Quantum Mechanics - Homework Assigment 1 Alejandro G´ omez omez Espinosa
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September 17, 2012
Shankar, Ex 1.4.2 Suppose Vn1 and Vn2 are two subspaces such that any element of Vn1 n V2 is is orthogonal to any element of Vn2 . Show Show that the dimensionality dimensionality of Vn1 n1 + n2 . 1
2
1
2
1
Let’s define the elements of Vn1 and 1
2
n
V2 2 :
n1
n1
|V = v |i
|W = w | j j
i
j
i=1
j =1
where i (i = 1,...,n1 ) and j j ( j = 1,...,n2 ) are orthonormal basis of Vn1 and Vn2 , respectively. Due to the orthogonality of the elements of the two subspaces, all states i and j j are orthogonal orthogonal as well. well. Therefore Therefore the set i , j j is an orthogonal basis of a vector space with dimension (n (n1 + n2 ).
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| |
|
1
2
||
On the other hand, we can set a linear combination of elements of Vn1
1
|Z = a|V + b|W where a and b are constants, and |V ∈ V , |W ∈ V n1
n2
1
2
n
V2 2 :
. Replacing with the previous
definitions:
n1
n2
n1
n2
|Z = a v |i + b w | j j = (av )|i + (bw )| j j i
i=1
2
j =1
j
1
i=1
j =1
Clearly we can set an orthogonal basis i , j j for this vector space which has a dimensionality (n (n1 + n2 ).
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2) Consider the set of 2.
G of all possible polynomials in x and y having maximum overall order
a) Show that this is a vector space and that its dimension is 6. Suppose that f, g and a6 , a5 , a4 , a3 , a2 , a1 and b6 , b5 , b4 , b3 , b2 , b1 are real. Then: f ( f (x, y) = a6 x2 + a5 xy + a4 y2 + a3 x + a2 y + a1
∈G
g (x, y) = b6 x2 + b5 xy + b4 y2 + b3 x + b2 y + b1 To show if is a vector space, we need to prove the 8 properties of a vector space. Let c and d be scalars:
G
1. (f + g )
∈G
f + g = a6 x2 + a5 xy + a4 y2 + a3 x + a2 y + a1 + b6 x2 + b5 xy +b4 y 2 + b3 x + b2 y + b1 = (a6 + b6 )x2 + (a (a5 + b5 )xy + (a (a4 + b4 )y2 + (a (a3 + b3 )x +(a +(a2 + b2 )y + (a (a1 + b1 )
∈ G 2. c(f + g) = cf + cg c(f + g) = c((a ((a6 + b6 )x2 + (a (a5 + b5 )xy + (a (a4 + b4 )y2 + (a (a3 + b3 )x +(a +(a2 + b2 )y + (a (a1 + b1 )) = c(a6 x2 + a5 xy + a4 y2 + a3 x + a2 y + a1 + b6 x2 + b5 xy + b4 y 2 +b3 x + b2 y + b1 ) = c(a6 x2 + a5 xy + a4 y2 + a3 x + a2 y + a1 ) + c(b6 x2 + b5 xy + b4 y2 +b3 x + b2 y + b1 ) = cf + cg 3. (c + d)f = cf + df (c + d)f = (c + d)[a )[a6 x2 + a5 xy + a4 y 2 + a3 x + a2 y + a1 ] = a6 (c + d)x2 + a5 (c + d)xy + a4 (c + d)y2 + a3 (c + d)x +a2 (c + d)y + a1 (c + d) = a6 cx2 + a5 cxy + a4 cy 2 + a3 cx + a2 cy + a1 c +a6 dx2 + a5 dxy + a4 dy2 + a3 dx + a2 dy + a1 d = c(a6 x2 + a5 xy + a4 y2 + a3 x + a2 y + a1 ) +d(a6 x2 + a5 xy + a4 y 2 + a3 x + a2 y + a1 ) = cf + df 4. c(df ) f ) = cdf cdf c(df ) f ) = c(a6 dx2 + a5 dxy + a4 dy2 + a3 dx + a2 dy + a1 d) = a6 cdx2 + a5 cdxy + a4 cdy2 + a3 cdx + a2 cdy + a1 cd = cd( cd(a6 dx2 + a5 dxy + a4 dy2 + a3 dx + a2 dy + a1 d) = cdf cdf 2
5. f + g = g + f f + g = a6 x2 + a5 xy + a4 y2 + a3 x + a2 y + a1 + b6 x2 + b5 xy +b4 y2 + b3 x + b2 y + b1 = b6 x2 + b5 xy + b4 y2 + b3 x + b2 y + b1 + a6 x2 + a5 xy +a4 y2 + a3 x + a2 y + a1 = g + f 6. Suppose h = c6 x2 + c5 xy + c4 y2 + c3 x + c2 y + c1 with c6 , c5 , c4 , c3 , c2 , c1 scalars. Then h + (f (f + g) = (h + f ) f ) + g h + (f (f + g) = c6 x2 + c5 xy + c4 y 2 + c3 x + c2 y + c1 +(a +(a6 x2 + a5 xy + a4 y2 + a3 x + a2 y + a1 +b6 x2 + b5 xy + b4 y2 + b3 x + b2 y + b1 ) (c6 x2 + c5 xy + c4 y2 + c3 x + c2 y + c1 +a6 x2 + a5 xy + a4 y2 + a3 x + a2 y + a1 ) +b6 x2 + b5 xy + b4 y2 + b3 x + b2 y + b1 = (h + f ) f ) + g 7. f + 0 = f f + 0 = a6 x2 + a5 xy + a4 y2 + a3 x + a2 y + a1 + 0 = a6 x2 + a5 xy + a4 y2 + a3 x + a2 y + a1 = f 8. f + ( f ) f ) = 0
−
f + ( f ) f ) = a6 x2 + a5 xy + a4 y2 + a3 x + a2 y + a1
−
+( a6 x2 = 0
−
2
− a xy − a y − a x − a y − a ) 5
4
3
2
1
Therefore, the set is a vector space. As we need six linear independent basis to write an element of , the dimension of this vector space is 6.
G
G
b) Now we add an inner product rule: we define f g to be the average value of f ∗ g on the unit circle, i.e., f g 21π u.c. f ∗ gdl. gdl . With this inner product product rule, does bec become an inner product product space?. space?. (Hint: (Hint: Be careful; careful; check all the conditions conditions for an inner product space.) To prove prove that is an inner inner product product space, space, we have have to chec check k the conditio condition n for such such space. space. In particul particular, ar, we can review review the condit condition ion f f 0 where 0 iff f = 0 . Let’s prove f = 0 0:
|
|
G
| | | | ⇒ 1 1 f |f = 2π f f dl = 2π 0dl = 0 Then, 0 ⇒ |f = |0. For this let’s consider this example:
G
|≥
∗
f ( f (x, y) = a6 x2 + a5 xy + a4 y2 + a3 x + a2 y + a1 3
where: a1 = (a4 + a6 )/2, a4 = a6 and a2 = a3 = a5 = 0. Hence: f = a6 x2 + a4 y2 + a1
f |f
1 2π 1 2π
= =
f f dl ∗
(a26 x4 + a24 y 4 + a21 )dl 2
where x = cosθ and y = sinθ
a + a |a − a | a + 2 + 8 4
=
6
4
6
2
1
applying applying our conditions, conditions, we found that f f = 0. It means means that there there is a nonzero polinomial with a inner product equal to zero, contrary of our postulate. Then, is not a inner product space.
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G
c) Let’s consider the 5-dimensional inner product space using the inner product of part (b) and spanned by the basis vectors 1 = 1, 2 = 2x, 3 = 2y, 4 = 2(x 2(x2 y 2 ), and 5 = 2 2xy. xy. Check that this is an orthonormal basis. We can prove prove that that this set of basis basis are orthonor orthonormal mal,, using using definiti definition on 10 of Shankar’s book. It tells us that A set of basis vectors all of unit norm, wich are pairwise orthogonal will be called an orthonormal basis. Then we have to prove that i j j = δ ij ij . (Remember that x = cos θ and y = sin θ ).
| √
−
|
√
|
√ |
|
√
|
2π
1 11 = 2π
1 | dl = dθ = 1 2π sin2θ 1 1 1 θ sin2θ 2|2 = 2π 2x dl = π cos θdθ = π 2 + 4 | = 1 sin2θ 1 1 1 θ sin2θ 3|3 = 2π 2y dl = π sin θdθ = π 2 − 4 | = 1 1 1 2(x + y ) dl = (x − 2x y + y )dl 4|4 = 2π 2(x π 1 1 = (cos θ − 2sin cos θ + sin θ)dθ = (4θ (4θ + sin sin 4θ)| = 1 π 8π 1 4 4 1 0
2π
2
2π 0
0
2π
2
2π 0
0
2
2π
2 2
4
2
2
4
0
2π
4
2
2
2π 0
4
0
5|5 = 2π As i| j j = δ
ij ij
4(2)x 4(2)x2 y2 dl =
2π
π
sin2 θ cos2 θdθ =
0
is true, this is an orthonormal basis.
4
π
32
(4θ (4θ
2π 0
sin4θ) | − sin4θ
=1
d) Find the matrix representation in the basis (i.e., write out the 5 the following linear operators:
× 5 matrix) for
M (”45 (”45◦ mirror” operation) associated with the replacement x y; To build this matrix, we will show the M i value and then find the matrix representation. M 1 = 1
| | | √ M |2 = 2y = |3 √ M |3 = 2x = |2 √ √ M |4 = 2(y 2(y − x ) = − 2(x 2(x − y ) = −|4 √ M |5 = 2 2yx = |5 2
2
2
↔
2
Then, the M matrix is:
1 0 M = 0 0
0 0 1 0 0 0
0 1 0 0 0
0 0 0 1 0
−
0 0 0 0 1
R (”rotation operation) associated with the replacement x
→ y, y → −x;
R1 = 1
| | √ R|2 = 2y = |3 √ R|3 = − 2x = −|2 √ √ R|4 = 2(y 2(y − (−x) ) = − 2(x 2(x − y ) = −|4 √ R|5 = −2 2yx = −|5 2
2
2
2
Then, the R matrix is:
1 0 R = 0 0
0 0 1 0 0
−
0
0 1 0 0 0
0 0 0 1 0
−
0 0 0 0 1
−
O associated with eliminating all even-order terms in the polynomial; O|1 = 0 √ O|2 = 2x = |2 √ O|3 = 2y = |3 √ O|4 = 2(0 − 0) = 0 √ O|5 = 2 2xy = |5 5
Then, the
O matrix is:
0 0 O = 0 0
0 1 0 0 0 0
0 0 1 0 0
0 0 0 0 0
0 0 0 0 1
S associated with the replacement x
x/2, y → y/2 y/2; → x/2 S |1 = |1 1 √ 1 S |2 = 2y = |2 2 2 √ 1 1 S |3 = 2x = |3 2 2 √ √ x y 2 1 4 = 2 = (x − y ) = |4 − S | 2 2 4 4 √ x y 1 S |5 = 2 2 = |5 22 2 2
2
2
2
Then, the S matrix is:
1 0 S = 0 0
0 0 0 0 1/2 0 0 0 1/2 0 0 0 0 1/4 0 0 0 0 0 1/2
W associated with the replacement f ( f (x, y)
Then, the W matrix is:
if (x, y). → if ( W |1 = i|1 W |2 = i|3 W |3 = i|2 W |4 = i|4 W |5 = i|5
i 0 W = 0 0
0 i 0 0 0 0
0 0 i 0 0
0 0 0 i 0
0 0 0 0 i
e) Which Which of the oper operato ators rs in part (d) are are Hermit Hermitian ian? ? Which Which are are Antih Antiherm ermiti itian? an? Which are unitary? Which are projection operators? Which are reflection operators? An operator V is Hermitian if V = V † = (V ∗ )T . Then, Then, it is easy easy to calculate calculate the transpose transpose conjugate conjugate matrix matrix of the previou previouss operator operators. s. I found found that M, and S are hermitian; R is unitary and W is anti-hermitian.
O
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f ) From among the operators in part (d), find a pair of Hermetian operators which form a ”complete ”complete set of commuting operators” operators” (see Shankar p.46) and find the orthonormal basis that simultaneously diagonalizes both operators. After several tries, we found that the pair of the operators R and S form a complete set of commuting operators. The eigenvalues of R are: +1, +1, 1, i, +i and for S are: 1/ 1/2, 1, 1/4. With this eigenvalues we can parametrized the set:
− −
1 0 |1, 1 = 0 ; 0 0
0 0 0 −i 1 |1/4, 1 = 0 ; |1/2, i = √ 2 1 ; 1 0 0
0
0 0 i 0 1 |1/2, −i = √ 2 1 ; |1/2, −1 = 0 ; 0 0 0
1
This represents the orthonormal basis that diagonalizes the two operators. Shankar, Ex 1.6.2 Given Ω and Λ are Hermitian what can you say about (1) ΩΛ; ΩΛ; (2) ΩΛ + ΛΩ; ΛΩ; (3) [Ω, [Ω, Λ]; Λ]; and (4) i[Ω, [Ω, Λ]? Λ]? From the definition of Hermitian: Ω = Ω† . Then: 1. (ΩΛ)† = Λ† Ω† = ΛΩ. Hence ΩΛ = ΛΩ, ΩΛ is not Hermitian.
2. (ΩΛ + ΛΩ) ΛΩ)† = (ΩΛ)† + (ΛΩ)† = Λ† Ω† + Ω† Λ† = ΛΩ + ΩΛ. Therefore this operator is Hermitian. 3. From the definitio definition: n: [Ω, [Ω, Λ] = ΩΛ ΛΩ. † † [Ω, [Ω, Λ] = (ΩΛ ΛΩ) = (ΩΛ)† (ΛΩ)† = ΛΩ Then, this operator is anti-Hermitian.
−
−
−
4. Similar Similar as (3): (3): † (i[Ω, [Ω, Λ]) = i(ΩΛ ΛΩ)† = i((ΩΛ)† [Ω, Λ]. This operator Hermitian. i[Ω,
−
−
−
[Λ, Ω] = −[Ω, [Ω, Λ]. − ΩΛ = [Λ, †
− (ΛΩ) ) = −i(ΛΩ − ΩΛ) = −i[Λ, [Λ, Ω] =
Shankar, Ex 1.6.3 Show that a product of unitary operators is unitary. Suppose U and V are unitary operators, hence U † U = 1 and V † V = 1. Then: (U V ) V )† U V = V † U † U V = V † V = I Therefore this product is unitary.
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Shankar, Ex 1.6.4 It is assu assume med d that that you know know (1) (1) what what a dete determ rmin inan antt is, is, (2) (2) that that T detΩ = detΩ (T denotes transpose), (3) that the determinant of a product of matric trices is the produc productt of the determin determinant ants. s. Prove Prove that the determin determinant ant of a unitar unitary y matrix is a complex number of unit modulus. Using the results from 1.6.3.: det(U det(U † U ) U ) = det(I det(I ) = 1 But, also: det(U det(U † U ) = det(U det(U † )det(U )det(U )) = det((U det((U T )∗ ) det( det(U U ) = det(U det(U ∗ ) det( det(U U ) = det U 2
|
|
Then: 2
| det U |
=1
| det U | = 1
⇒
. Shankar, Ex 1.9.2 If H is a Hermitian operator, show that U = eiH is unitary. (Notice the analogy with c numbers: if θ is real, u = eiθ is a number of unit modulus.) Suppose U = eiH is not unitary. As a result, the product U † U = 1. Then,
U † U = e−iH eiH = e−iH +iH = e0 = 1 That is against our supposition. Then U must be unitary. Shankar, Ex 1.9.3 For the case above, show that det U = eiTr H It is known that any Hermitian matrix can be diagonalized by a unitary transformation. If we diagonalize H and take ε1 ,...,εm as the eigenvalues of H, then: det U = det(e det(eiH ) = ei
8
m
j =1
εj
= eiTrH
