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I am just curious about the formalism of basic Quantum Mechanics. Lets take for instance the system of a spin-$\frac{1}{2}$ particle. The state of the particle is described by a vector in an abstract Hilbert space that is two dimensional (say $\mathcal{H}$). The set of Endomorphisms on $\mathcal{H}$ form a group (which i hope will be the $SU(2)$ group). Now I will just define an abstract Endomorphic map in $\mathcal H$, such that $$ \hat\sigma_z : \left|+\right> \rightarrow \left|+\right> \qquad || \qquad \left|-\right> \rightarrow -\left|-\right>$$ where $\left|+\right>,\left|-\right> \in \mathcal H$

Clearly, the operator $\hat \sigma_z$ is Hermitian and the eigenvectors are orthonormal and hence can be chosen as a basis set. Hence any arbitrary vector can be expanded about this. $$ \left|\psi\right> = c_+\left|+\right> + c_-\left|-\right> \qquad ~~{where}~~\qquad \mathbf C \ni c_\pm = \left<\pm|\psi\right> $$ Now from what I have learnt so far, I sort of see that I can construct an a map called Representation $\mathcal R$ such all the elements for $\mathcal H$ gets mapped to $\mathbf C^2 $

$$ \mathcal R : \mathcal H \rightarrow \mathbf C^2 \qquad | \qquad \mathcal R\big(\left|\psi\right> \big) = \begin{pmatrix} c_+\\ c_-\\ \end{pmatrix} $$

This representation map preserves the inner product also I believe. For instance,

$$ \left<\phi|\psi\right> \rightarrow \begin{pmatrix} d_+ & d_-\\ \end{pmatrix} \begin{pmatrix} c_+\\ c_-\\ \end{pmatrix} \in \textbf C $$

Further the operators can also be mapped by this representation map, where the abstract operators get mapped to square matrices.

$$ \mathcal R : \text{End}(\mathcal H) \rightarrow \text{End}(\mathbf C^2) \quad|\quad \mathcal R(\hat A) = \begin{pmatrix} \left<+\right|\hat A\left|+\right> & \left<+\right|\hat A\left|-\right>\\ \left<-\right|\hat A\left|+\right> & \left<-\right|\hat A\left|-\right>\\ \end{pmatrix} $$

With this setup, the Pauli matrices and the vector's 2-D irrep all correspond this map $\mathcal R$ right ? So all those things correspond to a representation constructed using the eigen vectors of $\sigma_z$ ?

I also wish to know how would one make this kind of a connection in the cases of position basis, especially between $\left|x\right>$ and $L_2$ spaces.

PS: I know this question is of least use to any particular community of research or even people learning, but this is just out of my curiosity. Pardon me if this is a very ridiculous question.

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1 Answer 1

For the relation between the abstract position basis and the $L^2$ spaces, I refer you to my answer here (read the other answers too, they're good ;) )

You are quite close with your understanding of the representations, but not quite there:

First of all, for the 2-dim spin-$\frac{1}{2}$ Hilbert space $\mathcal{H}_{\uparrow\downarrow}$, the set of endomorphisms $\mathrm{End}(\mathcal{H}_{\uparrow\downarrow})$ is not $\mathrm{SU}(2)$, but the whole of the 2D matrices, i.e. $\mathbb{C}^{2\times2}$. This is because every finite-dimensional Hilbert space of dimension $n$ is first and foremost a complex vector space, and all these are isomorphic to $\mathbb{C}^n$.

Now, a representation of a given group $G$ on any space $V$ is just a homomorphism $\rho : G \rightarrow \mathrm{Aut}(V)$. Since we have the inclusion $\mathrm{SU}(2) \subset \mathbb{C}^{2\times2}$, the space $\mathcal{H}_{\uparrow\downarrow}$ comes prequipped with a representation of $\mathrm{SU}(2)$. Since $\mathrm{SU}(2)$ is a Lie group, it has generators which lie in its Lie algebra, and every representation $\rho$ of the Lie group induces a representation $\mathrm{d}\rho : \mathrm{LieAlg}(G) \rightarrow \mathrm{End}(V)$ of the algebra (and vice versa, with a few caveats).

[Lie groups are amazing things, and very fundamental to theoretical physics, especially the understanding of symmetries. I advise you try to learn more things about them than I will say here.]

The three generators of $\mathrm{SU}(2)$ are canonically denoted $\sigma_{x},\sigma_y,\sigma_z$. You may now pick eigenvectors of (e.g.) $\mathrm{d}\rho(\mathrm{\sigma_z})$ on $\mathcal{H}_{\uparrow\downarrow}$ and use them as your basis. If you call these eigenvectors $|\pm\rangle$ (it is no accident that the eigenvalues of $\sigma_{x,y,z}$ in the fundamental representation (that is what this is) are $+1$ and $-1$), you have defined the same basis you have in your OP.

Of course, in this concrete example where the target space $\mathcal{H}_{\uparrow\downarrow}$ is just isomorphic to $\mathbb{C}^2$, the space on which $\mathrm{SU}(2)$ is natively defined, $\mathrm{d}\rho$ and $\rho$ are just identity (more precisely: inclusion) maps.

Now, everything else you called a representation in your question is just "ordinary" change of basis, this is what it means to choose the eigenvectors of $\sigma_z$ as a new basis for the vector space $\mathbb{C}^2$.

Feel free to ask for clarifications/additions if I have missed the point of your question or you didn't understand something.

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You are totally right about the group, The SU(2) is just a subset that preserves the norm upon transformation. My bad ! Thanks –  user35952 Jul 1 at 17:27
    
@user35952: Correct. Though note that the $\mathrm{U}(2)$ would be the full group of norm-preserving transformations, but we do not want that for our spinors, since it includes "reflections" as well as "rotations", while the $\mathrm{SU}(2)$ only has "rotations" (it is also orientation-preserving). (Also, it has nicer properties) –  ACuriousMind Jul 1 at 17:31
    
The change of basis in the space $\mathbb C^2$ is different. I am talking about the fact that if we chose the basis vectors of $\sigma_x$, the Pauli matrices will themselves change. So the Pauli matrices are basically the operators only if we construct the representation map using the basis vectors of $\sigma_z$ right ? –  user35952 Jul 1 at 18:16
    
If you change into the eigenbasis of a matrix, that matrix will have diagonal form in that basis, so the Pauli matrices will change if you change the basis. The "canonical" form of the Pauli matrices is w.r.t to some pre-chosen basis of the $\mathbb{C}^2$ in question and then taking the inclusion map as the representation. If you choose another basis of $\mathbb{C}^2$ beforehand, you will get a superfically different, but isomorphic representation. –  ACuriousMind Jul 1 at 18:28

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