    Which textbook? Which page? – Qmechanic Apr 01 '21 at 21:13
  • @Qmechanic Sorry, I should have included that. Modern Particle Physics by Mark Thomson page 212. – Jbag1212 Apr 01 '21 at 21:15
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    The algebra does not include the identity matrix. The group definitely does. The algebra contains its own identity, which is the zero matrix. – AccidentalFourierTransform Apr 01 '21 at 21:26
  • @AccidentalFourierTransform So is the author just being imprecise when he says "The remaining three unitary matrices form a ... group" ? – Jbag1212 Apr 01 '21 at 21:30
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    I don't know what it means for matrices to "form" a group. This is non-standard terminology. The pauli matrices are generators, they are not required to be unitary (although by sheer coincidence they are). Generators are hermitian. The span an algebra. They generate a group. – AccidentalFourierTransform Apr 01 '21 at 21:39
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    Yes, he is being sloppy. He means to say "generate" instead of "form". The three hermitian traceless matrices in the exponent form a Lie algebra, closed under commutation, and it is the exponentials of these algebra elements, and the products of the exponentials, which form the group. Obviously each exponential contains the identity, for small angles α. – Cosmas Zachos Apr 01 '21 at 21:45

1 Answers1

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You have the group of unitary matrices $\mathrm U(2)$, and you have the generators of that group, which constitute an algebra. It's important not to get these things confused.

$\mathrm U(2)$ consists of all $2\times 2$ complex matrices $ U$ such that $ U^\dagger U = U U^\dagger = \mathbb I$. The set of such matrices forms a (non-abelian) group with the group operation being ordinary matrix multiplication.

The set of generators of this group, which we denote $\mathfrak u(2)$, consists of all $2\times 2$ complex matrices $g$ which are antihermitian, i.e. $g^\dagger = -g$. The set of such matrices constitutes a vector space, and when we equip it with the product operation $[g,h]:= gh - hg$, it becomes an algebra (in fact, a Lie algebra).

A suitable basis for $\mathfrak u(2)$ is given by $\tau_0 = \pmatrix{i & 0 \\ 0 & i}$ together with the Pauli matrices multiplied by the imaginary unit, $$\tau_1 = i\sigma_1 = \pmatrix{0 & i \\ i & 0} \quad \tau_2 = i\sigma_2 = \pmatrix{ 0 & 1 \\ -1 & 0} \quad \tau_3 = i\sigma_3 = \pmatrix{i & 0 \\ 0 & -i}$$

A generic element of the algebra $\mathfrak u(2)$ is some linear combination $g = \sum_n \alpha_n \tau_n = \sum_n i\alpha_n \sigma_n$, and if $g$ is exponentiated we obtain an element of $\mathrm U(2)$, namely $$U = \exp[g] = \exp\left[\sum_n i\alpha_n \sigma_n\right]$$

One of the generators can be identified as $$ \hat U=\pmatrix{1 & 0 \\ 0 & 1}e^{i\phi}$$

This is either a typo or extremely sloppy. This $\hat U$ is equal to $\exp[\phi \tau_0]$; the set of all matrices of this form is a subgroup of $\mathrm U(2)$ generated by the 1D algebra consisting of constant multiples of $\tau_0$.

The remaining three unitary matrices have the property $\mathrm{det}(U)=1$.

This is also not right. The subgroup of $\mathrm{U}(2)$ which is generated by the remaining three generators $(\tau_1,\tau_2,\tau_3)$ has this property, but that's a very different statement.

But how can these form a group if they do not include the identity? I am assuming the group operation is matrix multiplication.

The group $\mathrm{SU}(2)$ does contain the identity. It is generated by $\tau_1,\tau_2,$ and $\tau_3$, and it is these matrices which do not include $\pmatrix{1 & 0 \\ 0 & 1}$ in their span.

Altogether I strongly dislike the passage you've quoted. The author flips back and forth between talking about the group $\mathrm U(2)$ and the algebra $\mathfrak u(2)$ which generates it, sometimes in the middle of a sentence.

J. Murray
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    Thank you for the great answer, as always. One thing: you define the generators of the group as being used like $U=\exp{g}.$ These $g$ are anti-hermitian. You multiplifed the pauli-spin matrices by $i$, making them anti-hermitian, and so $U=\exp{i\alpha^n \sigma_n}.$ Meanwhile, the author defines generators to be Hermitian, so that they correspond to observables, and says that we use them $U=\exp(ig).$ I am assuming this is just a difference in convention? – Jbag1212 Apr 01 '21 at 23:09
  • The generators of a group do not in general form an algebra. By group generator one usually means any element is a word in those generators subject to relations. In this context one has an algebra, the associated Lie algebra to the group, which has its generators, which are separate from the notion of generators of the group. – JamalS Apr 01 '21 at 23:22
  • The generators of $\mathfrak{u}(2)$ are the generators of the tangent space at the identity of the group, but they are not generators of the group since they are not even group elements - they can be taken into the group through the exponential map. – JamalS Apr 01 '21 at 23:24
  • @JamalS For a generic group, sure. But when we talk about Lie groups, at least in the convention with which I am familiar, we call the elements of some chosen basis for the Lie algebra the "(infinitesimal) generators" of the group, c.f. here and here and the first bullet point under "Differential equations" here. – J. Murray Apr 01 '21 at 23:36
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    @Jbag1212 Yes, it's exactly as you say. The physics convention is to take $U=\exp[ig]$ with $g$ Hermitian, and the mathematics convention is to take $U=\exp[g]$ with $g$ antihermitian. I usually stick with the latter out of personal preference, but it all comes down to taste. You may find one easier to work with than the other. – J. Murray Apr 01 '21 at 23:44
  • @J.Murray Well, you've completely changed terminology, indeed I wouldn't have corrected you had you said "infinitesimal" generators of the group, but "generators of the group" is a totally different meaning. – JamalS Apr 02 '21 at 15:16
  • @JamalS As pointed out in my previous comment, it is extremely common among physicists to leave out the word "infinitesimal" and just refer to them as generators. You are welcome to have the opinion that this terminology is unacceptable; I'm merely pointing out that it is nearly universal in the physics literature, and that I have never seen anybody use "generators" to refer to a generating set of a Lie group in a physics context. – J. Murray Apr 02 '21 at 15:24
  • @JamalS Upon reflection, the author refers to the generators as linearly independent, implying that he's referring to elements of the Lie algebra; in the next line he refers to $\pmatrix{1 & 0 \ 0 & 1}e^{i\phi}$ as a generator, meaning that he's back to $\mathrm U(2)$. I think I will stick with my original interpretation that this passage is highly ambiguous. – J. Murray Apr 02 '21 at 15:36