Research: Science and Education
Advanced Chemistry Classroom and Laboratory
Complex Characters Made Simple
edited by
Joseph J. BelBruno
Dartmouth College Hanover, NH 03755
Sidney F. A. Kettle School of Chemical Sciences and Pharmacy, University of East Anglia, Norwich NR4 7TJ, United Kingdom;
[email protected] A common topic in introductory courses on group theory in chemistry is that of optical activity. For a molecule to be optically active its symmetry must be such that a translation and rotation about a given axis have the same symmetry characteristics (transform as the same irreducible representation). More accurately, that for at least one electronic transition there must be parallel electric and magnetic dipoles (the transformations of electric and magnetic dipoles are isomorphic to translations Table 1. The C4v Character Table C4v
E
2C4
C2
2σv
2σv’
A1
1
1
1
A2
1
1
1
1 –1
1 –1
B1
1
–1
1
B2
1
E
2
0
1 –2
1 –1
–1
–1
0
0
σv′(2)
C4
∙
σv(2)
C2 σv(1)
E ∙
σv′(1)
C4
A1
∙ ∙
B1
∙
∙
∙
∙
∙ ∙ ∙
∙
∙
∙
∙ ∙ ∙
A2
∙
∙
∙
1
∙
∙
∙
∙
∙
B2 ∙
∙
∙
∙
E ∙
∙
∙
∙
∙ ∙ ∙
∙
∙
∙
∙ ∙ ∙
∙
∙
∙
∙ ∙ ∙
∙
∙
Figure 1. Diagrams of the irreducible representations of C4v with the nodal planes emphasized. Although the notation C4 and C43 is now generally preferred, in this text we have reverted to the use of C4+ and C4−. This is because the meanings of the latter pair are more visually apparent to the reader (as clockwise and anticlockwise) and because their counterparts in other groups, such as C3 (and so C3+ and C3−) are immediately evident.
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and rotations, respectively). This requirement means that the molecule is such that charge displacements within it can follow a helical path; left- and right-handed helices cannot be superimposed and neither can optical isomers. The topic is one that offers a very simple and immediate explanation of the rule that only molecules with symmetries lacking any improper rotation operation are optically active. However, it suffers from one major disadvantage. It is difficult to extend the discussion to include the character tables of such groups. The reason is that almost all of the relevant character tables contain complex numbers, and, commonly, students find complex numbers difficult. Not surprisingly, complex characters have featured in several articles in this Journal (1–3). However, none of them have considered the physical meaning of complex characters; in the present article we seek to remedy this absence. We are motivated by an interpretation that we have recently emphasized for the character tables of point groups. Irreducible representations simply distinguish distinct nodal patterns (4). It follows that it is possible to draw pictures of irreducible representations, something that has several consequences. First, it enables a quite different, more pictorial, approach to the teaching of group theory in chemistry (5). Second, it makes accessible topics that might be considered too advanced. Double groups (applicable to all odd-electron systems) and spin–orbit coupling are just two examples (6). Triple and higher groups, although currently without any scientific application, are no less accessible (7). But so far, all of the applications have been to groups for which the character tables contain only real characters. What of those with complex characters? Although the approach that has been developed is applicable to all point groups, it is most readily understood for axial point groups, and we will consider C4v as an example. Its conventional character table is given in Table 1. The pictorial interpretation of this table is detailed in Figure 1, where at the top, are shown the effects of the symmetry operations of the C4v group on any arbitrary point in the octant labeled E (the arbitrary point can be taken to represent any point within a C4v molecule; the octant results from the fact that it is necessary to represent seven other similar sets within the same diagram). This diagram is akin to the stereographic projections of crystallographers, but symmetry operations, not elements, are the focus of attention. One can find a hint at such diagrams in an early article in this Journal (1). The diagrams of the irreducible representation of the C4v point group are shown in Figure 1. Characters of 1 appear as + and those of ‒1 as −. The characters of the E irreducible representation in Table 1 are the sum of those in Figure 1, using the same conversion notation. Using these diagrams it is easy to show that each of the individual representations is normalized (square each element and add, using (+)2 = (−)2 = 1, then divide by the order of the group, 8) and that they are also orthogonal (multiply corresponding entries together, using (+) × (−) = (−) ≡ ‒1, and add). Direct products are similarly obtained by multiplying together the phases of the appropriate patterns to give a resultant pattern (8). To emphasise the fact
Journal of Chemical Education • Vol. 86 No. 5 May 2009 • www.JCE.DivCHED.org • © Division of Chemical Education
Research: Science and Education
that the diagrams in Figure 1 show nodal patterns, the nodal planes are shown bold. A1 is nodeless (by symmetry; one could draw circular nodes about the center of the A1 figure without changing its basic pattern). A2 has four nodal planes (which explains why no basis function is ever given for it—the simplest basis function would have to be fourth power in the Cartesian coordinates). B1 and B2 have two nodes (but which behaves like xy and which like x2 − y2 depends on the choice of orientation of x and y). Finally, the E diagrams each have one nodal plane. These two diagrams help point the way to the complex character problem. In that each represents the way that an individual E function behaves, one can ask the question “what happens to them under the operations of the group?”, of which the C4 operations are of particular interest to us. Applying the C4+ (or C4−) operation to either of the E patterns simply generates the other (possibly with all signs changed); the operations interchange the two functions. It is the fact that no such interchange occurs in the point group C4, to which we now turn, that is at the heart of the complex character problem. The character table for the C4 group is given in Table 2. In a sense it is simpler than that for C4v, in that it contains only half of the operations. The mirror plane reflection operations of C4v are absent in C4. In C4v they serve to interrelate the clockwise and anticlockwise C4 rotation operations, and so, in C4 these rotations become distinct. The consequence is seen in Table 2. The improper rotations of C4v have been lost, but the square root of ‒1, i, has appeared. Nonetheless, the A and B irreducible representations appear without i, so we can draw these two without difficulty. We do so in Figure 2, which is the beginning of the C4 equivalent of Figure 1. It is as soon as we attempt to picture the E pair that we run into trouble. Simply taking the recognizable characters from Table 2 and inserting them in Figure 2 gives two (apparently) identical diagrams for E! However, the quantities indicated by ? have yet to be inserted and, clearly, there can be no requirement that they are all the same—the two E diagrams must end up different. How are the ? quantities to be determined? The way forward lies in a return to the helix. A helical motion along the C4 axis presents no problem. A helix is a superposition, a combination, of a translation and a rotation. A C4 rotation will leave both a translation and a rotation about this axis unchanged; both transform as A. So, a helical motion along the z axis, C4, has A symmetry. It is in the plane perpendicular to the z axis that problems arise. It is not possible to define symmetry-correct x and y axes in the traditional way (the transformations of correct basis functions generate the characters in the character table), but let us ignore this and consider either a translation or a rotation about some arbitrary x axis. A C4 rotation will convert these into a corresponding translation or rotation about the corresponding y axis. The simplicity of a helical motion along the z axis no longer holds. What do we mean by a helical motion in the xy plane (a plane that can readily be defined, even if x and y axes cannot; it is a plane perpendicular to the z axis)? Consider the helix in Figure 3: a helix in the xy plane. It will not do. It suffers from the defect that our rotation operations are cyclic, so that (C4)4 = E, and the helix in Figure 3 does not have this property. It can be given the required closure property if the whole helix is deformed into a circle to give an annulus (a donut) with the two ends joined smoothly. Interestingly, the relevant shape was created several years ago by a sculptor–chemist, Béla Vízi (9). A photograph of his steel “Planar Electron” is shown in Figure 4. But, alas, although
Table 2. The C4 Character Table C4
E
C4
C2
C43
A
1 1
1 –1
1
B
1
1 –1
1
i
–1
–i
1
–i
–1
i
E
∙
C2
C4
∙
C4
E
A
B ∙
∙
∙
∙
∙
∙
∙
∙
∙
?
∙
?
?
∙
?
∙
E
Figure 2. An incomplete set of diagrams of the irreducible representations of C4 with some nodal planes emphasized. The ? are discussed in the text.
Figure 3. A 7-turn helix. This has to be twisted into a circle and the ends joined, to give a continuous, annular (donut), shape before it can be used to explain complex characters.
Figure 4. A sculpted version of the annular (donut) shape referred to in the caption to Figure 3. The sculpture, Tranzverzális Sík Elektron (Planar Electron), was first shown in an exhibition “Chemistry in Sculptures” at the University of Veszprém in 1993. The author is indebted to Professor Béla Vízi (http://www.vizibela.hu/ accessed Mar 2009) for permission to reproduce this photograph.
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Research: Science and Education ∙
C2
C4
∙
C4
E
Figure 5. A 1-turn helix that connects the results of the operations of the C4 group on a unit vector (arrowed).
A
∙ ∙
?
?
Figure 6. The same as Figure 5 but with the amplitude pattern of one of the E functions of Figure 2 added.
B ∙
∙
∙
∙
∙
∙
∙
∙
E ∙
∙ ∙
?
∙
∙
∙
?
Figure 7. As Figure 6 but for a helix of the opposite hand to that of Figure 5, drawn from a slightly different viewpoint.
Figure 8. The completed version of Figure 2. No nodal planes are shown for E; for each diagram a single nodal plane would run bottom left to top right, the distinction (and othogonality) between the diagrams being given by the arrows.
the helix in Figures 3 and 4 is useful for illustration it would only become relevant for groups Cn with n ≥ 15, scarcely of chemical interest. Of more chemical interest (but much less useful in getting the idea across!) is the case with a single turn to the helix. This is shown in Figure 5, applied to the C4 case. Consider the arrowed bar, with a length that is the internal radius of the helix and that we shall take to be unity. Under the C4+ rotation it follows a helical path, ending at the radius vector at the right of the diagram. As an aid, the transformation has been indicated by a thin line (this line in itself is almost meaningless, its intermediate points mean nothing, although it is part of the single-turn helix that we are discussing). With three more such steps, the original position of the vector is regained. Let us now add to Figure 5 the picture of the E irreducible representation in Figure 2. The result is in Figure 6. This figure enables us to see the problem rather clearly. With the original vector (the short rod) as reference, that connected to it by two steps is clearly to be represented as “–” because it is directed in the opposite direction. We also see the problem posed by the ? more clearly. The ? are associated with vectors that are perpendicular to that with which we started. Our immediate reaction would be to represent this by a character of 0 (it contains none of the starting vector), and, in a sense, this is correct. It may be correct, but it is also incomplete. There are two reasons for this, of which we can give one immediately. This is that if the helical motion is carried out twice, not once, we end with a vector that is the negative of that with which we started. That is, (C4+)2 = C2. As a consequence of this, the character of the C2 rotation, which we know to be ‒1, is the square
of the character of that of C4+. It follows that the character of the C4+ operation is i, because i2 = ‒1 (although, of course, ‒i would do just as well, but we will stay with i for the moment). Written in full, the character of the C4+ operation is 0 + i, where, of course, the 0 is always dropped. We shall elaborate on this shortly. We have found the value of one of the ?’s in Figure 6, what of the other? At first sight we might conclude that it, too, is i because the corresponding vector seems to point in the same direction as the first. But this is deceptive. One vector points radially outwards from the reference circle, the other points towards the center. These are opposite directions relative to the circle with which we are working, so the one is the negative of the other. Alternatively, one can recognize that if one of the “horizontal” vectors is obtained by a clockwise helical rotation from our reference, the other is obtained by an anticlockwise rotation, the negative of the first. We have our answer, one ? is i, the other is ‒i. But what of the second E function? Another diagram shows the answer. This diagram is shown in Figure 7, which may be compared with that in Figure 6. They are similar, except that one shows a right-hand helix and the other a left-hand helix. Together, they lead to the two components of the E irreducible representation. As far as characters are concerned, where the one has a character of i, the other has a character of ‒i for the same operation. The appearance of i and ‒i in Table 2 is thus explained, but what of Figure 2? Given the above explanation, perhaps the most informative, and simplest, pictorial representation is that of Figure 8. In this, the rotation to a perpendicular position is represented by a 90° arrow, either clockwise or anticlockwise.
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Research: Science and Education
The combination of two arrows in the same sense give rise to – (a character of ‒1) while two in opposite directions give + (the original arrangement, a character of 1). Let us use this to check that the E functions that we have drawn are normalized. Normally, this means that we simply square characters [or, here, use (+) × (+) = (+) ≡ 1; (−) × (−) = (+) ≡ 1] and divide by the order of the group (here, 4). But not this time. While squaring the + and ‒ give a total of 2, the combination of the arrows gives ‒2, leading to a total of 0. We are invited to conclude that the function does not exist, clearly ridiculous. To get the correct answer, we have to combine an arrow in one sense with one in the other, so that their combination is +, 1. The two E functions are complex conjugates and we have to combine each with its complex conjugate (the other!) when normalizing. We thus obtain the expected answer, the functions are properly normalized. Note that had we settled for a character of 0 for the E functions under the C4 rotations, then they would not have normalized properly. The reason is simple. A character of 0 would imply that the starting vector of Figure 5 had vanished on C4 rotation. It had not, it had rotated to a perpendicular direction. This aspect becomes clearer when a group other than C4 is considered. Consider the group C3, the character table of which is given in Table 3. It will be seen that it contains the characters ε and ε* (= ε2) where these two are complex conjugates. As is normal, the character ε is explained somewhere close to the table—it is exp(2πi/3). With an i at the heart of it, this seems to be a character with no real component. Not so; the identity
exp(i x ) = cos (x ) + i sin(x )
means that here wherever we read ε we can replace it with
cos (120 ) + i sin (120 )
[and ε *, exp(‒2 π i/3), by cos(120) – i sin(120), since cos(‒x) = cos(x) but sin(‒x) = ‒sin(x)]. And, of course, cos(120) is a real quantity. In general, for a group Cn, where 360/n = Θ, there will be characters cos(Θ) ± isin(Θ).1 When Θ = 90, the C4 group, cos(Θ) = 0 and this is the reason that we were able to say that the character for the components of the E irreducible representation contain the number 0, even if it is always ignored. When talking of the C4 group, it was convenient to focus on a (unit) vector (arrowed in Figure 5). As it is rotated about a helical path it does not change in magnitude, only direction. The magnitude of the component along its “original” direction is given by the cosine component of the character, it is cos2(Θ). But the ultimate answer must be 1, the vector does not change in magnitude, only direction. So, to the cos2(Θ) we have to add sin2(Θ) because their sum is always 1, no matter the value of Θ.
Table 3. The C3 Character Table C3
E
C3
C32
A
1 1
1 ε
1 ε*
1
ε*
ε
E
Note: ε = exp(2πi/3).
And this is exactly what the complex characters do. In addition to a cos(Θ) term, that which one would regard as the “normal” character, there is also an isin(Θ) term, which has the effect of preserving the vector length [when complex conjugates are multiplied together we get sin2(Θ)]. In retrospect, one can see that the case considered in this article, the C4 group, is the simplest possible. In general, in drawing diagrams akin to that of Figure 8, one has to show both components of the exponential. It is not immediately obvious how to do this in a simple way. Any suggestions? Note 1. This is for a single turn helix. For an n turn helix, Θ has to be replaced by nΘ.
Literature Cited
1. 2. 3. 4. 5.
6. 7. 8. 9.
Theobald, F. J. Chem. Educ. 1982, 59, 277–278. Carter, R. L. J. Chem. Educ. 1993, 70, 17–19. Baraldi, I.; Vanossi, D. J. Chem. Educ. 1997, 74, 806–809. Kettle, S. F. A. J. Chem. Educ. 1999, 76, 675–678. Kettle, S. F. A. Symmetry and Structure, 3rd ed.; John Wiley & Sons: Chichester, U.K., 2007. Kettle, S. F. A. Symmetry and Structure, 3rd ed.; John Wiley & Sons: Chichester, U.K., 2007; Chapter 12. Kettle, S. F. A. Spectrochimica Acta A 1998, 54, 1633–1643. Kettle, S. F. A. Spectrochimica Acta A 2001, 57, 1941–1949. Vízi, B. Kémia és Teremtés; Püski Kiadó: Budapest, 2002; p 32.
Supporting JCE Online Material
http://www.jce.divched.org/Journal/Issues/2009/May/abs634.html Abstract and keywords Full text (PDF) Links to cited JCE articles JCE Cover for May 2009 This article is featured on the cover of this issue. See p 531 of the table of contents for a detailed description of the cover.
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