What is an electron, a proton, a pion or a quark? Little balls of different size and mass? Strings vibrating in different modes?
Ignoring for the moment String Theory and other theories trying to unify Gravitation with Quantum Field Theory (QFT), I wanted to suggest you a simple exercise in algebra which illustrates what is understood as an elementary particle within the framework of QFT. And, as you surely guessed it, this has to do with our game csɔ (read as /kasak/; classification, symmetry, ɔonservation, or the other way around ;-).
This example illustrates as well how the patterns that help us understand nature emerge in Physics.
Let me explain this last sentence by recalling the world view that once had Heraclitus (or even better here). We know about him from the writings of Plato. According to him, Heraclitus claimed that everything was in permanent change, or, as is probably better known, you cannot step twice into the same stream. Namely, the current of water you see today is not the same as that from yesterday, whence it is not the same river, Heraclitus would probably say.
What has this to do with current Physics? Probably nothing. But it amuses me thinking that Physics, not without some bit of irony, rephrases Heraclitus when assuming that what is real is that which stays invariant when performing a change, a transformation; that what then gets conserved is the only thing meaningful to talk about, the rest being superfluous.
And that's the idea of our game csɔ, you remember? Let's see. Consider the ammonia molecule \(NH_3\). It has a tetrahedral configuration with a triangular base.
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| Balls & stick model of the ammonia molecule \(NH_3\). Nitrogen atom in blue; hydrogen atoms in grey. Figure retrieved from Wikipedia. |
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| Top view of the positions of atoms in a \(NH_3\) molecule. |
Question 1: What are the transformations that leave this structure invariant?
Try to answer before reading on...
Ok, the answer is the group of 6 transformations called \(C_{3v}\), which can be obtained considering the following symmetries: A) 120 deg. rotations around an axis perpendicular to the plane of the figure and passing through the N atom, and B) 3 reflections through each of the 3 perpendicular planes that contain N and each one of the H atoms.
If we label the rest of the positions by 1, 2 and 3, see the figure, we can use following notation. We'll call S1 to the reflection through the plane containing 0 & 1, that is, the transformation which swaps whatever is in locations 2 & 3; S2 the one containing 0 & 2, and therefore, only swapping whatever is in 1 & 3; and S3 the one swapping 1 & 2 -notice how we keep abusing of our language one step more every sentence!
[ Remark: It becomes somewhat tedious to keep repeating "whatever is in..." and usually it's not said, but still that's what's meant! ]
From the three possible rotations, that with an angle of \(360^\circ\) is the same as not doing anything, that is, it's the identity, and we'll call it \(E\). To the \(120^\circ\), anti-clockwise rotation, we'll call it \(C_1\), and the \(240^\circ\), anti-clockwise one, \(C_2\). Thus, the group of symmetries consists in \(C_{3v}\,=\,\{E, C_1, C_2, S_1, S_2, S_3\}\)
[Remark: Technically speaking, one says that the structure has the symmetry $C_3v$, and each transformation of the group is called a symmetry element.]
[Remark: Here we are using the word "group" in a quite informal way as a "collection of objects". We'll later see that it has a very precise mathematical definition. But we'll ignore that for moment -let's see how far we get :) ]
Question 2: If we denote by $C_2\,\cdot\,C_1$ the composition of transformations (in analogy to the usual composition of functions $f\circ g$), that is, first apply $C_1$ and, to the result, apply then $C_2$, convince yourself of the following identities:
\[ C_2\,\cdot\,C_1 = C_1\,\cdot\,C_2 = E \\
S_1\,\cdot\,S_1 = S_2\,\cdot\,S_2 = S_3\,\cdot\,S_3 = E \]
Question 3: A possible first notation for these 6 transformations is that of permutations. It works as follows.
For each transformation, list the four atoms sorted on one line. The sorting of each atom is given by its position. On a line beneath that one, list again the atoms after the transformation while respecting the same sorting of positions. In this way, we obtain
\[ E = \begin{bmatrix}N&H_1&H_2&H_3\\N&H_1&H_2&H_3\end{bmatrix} \\
C_1 = \begin{bmatrix}N&H_1&H_2&H_3\\N&H_3&H_1&H_2\end{bmatrix}\]
For instance, for $C_1$ this reads as follows: "the transformation $C_1$ leaves the object located on $0$ on the same location; that which was on location $1,\,H_1$, it moves it to location $2$; the one located on $2,\,H_2,$ is moved to $3$; and the one which was located on $3,\,H_3$, gets moved to location $1$".
As the first reference line will always be the same, we can omit it and thereby save some work! In this way, we'll write
\[ E = \left[ N\, H_1\, H_2\, H_3 \right] \\
C_1 = \left[ N\, H_3\, H_1\, H_2 \right] \\
C_2 = \left[ N\, H_2\, H_3\, H_1 \right] \]
Write the permutations corresponding to $S_1,\,S_2$ and $S_3$.
Question 4: We can represent these transformation as matrices as well. Consider the same sorting for columns and rows as that one above for positions. Again this sorting denotes thus the locations 0,1,2 and 3. We may then read each column, say column 'c' as "that which was on position 'c' gets moved to position..." and the answer is the location corresponding to the row where we put a "1" on that same column. We have then, e.g.,
$ E = \begin{pmatrix} 1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}$
and$C1 = \begin{pmatrix} 1&0&0&0\\0&0&0&1\\0&1&0&0\\0&0&1&0\end{pmatrix}$.
Write the matrices corresponding to $C_2,\, S_1,\, S_2$ and $S_3$.
Verify the following identities by doing the corresponding matrix multiplications:
\[ C_1\,\cdot\,C_2 = C_2\,\cdot\,C_1 = E \\
S_1\,\cdot\,S_1 = E \\
S_1\,\cdot\,S_2 = C_2 \]
Question 5: The inverse of a transformation has to be one such that when multiplying by the said transformation one obtains the identity. Write as a matrix the inverse of the transformation $C_1$, i.e., $(C_1)^{-1}$, and verify that $C_1\,\cdot\,(C_1)^{-1} = (C_1)^{-1}\,\cdot\,C_1 = E$.
Do the same for $S_1$. Can you write down the matrices corresponding to inverses of $S_2$ and $S_3$ without calculating them? Actually, it's easy to write down all inverses without calculating the inverse matrix. It suffices to understand the notation and realize what changes each inverse transformation does.
Question 6: The matrices we obtained in Q4 above are the matrices written on the basis of vectors given by $B_0\,=\,\{e_0,\, e_1,\,e_2,\,e_3\}$, as given on the big slide at the bottom. This basis represents the four positions as vectors in a virtual vectorial space (nothing to do with the 3-dimensional space where we can construct the triangular tetrahedron, nor the one where the ammonia molecule lives in).
Write the corresponding matrices in the basis $B_z$ as given by the slide at the bottom.
The final result is as follows. In the basis $B_z$, the symmetry $C_{3v}$ has the following matricial representation:
$ E^z \,=\, \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1& 0 & 0\\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{pmatrix} \, C_1^z\,=\,\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\0 & 0 & -\frac{1}{2} & \frac{1}{2}\\0 & 0 & -\frac{3}{2} & -\frac{1}{2}\end{pmatrix} \, C_2^z\,=\,\begin{pmatrix} 1 & 0 & 0 & 0\\0 & 1 & 0 & 0\\0 & 0 & -\frac{1}{2} & -\frac{1}{2}\\0 & 0 & \frac{3}{2} & -\frac{1}{2}\end{pmatrix} $
$ S_1^z\,=\,\begin{pmatrix} 1 & 0 & 0 & 0\\0 & 1 & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & -1\end{pmatrix} \, S_2^z\,=\,\begin{pmatrix} 1 & 0 & 0 & 0\\0 & 1 & 0 & 0\\0 & 0 & -\frac{1}{2} & \frac{1}{2}\\0 & 0 & \frac{3}{2} & \frac{1}{2}\end{pmatrix} \, S_3^z\,=\,\begin{pmatrix} 1 & 0 & 0 & 0\\0 & 1 & 0 & 0\\0 & 0 & -\frac{1}{2} & -\frac{1}{2}\\0 & 0 & -\frac{3}{2} & \frac{1}{2}\end{pmatrix} $
Notice how the 6 matrices get structured in blocks around the diagonals (in linear algebra, this is similar to what's called the Jordan normal form, or Jordan canonical form of a matrix, which is the simplest representation we can get when a matrix cannot be diagonalized). We can distinguish two blocks of 1x1 size and one of size 2x2 on all six matrices!
Yes, for $E$ and $S_1$ that may sound a bit strange, but it isn't. The said 2x2 block is just $\begin{pmatrix} 1&0\\0&1 \end{pmatrix}$ for $E$ and $\begin{pmatrix} 1&0\\ 0& -1 \end{pmatrix}$ for $S_1$.
Conclusion:
As you can see, none of the transformations "mix" the vectors $Z_0$ and $Z_1$, neither among them, nor with $Z_2$ or $Z_3$. However, vectors $Z_2$ and $Z_3$ do get mixed under the transformations $C_1,\,C_2,\,S_2$ and $S_3$. Under $S_1$, $Z_2$ stays invariant, but $Z_3$ reverses its orientation. Whence, in some sense, $Z3$ does mix as well in this case, even if it´s with himself!
Indeed, the fact that there are blocks that do not mix means that these blocks present some "identity", some reality; it means it is something that gets preserved, conserved, under any transformation. It is, thus, something we may call a (symmetry) pattern.
In the following figure you can see my attempt to graphically represent the different patterns one may distinguish. Not all are obtained within a single basis, as happens with the patterns given by the vectors $e_0, V_0$ and $Z_1$.
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| Four basic patterns of the $C_{3v}$ symmetry: I) everything collapsed on a dot, II) a triangular tetrahedral frame as a whole; III) a triangular frame corresponding to the base; IV) two more abstract, intertwinned patterns related to the orientation of location 0 relative to the plane formed by 1, 2 and 3. |
The job of the physicist is then to apply this language in the description of physical systems in a way that helps us understand them better. For instance, these ideas may provide an easier argument explaining why the electrostatic field is perpendicular to the surface of a conductor, or why the magnetic field of a current through a wire has only an angular component around that wire.
But the ideas of symmetry are much more powerful. Physics considers the elementary particles, and, for what it matters, any other particle as well, in an analogous way as we did here, namely as symmetry patterns analogous to the examples shown here. More exactly, an elementary particle is but a component of an irreducible representation of a symmetry group. By irreducible representation it is meant the blocks we obtained above; a component is like a vector defining such a block, i.e., like $\mathbf{e}_0$ or like $\mathbf{Z}_2$.
Furthermore, the concept of symmetry allows the physicist to classify things, like, say, atoms :-) . This is just the familiar Mendeleev's periodic table of elements: the concept of symmetry, as we did here, is used to identify recurrent patterns in the configuration of the electrons of the atoms. Once you can identify different patterns, you can classify things based on who shows which pattern, in much the same way we all have done often as kids. Thus, each column of the periodic table groups together elements with the same irreducible representation of the group of rotations (rotations by an arbitrary angle; see next paragraph). This is the symmetry manifested by the valence electrons (those farthest from an atom's nucleus). A version of the periodic table of elements that best reflects this classification is Janet's left-step periodic table.
The difference in QFT is that the corresponding symmetry groups are very different. In particular, the may represent groups of continuous transformations, i.e., they accept rotations by an arbitrary angle, and not only by $120^\circ$; or each transformation may depend on the location in space; or they may refer to transformations in a virtual space that has nothing to do with swapping balls around some locations in space.
Thus, as we can see, the concept of symmetry has been very useful for the physicist in describing nature. It has allowed her understand which quantities have to be conserved, defining the essential concept of elementary particle, and also to classify systems. But there is more: physics shows us that the four fundamental forces are uniquely determined by the symmetries that characterize them. That is, once the symmetries are given, so are the forces! The theory of liquid crystal, in condensed matter, is another clear example where the symmetry of the constituent particles uniquely determines the dynamics, i.e., the response of the liquid crystal.
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| The Emergence of Patterns I |
Question 1: Evidently, $C_{3v}$ is a symmetry group of the ammonia molecule $NH_3$. Any of its symmetry elements transforms the molecule into a state that is indistinguishable from the initial state before the transformation. We could end the discussion here, as the solution to the question is already in the above text: It's enough checking that $C_{3v}$ is indeed a symmetry!
However, if we care for the details, we could ask, how we know for sure that these are all the symmetries of $NH_3$. That is, aren't we missing a symmetry transformation of $NH_3$ that is not included in $C_{3v}$? For the sake of completeness of the discussion in this post, we will now try to respond to this question here.
For example, we could swap the atoms $N$ and $H_1$. Clearly, this is not a symmetry, as we would immediately realize that somebody moved the molecule. However, if we then rotate the molecule within the plane containing atoms $N$, $H_1$ and $H_3$ we could bring $N$ atom back on the top position. Isn't this a symmetry transformation? Isn't this, therefore, a transformation different that those contained in $C_{3v}$? And from a more general point of view, why did we consider rotations only around the $N$ atom? why not also rotations around each and every hydrogen atom? And, what about reflections including $N$ as well? Couldn't there be combinations of such kind of rotations and reflections that lead to additional symmetries of $NH_3$?
That is our mindset now. These are the questions we are trying to clarify. Fine. Let's look into it in detail now.
For the sake of simplifying our discussion, let's first consider only transformations that leave invariant hydrogen atom $H_2$. For those leaving invariant $H_1$ or $H_3$ the argument should be similar.
Fine. If $H_2$ is invariant, we just need to repeat the same arguments in the text above, but considering $H_2$ as if it were the nitrogen atom. That is, we will have 3 reflections and 3 rotations in the plane containing atoms $N$, $H_1$ and $H_3$. We will call the first three $\sigma_0$, $\sigma_1$ and $\sigma_3$, with the indices referring to the positions left invariant, and we'll call the rotations $D_3$, $D_3^2$ and $D_3^3\equiv E$, the last one being the identity. We will also call the group of transformations formed by these elements $C_{3v}^{II}$. Notice how, $\sigma_0$ is identical to the reflection $S_2$ of our previous group $C_{3v}$. This will be the key for why we don't need to consider any additional transformations to those already discussed.
Evidently, $C_{3v}^{II}$ is not a group of symmetries of the ammonia molecule $NH_3$. It's easy to see that none of the elements $D_3$, $\sigma_1$ or $\sigma_3$ is a symmetry element -and if $D_3$ isn't, so is $D_3^2$, isn't? But, what about combinations of these transformations? In order to set up a combination resulting in a symmetry, we must manage to get the $N$ atom back on position $0$ at the end of such a transformation. We will see now that this is that strong a constraint that it limits the answer to only one possibility -besides the trivial one, $E$.
Let's use the notation corresponding to permutations -it's more concise and this text is becoming already very long. Let's consider now the first combination we mentioned above, namely, $D_3^2\cdot \sigma_3$, that is, (fasten your seat belts now!) swapping whatever are in locations $0$ and $1$, followed by a counter-clockwise rotation of $240^\circ$ around an axis containing $2$ and perpendicular to the plane $0$, $1$ and $3$. (Side note: one day I'll draw a picture and just refer to it.)
Ok. The reflection is easy to write down, namely $S_3\,=\,\left[1,\,0,\,2,\,3\right]$. The other isn't much more difficult: $D_3^2\,=\,\left[1,\,3,\,2,\,0\right]$. We might have applied twice $D_3\,=\,\left[3,\,0,\,2,\,1\right]$ as well, instead of guessing it directly. Fine, we then have
$ D_3^2\cdot\sigma_3\,=\,D_3^2\,\left[1,\,0,\,2,\,3\right]\,=\,\left[0,\,3,\,2,\,1\right]\,=\,\sigma_0$
It's relatively easy to convince oneself that, no matter with which transformation $T\in C_{3v}^{II}$ we start with. If we want that the next transformation $T'$ leaves the $N$ atom on position $0$, the resulting combination will always end up being $\sigma_0$! In other words, now all the symmetries of $NH_3$ are the transformations contained within the subgroup $G\,\equiv\,\{E,\,\sigma_0\}\,\supset\,C_{3v}^{II}$. But, as $\sigma_0\,=\,S_2$, $G$ is but a subgroup of $C_{3v}$.
We would have arrived to the same conclusion if we would have fixed atom $H_1$ or $H_3$ instead of $H_2$. We are shown, therefore, that the group $C_{3v}$ contains all possible symmetry elements of the ammonia molecule $NH_3$. QED :-)
Question 2: $C_1$ is a transformation that rotates each and every one of the hydrogen atoms $120^\circ$ counter-clockwise around an axis containing $N$. $C_2$ is a rotation in the same sense and around the same axis, but by angle of $240^\circ$. Thus, if we first rotate all $H$ atoms $120^\circ$ and right afterwards we add a $240^\circ$ rotation to it, the net effect will be like rotating all those atoms $360^\circ$, and therefore, like not doing anything. In mathematical jargon, the combined effect of both transformations is identical to the identity transformation! The same happens if we first apply $C_2$ and then $C_1$. It therefore holds that $C_2\cdot C_1\,=\,E\,=\,C_1\cdot C_2$.
It's easier to see it for reflections: Consider a sorted list of characters, a, b, c,... Now swap two, and only two, characters, e.g., the first and the third characters. The list would then turn into c, b, a, ... If we now apply the same transformations again, that is, we swap again the first and third characters, we get a, b, c, ... In other words, it's like we didn't do anything. Thus, applying twice a transposition -which is the technical name for such a swap when we make no reference to space- is the same as the identity. As each of $S_1$, $S_2$ and $S_3$ are transpositions, it readily follows the three identities $S_1\cdot S_1\,=\,S_2\cdot S_2\,=\,S_3\cdot S_3\,=\,E$.
Question 3: It's easy to convince yourself that the solutions are
$ S_1\,=\,\left[N,\,H_1,\,H_3,\,H_2\right]$,
$ S_2\,=\,\left[N,\,H_3,\,H_2,\,H_1\right]$,
$ S_3\,=\,\left[N,\,H_2,\,H_1,\,H_3\right]$.
Question 4: As we already know, $C_2$ leaves $N$ invariant, but rotates all hydrogens by $240^\circ$.
Thus, it makes $H_1$ hop twice: from its initial position, $1$, to $2$ and from there to $3$. In the same way, it moves $H_2$ to the position were initially we had $H_1$, i.e., $1$, etc. Therefore, we can directly write the corresponding matrix
$C_2\,=\,\begin{pmatrix}1& 0 & 0 & 0 \\0& 0 & 1 &
0 \\0& 0 & 0 & 1 \\0& 1 & 0 & 0
\\\end{pmatrix}$
The reflections are even easier and we can directly write them down
$ S_1\,=\,\begin{pmatrix}1& 0 & 0 & 0 \\0& 1 & 0 &
0 \\0& 0 & 0 & 1 \\0& 0 & 1 & 0
\\\end{pmatrix}$
$ S_2\,=\,\begin{pmatrix}1& 0 & 0 & 0 \\0& 0 & 0
& 1 \\0& 0 & 1 & 0 \\0& 1 & 0 & 0
\\\end{pmatrix}$
$ S_3\,=\,\begin{pmatrix}1& 0 & 0 & 0 \\0& 0 & 1
& 0 \\0& 1 & 0 & 0 \\0& 0 & 0 & 1
\\\end{pmatrix}$
Question 5: Let's determine the inverse matrix of $C_1$ without calculating it. Remember that such an matrix must "undo" what $C_1$ does. For instance, $C_1$ move an atom at $1$ to position $2$. It's inverse must thus bring to location $1$ whatever is at $2$. Thus, its column number $2$ will have a one at row $1$ and the rest will be zeros. Furthermore, as $C_3$ brings what is at $3$ to $1$, its inverse must bring what is at $1$ to $3$, and therefore its column $1$ will all be zeros except at row $3$ where it must have a one. Similarly, we can see the its column $3$ will have a single 1 only at row $2$.
$ \left(C_1\right)^{-1}\,=\,\begin{pmatrix}1& 0 & 0 & 0
\\0& 0 & 1 & 0 \\0& 0 & 0 & 1 \\0& 1 & 0
& 0 \\\end{pmatrix}$
As we can see, this matrix is just transformations $C_2$! In short, it is $C_2\,=\,\left(C_1\right)^{-1}$.
Again, the case of the reflections is easier:
$ \left(S_1\right)^{-1}\,=\,\begin{pmatrix}1& 0 & 0 & 0
\\0& 1 & 0 & 0 \\0& 0 & 0 & 1 \\0& 0 & 1
& 0 \\\end{pmatrix}\,=\,S_1$
$ \left(S_2\right)^{-1}\,=\,\begin{pmatrix}1& 0 & 0 & 0
\\0& 0 & 0 & 1 \\0& 0 & 1 & 0 \\0& 1 & 0
& 0 \\\end{pmatrix}\,=\,S_2$
$ \left(S_3\right)^{-1}\,=\,\begin{pmatrix}1& 0 & 0 & 0
\\0& 0 & 1 & 0 \\0& 1 & 0 & 0 \\0& 0 & 0
& 1 \\\end{pmatrix}\,=\,S_3$
Question 6: The basis $B_z$ is composed of the vectors
$\mathbf{Z}_0\,=\,\mathbf{e}_0$, $\mathbf{Z}_1\,=\,\mathbf{e}_1\,+\,\mathbf{e}_2\,+\,\mathbf{e}_3$, $ \mathbf{Z}_2\,=\,2 \mathbf{e}_1\,-\,\left(\mathbf{e}_2\,+\,\mathbf{e}_3\right)$ y $\mathbf{Z}_3\,=\,\mathbf{e}_3\,-\,\mathbf{e}_2$
Let's construct the matrices with the columns and rows sorted in the same way, namely, Z0, Z1, Z2 and Z3. We can follow to ways to obtain these matrices. One path is the "mechanical" one, which is longer but doesn't require "to think" much. This path consists in writing the matrix change of basis, let's call it $M$, given by the previous definitions of the $B_z$ basis, and apply $M$ as a similarity transformation to the matrices we obtained earlier. For instance, if call the matrix representation of $C_1$ in the basis $B_0$ as $C_1^0$, and that one in the basis $B_z$ as $C_1^z$, it holds that $C_1^z\,=\,M^{-1}\cdot C_1^0\cdot M$.
Painful! It's convenient to know this way of calculating it, and sometimes it might be the only approach available; we also need to master matrix algebra, of course. But, if we clearly understand what a matrix is, we have another way of solving this question which is way easier.
[Now comes a side note on elementary linear algebra.
Let's recall what a matrix is. The column $j$ of an arbitrary matrix $A$ is the result of applying $A$ on the basis vector $b_j$, i.e., the column $j$ consists on the vector $Ab_j$. An arbitrary component $i$ of this vector is thus $\left(A b_j\right)^i\,=\,A^{ij}$ (strictly speaking it would be $A^{i\phantom{j}}_{\phantom{i} j}$ as it is a 1-covariant, 1-contravariant tensor. Will we ignore, however, such subtleties as we are not considering the effects of having any particular metric around). In other words, the matrix element $A^{rs}$ of a matrix $A$ is the component $r$ of the image of the basis vector $b_s$ given by the linear application $A$.
For the case of a matrix change of basis $M$ the interpretation changes only slightly. This takes a vector of basis $B_z$, say $\mathbf{Z}_1$, and assigns it as image a linear combination of vectors of basis $B_0$, namely $\mathbf{e}_1\,+\,\mathbf{e}_2\,+\,\mathbf{e}_3$. Sorting the columns as Z0, Z1, Z2 and Z3, and the rows as e0, e1, e2 and e3, the previous change of basis, $M$ has the matrix representation given by
$ M\,=\,\begin{pmatrix}1&0&0&0\\0&1&2&0\\0&1&-1&-1\\0&1&-1&1\end{pmatrix}$.
End of review of linear algebra.]
Therefore, in order to construct the corresponding matrices, we just need to build the images of Z0, Z1, Z2 and Z3 as given by each and every one of the 6 symmetry elements of $C_{3v}$.
Immediately we see that $\mathbf{Z}_0$ is invariant under all transformations -as it should, for it corresponds to position $0$. But also that $\mathbf{Z}_1$ is invariant under $C_{3v}$ as well. This is so because $\mathbf{Z}_1$ is a symmetric combination of all three positions $1$, $2$ and $3$, i.e., it considers them all together without making any distinctions among them. Any rotation or reflection will thus change only the order of the summands in $\mathbf{Z}_1$. But this doesn't change $\mathbf{Z}_1$ at all!
There is still left to see what happens with $\mathbf{Z}_2$ and $\mathbf{Z}_3$. These are not invariant, but transform among each other! Let's look in detail the case of $S_2$. In this case we obtain following results:
$ S_2 \mathbf{Z}_2 \,=\, 2 e_3\,-\,\left(e_2\,+\,e_1\right) \,=\,-\,\frac{1}{2}\mathbf{Z}_2\,+\,\frac{3}{2}\mathbf{Z}_3$
$ S_2 \mathbf{Z}_3 \,=\, e_1\,-\,e_2\,=\,\frac{1}{2}\mathbf{Z}_2\,+\,\frac{1}{2}\mathbf{Z}_3$
Therefore, the matrix corresponding to $S_2$ in the basis $B_z$ is
$ S_2^z\,=\,\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&-\frac{1}{2}&\frac{1}{2}\\0&0&\frac{3}{2}&\frac{1}{2}\end{pmatrix}$
You can follow the same method to obtain the matrices of the remaining transformations.
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