Quinn J.J., Yi K.-S. Solid State Physics: Principles and Modern Applications
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Uncorrected Proof
BookID 160928 ChapID 05 Proof# 1 - 29/07/09
140 5 Use of Elementary Group Theory in Calculating Band Structure
Fig. 5.6. Empty lattice bands of a square lattice along the line Γ → Δ → X.Energy
is measured in units of
h
2
2ma
2
.Wehavedrawnstraight lines connecting E
l
1
l
2
(Γ) to
E
l
1
l
2
(X) for the sake of simplicity. The pair of integers (l
1
,l
2
) is indicated for each
band, and the band degeneracy is given in the parenthesis. In fact the energy along
Δvariesasξ
2
+2l
1
ξ + l
2
1
+ l
2
2
5.4 Use of Irreducible Representations 228
It is apparent from the V (r) = 0 empty lattice structure that at some points 229
in the Brillouin zone more than one band has the same energy. If we refer 230
to the two-dimensional square lattice, we find that at E(X)=
1
4
h
2
2ma
2
there 231
are two degenerate bands, viz. (l
1
,l
2
)=(0, 0) and (−1,0). At E(Γ) =
h
2
2ma
2
232
there are four degenerate bands (−1,0), (1,0), (0,1), (0,−1). The vector space 233
formed by the degenerate bands at E(k) is invariant under the operations 234
of the group of the wave vector k. This means that the space of degenerate 235
states at a point k in the Brillouin zone provides a representation of the 236
group of the wave vector k. We can decompose this representation into its 237
irreducible components and use the decomposition to label the states. This 238
process does not change the empty lattice band structure that we have already 239
Uncorrected Proof
BookID 160928 ChapID 05 Proof# 1 - 29/07/09
5.4 Use of Irreducible Representations 141
t6.1 Table 5.6. Empty lattice bands of the group 4mm
t6.2 l
1
, l
2
2ma
2
h
2
E
l
1
l
2
(Γ) = l
2
1
+ l
2
2
2ma
2
h
2
E
l
1
l
2
(X)=(l
1
+
1
2
)
2
+ l
2
2
t6.3 0 0 0
1
4
t6.4 −10 1
1
4
t6.5 1 0 1
9
4
t6.6 −20 4
9
4
t6.7 2 0 4
25
4
t6.8 0 ±11
5
4
t6.9 0 ±24
17
4
t6.10 −1 ±12
5
4
t6.11 −1 ±25
17
4
obtained. It is simply a convenient choice of basis functions for each of the 240
spaces of degenerate energy states. However, once the periodic potential V (r) 241
is introduced, it is immediately seen that band gaps appear as a consequence 242
of the decomposition of degenerate states into irreducible components. We 243
shall see that states belonging to different IR’s do not interact (i.e. they are 244
not coupled together by the periodic potential). 245
The periodic potential V (r) can be expressed as a Fourier series 246
V (r)=
K
V
K
e
iK·r
, (5.30)
where K is a reciprocal lattice vector. For the two-dimensional square lattice
247
we can write 248
V (r)=
l
1
l
2
V
l
1
,l
2
e
2πi
a
(l
1
x+l
2
y)
. (5.31)
249
Because V (r) is invariant under the operations of the point group, it is not 250
difficult to see that 251
V
l
1
,l
2
= V
−l
1
,l
2
= V
l
1
,−l
2
= V
−l
1
,−l
2
= V
l
2
,l
1
= V
−l
2
,l
1
= V
l
2
,−l
1
= V
−l
2
,−l
1
.
(5.32)
In our previous discussion of the effect of the periodic potential, we were able
252
to obtain an infinite set of coupled algebraic equations, (4.40), which could be 253
written 254
E − V
0
−
¯h
2
2m
(k + K)
2
C
K
=
H=0
V
H
C
K−H
. (5.33)
Uncorrected Proof
BookID 160928 ChapID 05 Proof# 1 - 29/07/09
142 5 Use of Elementary Group Theory in Calculating Band Structure
Here, C
K
was the coefficient in the expansion of u(r), the periodic part of 255
the Bloch function, in Fourier series. This infinite set of equations could be 256
expressed as a matrix equation, (4.41). The off-diagonal matrix elements are 257
of the form K
i
|V (r)|K
j
= V
K
i
−K
j
. When the degeneracy of a particular 258
energy state becomes large (e.g. at E(Γ) = 5
h
2
2ma
2
the degeneracy is 8), the 259
degenerate states must be treated exactly and there is no reason to suppose 260
that any off-diagonal matrix elements vanish. However, when we classify the 261
degenerate states according to the IR’s of the group of the wave vector, we 262
are able to simplify the secular equation by virtue of a fundamental theorem 263
on matrix elements. 264
Theorem on Matrix Elements (without proof) 265
Matrix elements of any operator which is invariant under all the operations 266
of a group are zero between functions belonging to different IRs of the group. 267
Matrix elements are also zero between functions belonging to different rows of 268
the same representation. 269
When one classifies the degenerate states according to the IR’s of the group 270
of the wave vector, many of the degenerate states will belong to different IRs 271
and therefore the off-diagonal matrix elements of V (r) between them will 272
vanish. 273
5.4.1 Determining the Linear Combinations of Plane Waves 274
Belonging to Different IRs 275
Let us begin by considering the states at Γ. The plane-wave wave functions 276
and energies are given, respectively, by 277
Ψ
l
1
l
2
(Γ) = e
2πi
a
(l
1
x+l
2
y)
E
l
1
l
2
(Γ) =
h
2
2ma
2
(l
2
1
+ l
2
2
).
(5.34)
278
Therefore, the energies at Γ, the bands corresponding to that energy, and the 279
degeneracy are as given in Table 5.7. 280
At E
Γ
= 0, there is a single state (let us measure E in units of
h
2
2ma
2
). The 281
wave function is given by 282
Ψ
00
(Γ) = 1 (5.35)
It is unchanged by every operation of G
Γ
, the group of the wave vector Γ. 283
Therefore, it belongs to the IR Γ
1
because every element of G
Γ
operating on 284
Ψ
00
gives +1 ×Ψ
00
or every operation is represented by the 1 ×1 unit matrix 285
D = 1. This, of course, is the Γ
1
representation. At E
Γ
= 1, there are four 286
states 287
Ψ
±1,0
(Γ) = e
±
2πi
a
x
,
Ψ
0,±1
(Γ) = e
±
2πi
a
y
. (5.36)
Uncorrected Proof
BookID 160928 ChapID 05 Proof# 1 - 29/07/09
5.4 Use of Irreducible Representations 143
t7.1 Tabl e 5.7. Empty lattice energy at Γ and its degeneracy for a two dimensional
square lattice
t7.2
2ma
2
h
2
E(Γ) Bands (l
1
,l
2
) Degeneracy
t7.3 0 (0,0) 1
t7.4 1 (±1, 0); (0, ±1) 4
t7.5 2 (−1, ±1); (1, ±1) 4
t7.6 4 (±2, 0); (0, ±2) 4
t7.7 5 (−1, ±2); (1, ±2); (−2, ± 1); (2, ±1) 8
The eight operations of G
Γ
change x into ±x and y into ±y or x into ±y and 288
y into ±x. From the four function Ψ
±1,0
and Ψ
0,±1
we can form the following 289
linear combinations 290
Ψ(Γ
1
)=cos
2πx
a
+cos
2πy
a
∝ Ψ
1,0
+Ψ
−1,0
+Ψ
0,1
+Ψ
0,−1
, (5.37)
291
Ψ(Γ
3
)=cos
2πx
a
− cos
2πy
a
∝ Ψ
1,0
+Ψ
−1,0
− Ψ
0,1
− Ψ
0,−1
, (5.38)
and
292
Ψ(Γ
5
)=
Ψ
(1)
(Γ
5
)
Ψ
(2)
(Γ
5
)
=
sin
2πx
a
sin
2πy
a
∝
Ψ
1,0
− Ψ
−1,0
Ψ
0,1
− Ψ
0,−1
. (5.39)
Because the cosine function is an even function of its argument, every opera-
293
tion of G
Γ
leaves (cos
2πx
a
+cos
2πy
a
) unchanged, and this function transforms 294
according to the IR Γ
1
. The function (cos
2πx
a
−cos
2πy
a
) is left unchanged by 295
operations (E,R
2
, m
x
,m
y
) which change x →±x, but it changes to minus 296
itself under operations (R
1
,R
3
, m
+
,m
−
) which change x →±y.Thusthe 297
operations of G
Γ
operating on (cos
2πx
a
−cos
2πy
a
)doexactlythesamethingas 298
multiplying by the matrices belonging to the representation Γ
3
. In a similar 299
way one can show that the operations of G
Γ
operating on the column vector 300
sin
2πx
a
sin
2πy
a
have exactly the same effect as multiplying by the set of matrices
301
forming the 2 × 2representationΓ
5
. 302
Exercise 303
The reader should determine the linear combinations of plane waves at E
Γ
=2 304
and E
Γ
= 4 belonging to the appropriate IRs of G
Γ
. 305
At E
Γ
= 5 there are eight states 306
Ψ
±1,±2
(Γ) = e
±
2πi
a
(±x±y)
,
Ψ
±2,±1
(Γ) = e
±
2πi
a
(±2x±y)
.
(5.40)
The simplest way to determine the linear combinations belonging to IRs of
307
G
Γ
is first to form sine and cosine functions like 308
Uncorrected Proof
BookID 160928 ChapID 05 Proof# 1 - 29/07/09
144 5 Use of Elementary Group Theory in Calculating Band Structure
Ψ
1
=cos
2π
a
x cos
2π
a
2y,
Ψ
2
=cos
2π
a
2x cos
2π
a
y,
Ψ
3
=sin
2π
a
x sin
2π
a
2y,
Ψ
4
=sin
2π
a
2x sin
2π
a
y,
Ψ
5
=cos
2π
a
x sin
2π
a
2y,
Ψ
6
=cos
2π
a
2x sin
2π
a
y,
Ψ
7
=sin
2π
a
x cos
2π
a
2y,
Ψ
8
=sin
2π
a
2x cos
2π
a
y.
It is easy to see how these Ψ
i
s are transformed by the operations of G
Γ
. 309
For example, all operations of G
Γ
which transform x →±x transform Ψ
1
310
into itself; all operations which transform x →±y transform Ψ
1
into Ψ
2
. 311
Therefore, the linear combination Ψ
1
+Ψ
2
is unchanged by every operation 312
of G
Γ
and belongs to Γ
1
. The linear combination Ψ
1
− Ψ
2
is unchanged by 313
the operations which take x →±x, but changed to −(Ψ
1
−Ψ
2
)byoperations 314
which take x →±y.ThusΨ
1
− Ψ
2
belongs to the IR Γ
3
. We find, by similar 315
analysis 316
Ψ(Γ
1
)=cos
2π
a
x cos
2π
a
2y +cos
2π
a
2x cos
2π
a
y =Ψ
1
+Ψ
2
,
Ψ(Γ
3
)=cos
2π
a
x cos
2π
a
2y − cos
2π
a
2x cos
2π
a
y =Ψ
1
− Ψ
2
,
Ψ(Γ
2
)=sin
2π
a
x sin
2π
a
2y − sin
2π
a
2x sin
2π
a
y =Ψ
3
− Ψ
4
,
Ψ(Γ
4
)=sin
2π
a
x sin
2π
a
2y +sin
2π
a
2x sin
2π
a
y =Ψ
3
+Ψ
4
,
Ψ
(1)
(Γ
5
)=
cos
2π
a
x sin
2π
a
2y
cos
2π
a
y sin
2π
a
2x
=
Ψ
5
Ψ
8
,
Ψ
(2)
(Γ
5
)=
sin
2π
a
x cos
2π
a
2y
sin
2π
a
y cos
2π
a
2x
=
Ψ
7
Ψ
6
.
(5.41)
5.4.2 Compatibility Relations
317
The character tables for 4mm and its principal subgroups are listed in Tables 318
5.8-5.11. Notice that under the operation m
x
, functions belonging to the IR’s 319
Uncorrected Proof
BookID 160928 ChapID 05 Proof# 1 - 29/07/09
5.4 Use of Irreducible Representations 145
t8.1 Tabl e 5. 8. Character table of the groups 4 mm and its principal subgroups
t8.2 Γ
1
= M
1
Γ
2
= M
2
Γ
3
= M
3
Γ
4
= M
4
Γ
5
= M
5
t8.3E 1 1112
t8.4 R
2
1 111−2
t8.5 R
1
, R
3
11−1 −10
t8.6 m
x
, m
y
1 −11−10
t8.7 m
+
, m
−
1 −1 −110
t9.1 Tabl e 5. 9. Character table of a principal subgroup G
X
t9.2 X
1
X
2
X
3
X
4
t9.3 E 1 1 1 1
t9.4 R
2
11−1 −1
t9.5 m
x
1 −11−1
t9.6 m
y
1 −1 −11
t10.1 Table 5.10. Character table of a principal subgroup G
Δ
t10.2 Δ
1
Δ
2
t10.3 E 1 1
t10.4 m
x
1 −1
t11.1 Table 5.11. Character table of a principal subgroup G
Σ
t11.2 Σ
1
Σ
2
t11.3 E 1 1
t11.4 m
+
1 −1
1. Γ
1
and Γ
3
are unchanged. 320
2. Γ
2
and Γ
4
change sign. 321
3. X
1
and X
3
are unchanged. 322
4. X
2
and X
4
change sign. 323
5. Δ
1
are unchanged. 324
6. Δ
2
change sign. 325
Because of this only a Δ
1
band can begin at an Γ
1
or Γ
3
band and only a 326
Δ
1
band can end at an X
1
or X
3
band. We call such restrictions compatibility 327
relations. For our purpose it is sufficient to know that 328
• AbandΔ
1
can connect Γ
1
, Γ
3
, Γ
5
to X
1
,X
3
. 329
• AbandΔ
2
can connect Γ
2
, Γ
4
, Γ
5
to X
2
,X
4
. 330
• AbandΣ
1
can connect Γ
1
, Γ
4
, Γ
5
to M
1
,M
4
,M
5
. 331
• AbandΣ
2
can connect Γ
2
, Γ
3
, Γ
5
to M
2
,M
3
,M
5
. 332
Uncorrected Proof
BookID 160928 ChapID 05 Proof# 1 - 29/07/09
146 5 Use of Elementary Group Theory in Calculating Band Structure
5.5 Using the Irreducible Representations 333
in Evaluating Energy Bands 334
Instead of labelling energy bands at particular symmetry points or along 335
particular symmetry lines by the integers (l
1
,l
2
) [or in three-dimensions 336
(l
1
,l
2
,l
3
)], it is possible to label the states by their energy and by the lin- 337
ear combination belonging to a particular IR of G
k
.Thus,attheΓpointof 338
E =1·
h
2
2ma
2
we may write the four states as 339
|l
1
,l
2
=
⎧
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎩
|1, 0 =e
2πi
a
x
|−1, 0 =e
−
2πi
a
x
|1, 0 =e
2πi
a
y
|0, −1 =e
−
2πi
a
y
(5.42)
or we can write (in units of
h
2
2ma
2
=1) 340
|E
Γ
=1, Γ
1
=cos
2πx
a
+cos
2πy
a
(5.43)
|E
Γ
=1, Γ
3
=cos
2πx
a
− cos
2πy
a
(5.44)
|E
Γ
=1, Γ
5
1
|E
Γ
=1, Γ
5
2
=
Ψ
(1)
(Γ
5
)
Ψ
(2)
(Γ
5
)
=
sin
2πx
a
sin
2πy
a
. (5.45)
There is a distinct advantage to using the basis functions belonging to IRs of
341
G
Γ
that results from the theorem on matrix elements. 342
Any matrix elements of the periodic potential (i.e. an operator with the 343
full symmetry of the point group) between states belonging to different IR’s 344
is zero. Thus, the secular equation becomes 345
ε
0
(Γ
1
) − E Γ
1
0 |V |1Γ
1
Γ
1
0 |V |1Γ
3
Γ
1
0 |V |1Γ
5
1
···
Γ
1
1 |V |0Γ
1
ε
1
(Γ
1
) − E Γ
1
1 |V |1Γ
3
Γ
1
1 |V |1Γ
5
1
···
Γ
3
1 |V |0Γ
1
Γ
3
1 |V |1Γ
1
ε
1
(Γ
3
) − E Γ
3
1 |V |1Γ
5
1
···
1
Γ
5
1 |V |0Γ
1
1
Γ
5
1 |V |1Γ
1
1
Γ
5
1 |V |1Γ
3
1
ε
1
(Γ
5
) − E ···
2
Γ
5
1 |V |0Γ
1
2
Γ
5
1 |V |1Γ
1
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
=0
(5.46)
Equation (5.46) reduces to
346
ε
0
(Γ
1
) − E Γ
1
0 |V |1Γ
1
00···
Γ
1
1 |V |0Γ
1
ε
1
(Γ
1
) − E 00···
00ε
1
(Γ
3
) − E 0 ···
000ε
1
(Γ
5
) − E ···
0000···
.
.
.
.
.
.
.
.
.
.
.
. ···
= 0 (5.47)
Uncorrected Proof
BookID 160928 ChapID 05 Proof# 1 - 29/07/09
5.5 Using the Irreducible Representations in Evaluating Energy Bands 147
Here, ε
n
(Γ
j
)=
h
2
2ma
2
n
2
+ Γ
j
n |V |Γ
j
n. There are two things to be noted: 347
1. The matrix elements of V between different IR’s vanish, so many off- 348
diagonal matrix elements are zero. This reduces the determinantal equation 349
to a block diagonal form. 350
2. The diagonal matrix elements Γ
j
n |V |Γ
j
n are, in general, different for 351
different IRs Γ
j
. This lifts the degeneracy at the symmetry points and 352
splits the fourfold degeneracy into nondegenerate states Γ
1
and Γ
3
and 353
one doubly degenerate state Γ
5
at E(Γ) 1 ·
h
2
2ma
2
. 354
When the energy bands along Δ and along X are classified according to the 355
IRs of the appropriate symmetry group, a band structure like that sketched 356
in Fig. 5.7 results. The degeneracies at Γ and X are lifted by the diagonal 357
matrix elements of the potential. The rare case when two different IRs have 358
the same value of the diagonal matrix element of V (r) is called as accidental 359
degeneracy. Two Δ-bands belonging to Δ
1
,ortwobelongingtoΔ
2
cannot 360
cross because they are coupled by the nonvanishing matrix elements of V (r) 361
between the two bands. However, a Δ
1
band can cross a Δ
2
band because 362
Γ
3
Fig. 5.7. Electronic energy bands of a square lattice with V = 0 along the line
Γ → Δ → X. The bands are schematic, showing where splittings and anticrossings
occur on the simplified diagram (Fig. 5.6) which connects E(Γ) and E(X) by straight
lines
Uncorrected Proof
BookID 160928 ChapID 05 Proof# 1 - 29/07/09
148 5 Use of Elementary Group Theory in Calculating Band Structure
Ψ
Δ
1
|V (r)|Ψ
Δ
2
= 0. Bands that are widely separated in energy (e.g. the 363
bands at E(Γ) = 1 and E(Γ) = 4) can be treated by perturbation theory as 364
was done in the nearly free electron model. One can observe that degeneracies 365
do not occur frequently for bands belonging to the same IR’s at Γ (or at X) 366
until the energies become high. 367
5.6 Empty Lattice Bands for Cubic Structure 368
5.6.1 Point Group of a Cubic Structure 369
Every operation of the cubic group will turn x into ±x, ±y, ±z.Itiseasyto 370
see that there are 48 different operations that can be listed as follows: 371
1. x →±x, y →±y, z →±z. 372
2. x →±x, y →±z, z →±y. 373
3. x →±y, y →±x, z →±z. 374
4. x →±y, y →±z, z →±x 375
5. x →±z, y →±y, z →±x 376
6. x →±z, y →±x, z →±y 377
Since there are ± signs we have two possibilities at each step, giving 2
3
=8 378
operations on each line or 48 operations all together. 379
We can also think of the 48 operations in terms of 24 proper rotations and 380
24 improper rotations: 381
Proper Rotations 382
E: Identity → 1operation 383
4: Rotation by ±90
◦
about x-, y-, or z-axis → six operations 384
4
2
: Rotation by ±180
◦
about x-, y-, or z-axis → three operations 385
2: Rotation by ±180
◦
about the six [110], [1
¯
10], [101], [10
¯
1], [011], [01
¯
1] axes → 386
six operations 387
3: Rotation by ±120
◦
about the four 111 axes → eight operations Hence, 388
we have 24 proper rotations in total. 389
Improper Rotations 390
Multiply each by J (inversion operator: r →−r) to have 24 improper rota- 391
tions. 392
393
The 24 improper rotations are obtained by multiplying each of the 24 394
proper rotations by J, the inversion operation (r →−r). Clearly there are 10 395
classes and 48 operations. Using the theorem 396
i=IR
l
2
i
= g
we can see that there are 10 IRs, four one-dimensional, and two two-
397
dimensional, and four three-dimensional ones. 398
Uncorrected Proof
BookID 160928 ChapID 05 Proof# 1 - 29/07/09
5.6 Empty Lattice Bands for Cubic Structure 149
2{1
2
+1
2
+2
2
+3
2
+3
2
} =48.
It is probably worthwhile to sit down with a simple cube and work out the
399
following rotations: 400
1. E:
⎛
⎝
x
y
z
⎞
⎠
→
⎛
⎝
x
y
z
⎞
⎠
.
401
2. Rotation of ±
π
2
about x, y, z axes, 402
C
x
(1):
⎛
⎝
x
y
z
⎞
⎠
→
⎛
⎝
x
z
−y
⎞
⎠
, C
x
(−1):
⎛
⎝
x
y
z
⎞
⎠
→
⎛
⎝
x
−z
y
⎞
⎠
403
C
y
(1):
⎛
⎝
x
y
z
⎞
⎠
→
⎛
⎝
−z
y
x
⎞
⎠
, C
y
(−1):
⎛
⎝
x
y
z
⎞
⎠
→
⎛
⎝
z
y
−x
⎞
⎠
404
C
z
(1):
⎛
⎝
x
y
z
⎞
⎠
→
⎛
⎝
y
−x
z
⎞
⎠
, C
z
(−1):
⎛
⎝
x
y
z
⎞
⎠
→
⎛
⎝
−y
x
yz
⎞
⎠
405
3. Rotation of π about x, y, z axes, 406
C
x
(2):
⎛
⎝
x
y
z
⎞
⎠
→
⎛
⎝
x
−y
−z
⎞
⎠
, C
y
(2):
⎛
⎝
x
y
z
⎞
⎠
→
⎛
⎝
−x
y
−z
⎞
⎠
407
C
z
(2):
⎛
⎝
x
y
z
⎞
⎠
→
⎛
⎝
−x
−y
z
⎞
⎠
408
4. Rotation of π about the six 110 axes, 409
C
x,y
(2):
⎛
⎝
x
y
z
⎞
⎠
→
⎛
⎝
y
x
−z
⎞
⎠
, C
x,−y
(2):
⎛
⎝
x
y
z
⎞
⎠
→
⎛
⎝
−y
−x
−z
⎞
⎠
410
C
y,z
(2):
⎛
⎝
x
y
z
⎞
⎠
→
⎛
⎝
−x
z
y
⎞
⎠
, C
y,−z
(2):
⎛
⎝
x
y
z
⎞
⎠
→
⎛
⎝
−x
−z
−y
⎞
⎠
411
C
z,x
(2):
⎛
⎝
x
y
z
⎞
⎠
→
⎛
⎝
z
−y
−x
⎞
⎠
, C
z,−x
(2):
⎛
⎝
x
y
z
⎞
⎠
→
⎛
⎝
−z
−y
−x
⎞
⎠
412
5. Rotation of ±
2π
3
about the six 111 axes, 413
C
x,y,z
(+):
⎛
⎝
x
y
z
⎞
⎠
→
⎛
⎝
y
z
x
⎞
⎠
, C
x,y,z
(−):
⎛
⎝
x
y
z
⎞
⎠
→
⎛
⎝
z
x
y
⎞
⎠
414
C
x,y,− z
(+):
⎛
⎝
x
y
z
⎞
⎠
→
⎛
⎝
−z
x
−y
⎞
⎠
, C
x,y,− z
(−):
⎛
⎝
x
y
z
⎞
⎠
→
⎛
⎝
y
−z
−x
⎞
⎠
415
C
x,−y,z
(+):
⎛
⎝
x
y
z
⎞
⎠
→
⎛
⎝
−y
−z
x
⎞
⎠
, C
x,−y,z
(−):
⎛
⎝
x
y
z
⎞
⎠
→
⎛
⎝
z
−x
−y
⎞
⎠
416