PH575: Introduction to Solid State Physics SPRING 2018 ...physics.oregonstate.edu/.../lib/exe/fetch.php?media=hw1soln_2018.pdf · PH575: Introduction to Solid State Physics SPRING

PH575: Introduction to Solid State Physics SPRING 2018 Homework # 1 SOLUTION

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1. Paragraph about a research article 2. Find an example of a solid in which (i) s (ii) p (iii) d orbitals are the most important for the material

properties. Give a short explanation. (i) Consider carbon (diamond) where the hybridization of s and p orbitals leads to equivalent bonds in all directions (in contrast to graphite – see ii), and to the insulating rather than conductive nature. (ii) Consider carbon (graphite) where the orbitals near the Fermi level are p-orbitals in a π-bonding configuration. They are responsible for the layered structure of graphite and the fact that it is conducting. (iii) Any number of transition metal solids have d-orbitals near the top of the valence bands. Cu and Ag owe their colors to the presence of d-bands. Within a few eV of the top of the valence band.

3. Expressing s, p, d functions as linear combinations of Ylm functions

(based on Sutton, problem #5) This problem gives you the experience of manipulating wave functions and expressing one orthogonal set (the s, p, d real spherical harmonics) in terms of another orthogonal set (the Ym θ,φ( ) complex spherical harmonics). Sutton problem #5 lists the (complex) spherical harmonic functions Ym θ ,φ( ) for angular momentum l = 0, 1, 2. (These are the angular parts of the solutions to the Schrödinger eigenvalue equation

H Rn r( )Ym θ ,φ( )ϕ r( )

= En Rn r( )Ym θ ,φ( )

ϕ r( )

for the H-atom in the position representation.) Sutton Eq. 1.22 lists the (real) functions that describe the usual five d-orbitals for the H-atom.

(i) The d states:

Sutton 1.22 has the d orbital wave functions:

dxy x, y, z( ) = 15

4πxyr2; dyz x, y, z( ) = 15

4πyzr2; dxz x, y, z( ) = 15

4πxzr2

dz2x, y, z( ) = 15

4π3z2 − r2

2 3r2; d

x2−y2x, y, z( ) = 15

4πx2 − y2

2r2

Here are plots from http://en.wikibooks.org/wiki/High_School_Chemistry/Shapes_of_Atomic_Orbitals:



(I) The dz2 orbital:

€

dz 2 =154π

3z2 − r2

2 3r2=

516π

3r2 cos2θ − r2

r2=

516π

3cos2θ −1( )

By inspection of the given forms of the functions, it is easy to see that this is exactly the same as the Y20 spherical harmonic!

dz2x, y, z( ) = Y20 r,θ ,φ( ) and dz2 = Y20

(II) The dx2-y2 orbital is an equal linear combination of the Y2-2 and Y22. We can see this by some simple algebra:

dx2 − y2

=154π

x2 − y2

2r2=

1516π

sin2θ cos2φ − sin2θ sin2φ( ) = 1516π

sin2θ cos2φ

dx2 − y2

=12

1532π

sin2θ e2iφ + e−2iφ( )

Now we identify the terms on the right with Y22 and Y2-2 to obtain:

dx2 − y2

=12Y2−2 +Y22( )

Weseethat

a21 = a22 =12

.

This problem is simple, so the “guess and check method” above works. To solve more complex problems, one should use the general method of projection:

dx2−y2

= a00 Y00 + a1−1 Y1−1 ...+ a20 Y20 + a21 Y21 + a2−1 Y2−1 + a22 Y22 + a2−2 Y2−2 ..

Project onto (for example) Y21

Y21 dx2−y2 = ...+ Y21 a20 Y20 + Y21 a21 Y21 + Y21 a2−1 Y2−1 + Y21 a22 Y22 + Y21 a2−2 Y2−2 .. Only one non-zero term on rhs:



Y21 dx2−y2 = Y21 a21 Y21 = a21 Y21 Y211

= a21

When you evaluate the integral on the lhs, just as in the example in the earlier part of the homework, you

will find it is 12

so a21 =12

.

Similarly you’ll find by projecting onto Y2−1 that a2−1 =12

.

Projection onto to any other bra gives zero for all other coefficients. ** Once you are confident about doing the integrals, use Mathematica or an equivalent program to help you do them. If you can write the code yourself, then you understand what you need to do and you are ready for the computer. If you’re not sure what to encode, then you may need a bit more practice. It’s like a calculator – when asked to add four 6-digit numbers, you use a calculator because it relieves tedium, not because you don't understand the concept of addition. (III) The dxy orbital is an equal linear combination of the Y2-2 and Y22 eigenstates of Lz.

dxy =154π

xyr2

=154π

r2 sin2θ sinφ cosφr2

=12

154π

sin2θ sin2φ( ) = − i2

1532π

sin2θ e2iφ − e−2iφ( )

dxy =i2Y2−2 −Y22( )

(IV) The dyz orbital is an equal linear combination of the Y2-1 and Y21 states.

dyz =i2Y2−1 +Y21( )

(V) The dxz orbital is an equal linear combination of the Y2-1 and Y21 states.

dxz =12Y2−1 −Y21( )



4. Expressing operators as matrices In this course, you will come across operators expressed as matrices and you will have to manipulate matrix elements. The most common example is the Hamiltonian operator, but this problem will use

angular momentum operators, L2 and Lz. For this problem, all you need to know is that the Y2m are

eigenstates of L2 and Lz:

L2 Y2m = +1( )2 Y2m ;

Lz Y2m = m Y2m

(a) Find the 5x5 matrix representation for the L2 and the Lz operators in the space described by the five

complex basis vectors Y2m .

(b) Do the same for the five real basis functions dj

(c) Show that the trace of the matrices is the same regardless of the basis (a specific example of #3, below). Useful info: A matrix element for an operator O is defined as

€

Oij = i O j

where the ket

€

j represents the state of a system described by a set of quantum numbers j, and the bra

€

j is its complex conjugate. So for example, in the case of the l=2 states of hydrogen. There are 5 basis states (10 if we include spin, but don't consider spin in this problem), the kets might be, in the

€

Y,m basis:

€

1 = = 2,m = −2 = Y2,−2 ; 2 = = 2,m = −1 = Y2,−1 ; 3 = = 2,m = 0 = Y2,04 = = 2,m = +1 = Y2,+1 ; 5 = = 2,m = +2 = Y2,+2

and in the

€

d j basis (where we use primes to remind us we’re counting in a different basis)

1' = dx2−y2

; 2 ' = dyz ; 3' = dz2; 4 ' = dxz ; 5 ' = dxy

SOLUTION:

It is important to know that the Ym functions are eigenstates of both the L2 and Lz operators and that

L2 Y,m = +1( )2 Y,m = 62 Y2,m for = 2

Lz Y,m = m Y,m m=-2,-1,0,1,2 for = 2.

(a) With the

€

Y,m basis states defined above:

L2 1 = L2 Y2,−2 = 62 Y2,−2L2 2 = L2 Y2,−1 = 62 Y2,−1L2 3 = L2 Y2,0 = 62 Y2,0L2 4 = L2 Y2,+1 = 62 Y2,+1L2 5 = L2 Y2,+2 = 62 Y2,+2 and

Lz 1 = −2 Y2,−2Lz 2 = − Y2,−1Lz 3 = 0 Y2,0Lz 4 = + Y2,+1Lz 5 = +2 Y2,+2 .

L2 operator in the Y,m basis: Take a representative example where the bra and ket are the same state:



L211 = 1 L2 1 = Y2,−2 L

2 Y2,−2 = 62 Y2,−2 Y2,−2 We pulled the constant out of the integral. Now make use of the orthonormality condition that Y,m Y,m =1 if BOTH quantum numbers in the ket match BOTH quantum numbers in the bra. Thus

L211 = 62 . The same is true for all diagonal elements.

Now we look at off-diagonal elements. One example will represent them all, so consider: L212 = 1 L

2 2 = Y2,−2 L2 Y2,−1 = 62 Y2,−2 Y2,−1

Now make use of the orthonormality condition that Y,m Y,m = 0 if EITHER quantum number in the ket is different from the corresponding one in the bra. Thus L212 = 0 . The same is true for all off-diagonal elements. The information can be summarized as

L2 = 62

1 0 0 0 00 1 0 0 00 0 1 0 00 0 0 1 00 0 0 0 1

!

"

######

$

%

&&&&&&

in the Y,m basis.

Lz operator in the Y,m basis: Take a representative example where the bra and ket are the same state:

Lz11 = 1 Lz 1 = Y2,−2 Lz Y2,−2 = −2 Y2,−2 Y2,−2 = −2.

using the same reasoning as before. The other diagonal elements give:

Lz2,2 = 2 Lz 2 = − Y2,−1 Y2,−1 = −

Lz3,3 = 3 Lz 3 = 0 Y2,0 Y2,0 = 0

Lz4,4 = 4 Lz 4 = + Y2,+1 Y2,+1 = +

Lz5,5 = 5 Lz 5 = +2 Y2,+2 Y2,+2 = +2

A representative off-diagonal element would be:

Lz12 = 1 Lz 2 = Y2,−2 Lz Y2,−1 = − Y2,−2 Y2,−1 = 0

The information can be summarized as

Lz =

−2 0 0 0 00 −1 0 0 00 0 0 0 00 0 0 1 00 0 0 0 2

"

#

$$$$$$

%

&

''''''

in the Y,m basis.



(b) Define the dj basis states as above:

1' = dx2−y2

; 2 ' = dyz ; 3' = dz2; 4 ' = dxz ; 5 ' = dxy

In this basis set, the matrix representations may no longer be diagonal because the d states may NOT be eigenstates of L2 and Lz. We have to check. We can use the expressions we found in problem 1 for the dj

basis states in terms of the Y,m 's to make it easy to find the matrix elements. The L2 operator in the dj basis

L211 = 1 L2 1 = d

x2−y2L2 d

x2−y2= d

x2−y2L2 1

2Y2,−2 + 1

2Y2,+2{ }

where we've made use of

dx2−y2

=12Y2−2 +Y22( )

Now we use the fact the Y,m are eigenstates of the operator L2

L211 = 12dx2−y2

62 Y2,−2 + 62 Y2,+2{ }= 1262 d

x2−y2Y2,−2 + Y2,+2{ }

Now we use the same substitution as before: L211 = 1

262 d

x2−x2Y2,−2 + Y2,+2{ }= 1

262 1

2Y2,−2 + 1

2Y2,+2{ } Y2,−2 + Y2,+2{ }

= 12 6

2 Y2,−2 Y2,−2 + Y2,−2 Y2,+2 + Y2,+2 Y2,−2 + Y2,+2 Y2,+2{ }= 1

2 62 1+ 0+ 0+1{ }

L211 = 62

We get the same result for L2 as in the Y,m basis. This turns out to be true for all the other diagonal elements, as you can easily check. The reason for this is that the d states are linear combinations of eigenstates of L2 with the SAME eigenvalues of L2. (This is not true for Lz as we will see.) We should check the off-diagonal elements. Are all zero like before? It's easy to see that elements like L2

14 are zero, because they project states onto each other that have different m eigenvalues.

L214 = 1 L2 4 = d

x2−y2L2 dxz .

L214 = dx2−y2

L2 dxz = 12Y2,−2 + 1

2Y2,+2{ }L2 1

2Y2,−1 − 1

2Y2,+1{ }

= 12 Y2,−2 + Y2,+2{ } − Y2,−1 − Y2,+1{ }

= − 12 Y2,−2 + Y2,+2{ } Y2,−1 + Y2,+1{ }

= − 12 Y2,−2 Y2,−10

+ Y2,−2 Y2,+1

0

+ Y2,+2 Y2,−10

+ Y2,+2 Y2,+1

0

"#$

%$

&'$

($

= 0

But what about elements like L215 = 1 L

2 5 = dx2−y2

L2 dxy

where we have different combinations of the same Y,m 's contributing to both states? Let's see ….



L215 = dx2−y2

L2 dxy = 12Y2,−2 + 1

2Y2,+2{ }L2 i

2Y2,−2 − i

2Y2,+2{ }

= i2 Y2,−2 + Y2,+2{ } 62 Y2,−2 − 62 Y2,+2{ }

= 62 i2 Y2,−2 + Y2,+2{ } Y2,−2 − Y2,+2{ }

= 62 i2 Y2,−2 Y2,−2

1

− Y2,−2 Y2,+20

+ Y2,+2 Y2,−2

0

− Y2,+2 Y2,+21

"#$

%$

&'$

($

= 0

So, the off-diagonal elements are still zero, and we conclude that

L2 = 62

1 0 0 0 00 1 0 0 00 0 1 0 00 0 0 1 00 0 0 0 1

!

"

######

$

%

&&&&&&

in the dj basis, too.

The Lz operator in the dj basis is not a diagonal matrix. Take a representative example where the bra and ket are the same state: Lzz11 = 1 L2 1 = d

x2−y2L2 d

x2−y2= d

x2−y2Lz 1

2Y2,−2 + 1

2Y2,+2{ }

= 12dx2−y2

−2 Y2,−2 + 2 Y2,+2{ }= 1

22 1

2Y2,−2 + 1

2Y2,+2{ } − Y2,−2 + Y2,+2{ }

= 12 2 − Y2,−2 Y2,−2

1

+ Y2,−2 Y2,+20

− Y2,+2 Y2,−2

0

+ Y2,+2 Y2,+21

"#$

%$

&'$

($

= 0

.

using the same reasoning as before. The other diagonal elements also give zero (you check this): An off-diagonal element that is clearly zero would be one where states of different ml quantum numbers are projected onto each other, like we saw for the L2 operator, above. Lz114 = d

x2−y2Lz dxz = 0

But we have to consider elements like Lz115 = d

x2−y2Lz dxy = 1

2Y2,−2 + 1

2Y2,+2{ }Lz i

2Y2,−2 − i

2Y2,+2{ }

= i2 Y2,−2 + Y2,+2{ } −2 Y2,−2 − 2 Y2,+2{ }

= −2 i2 Y2,−2 + Y2,+2{ } Y2,−2 + Y2,+2{ }

= −2 i2 Y2,−2 Y2,−2

1

− Y2,−2 Y2,+20

+ Y2,+2 Y2,−2

0

+ Y2,+2 Y2,+21

"#$

%$

&'$

($

= −2i

Also (with HC = Hermitian conjugate for those who know; do the long form if you don't) Lz551 = dxy Lz dx2−y2 = Lz115

HC

= 2i



The other non-zero off-diagonal elements are Lz24 = yz Lz xz and its Hermitian conjugate Lz442 = Lzz24HC .

Lzz24 = dyz Lz dxz = −i2Y2,−1 + −i

2Y2,+1{ }Lz 1

2Y2,−1 − 1

2Y2,+1{ }

= − i2 Y2,−1 + Y2,+1{ } − Y2,−1 − Y2,+1{ }

= i2 Y2,−1 + Y2,+1{ } Y2,−1 + Y2,+1{ }

= i2 Y2,−1 Y2,−1

1

− Y2,−1 Y2,+10

+ Y2,+1 Y2,−1

0

+ Y2,+1 Y2,+11

"#$

%$

&'$

($

= i

Lz442 = −i The information can be summarized as

Lz = i

0 0 0 0 −20 0 0 1 00 0 0 0 00 −1 0 0 02 0 0 0 0

"

#

$$$$$$

%

&

''''''

in the dj basis.

(c) Notice that the sum of the diagonal elements does NOT depend on which basis you use to generate the matrix! This is a trivial result for the L2 operator because the matrices are the same in both bases, but for the Lz operator, it is not trivial: Tr Lz( ) = (−2−1+ 0+1+ 2) = 0 in the Y,m basis

Tr Lz( ) = (0+ 0+ 0+ 0+ 0)i = 0 in the dj basis 5. Sutton, problem #8. This problem makes use of an identity that appears often in Quantum Mechanics. It is given in Sutton Eq 2.4

n nn∑ = I (1)

where the kets n form a basis to represent any functions in the space of interest. The sum is over all the

states (and may be infinite). I is the identity operator. If you don’t understand this, talk to me – it takes less than 3 minutes to explain. Eq. 1 is often called the “completeness relation” (that’s part of the 3-minute explanation). Suppose we have two sets of basis kets. One is represented by Roman letters, and these kets are orthonormal among themselves. The other is represented by Greek letters and these are also orthonormal among themselves. (But the Greeks and the Romans are not orthogonal to one another!) Either the Greek or the Roman set would obey Eq. 1. The matrix representation of an operator would be different in the two different bases, but the trace of that matrix (the sum of the diagonal elements) is invariant. We are asked to show this. The (m,n) matrix element is m H n and the trace is therefore n H n

n∑ . Thus we must

prove: n H nn∑ = β H β

β∑ .

Start with the trace in the Roman basis and insert the identity, a perfectly harmless thing to do because it leaves the operator unchanged (similar to “multiplying by 1”):



n H nn∑ = n HI n

n∑ .

Now use the completeness relation (Eq. 1): n H n

n∑ = n HI n

n∑ = n H β β n

β∑

n∑

Now we note that the quantity on the right is the sum of the product of 2 numbers and the order of the numbers doesn't matter, so we'll interchange them:

n H nn∑ = n HI n

n∑ = β n n H β

β∑

n∑

Now recognize the identity operator again, but this time with the Roman kets:

n H nn∑ = β n n H β

β∑

n∑ = β IH β

β∑

and the result is proven: n H n

n∑ = β H β

β∑

What does it mean? This trace invariance shows up in many places in physics. You find it in the inertia tensor in 3-d rotations, in the tensors representing field gradients in various spectroscopies, and here in quantum mechanics. It is related to the fact that any operator represents a real, measurable quantity and thus the eigenvalues are the same, regardless of the representation.

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PH575: Introduction to Solid State Physics SPRING 2018 ...physics.oregonstate.edu/.../lib/exe/fetch.php?media=hw1soln_2018.pdf · PH575: Introduction to Solid State Physics SPRING