Exercise 1 (Function vector spaces and operators)
Let us assume the vector space of polynomial functions defined on [−1,1] such that the degree of the polynomials are at most 2 i.e. the P(3) space.
An arbitrary function in this space can be written as
∣f⟩=c0+c1x+c2x2. This means we can take the functions
∣p0⟩∣p1⟩∣p2⟩=1=x=x2 as a set of basis functions for this space. Although any vector from the space can be written as a linear combination of ∣pj⟩, they are unfortunately not orthonormal. In this problem, we see how we can orthonormalize them using the Gram–Schmidt procedure. Using this procedure, we generate three vectors ∣e0⟩, ∣e1⟩, and ∣e2⟩ based on ∣pj⟩, such that they are orthonormal. In the rest of this problem, assume the inner product is defined as
⟨f∣g⟩=∫−11f∗(x)g(x)dx and the norm (or length) of a vector is defined as
∥∣f⟩∥=⟨f∣f⟩ The Gram-Schmidt procedure starts by normalizing the first basis
∣e0⟩=∥∣p0⟩∥∣p0⟩. This vector is normalized because
⟨e0∣e0⟩=(∥∣p0⟩∥1)2⟨p0∣p0⟩=⟨p0∣p0⟩1⟨p0∣p0⟩=1 Derive e0(x) using Eq. 5.
Solution
We should calculate the normalization factor
⟨p0∣p0⟩=∫−11(1)(1)dx=2 This means, the first basis function is
∣e0⟩=21
To find the next basis vector ∣e1⟩, we cannot just choose it to be proportional to ∣p1⟩ as we did for ∣e0⟩ because ∣p1⟩ might not be orthogonal to ∣p0⟩ which then means ⟨e0∣e1⟩∝⟨p0∣p1⟩=0.
In order to avoid this issue, we first construct a function ∣e1′⟩ by removing any component of ∣e0⟩ there is in ∣p1⟩ as
∣e1′⟩=∣p1⟩−⟨e0∣p1⟩∣e0⟩ and then normalize this new function
∣e1⟩=∥∣e1′⟩∥∣e1′⟩ to get our next basis function e1(x). This works because ∣e1′⟩ is constructed in a way that is orthogonal to ∣e0⟩
⟨e0∣e1′⟩=⟨e0∣p1⟩−⟨e0∣p1⟩1⟨e0∣e0⟩=0. Derive e1(x).
Solution
We first calculate
⟨e0∣p1⟩=∫−11(21)(x)dx=0 This means the non-normalized basis function is e1′(x)=x−(0)(21)=x. The normalization factor will be
⟨e1′∣e1′⟩=∫−11(x)(x)dx=32 And finally, the normalized basis function is
e1(x)=23x
Using a similar logic, we can construct
∣e2′⟩=∣p2⟩−⟨e0∣p2⟩∣e0⟩−⟨e1∣p2⟩∣e1⟩ and our final basis vector will be
∣e2⟩=∥∣e2′⟩∥∣e2′⟩ by normalization.
Show that the ∣e2′⟩ as defined in Eq. 15 is orthogonal to the previous basis vectors we had (∣e0⟩ and ∣e1⟩) by showing ⟨e0∣e2′⟩=0 and ⟨e1∣e2′⟩=0. (You do not need to calculate the integrals for this. Use steps similar to Eq. 11)
Solution
We have
⟨e0∣e2′⟩=⟨e0∣p2⟩−⟨e0∣p2⟩1⟨e0∣e0⟩−⟨e1∣p2⟩0⟨e0∣e1⟩=⟨e0∣p2⟩−⟨e0∣p2⟩=0. Similarly,
⟨e1∣e2′⟩=⟨e1∣p2⟩−⟨e0∣p2⟩0⟨e1∣e0⟩−⟨e1∣p2⟩1⟨e1∣e1⟩=⟨e1∣p2⟩−⟨e1∣p2⟩=0. Derive e2(x).
Solution
First, we calculate the inner products
⟨e0∣p2⟩=∫−11(21)(x2)dx=2132 ⟨e1∣p2⟩=∫−11(23x)(x2)dx=(23)(0)=0 Which leads to
e2′(x)=x2−(0)(23x)−(2132)(21)=x2−31 Now we have to normalize this
⟨e2′∣e2′⟩=∫−11(x2−31)(x2−31)dx=∫−11x4−32x2+91dx=52−3232+92=458 Which means the normalized basis function is
e2(x)=845(x2−31)=85(3x2−1)
Final solutions
The final functions are
e0(x)=21,e1(x)=23x,e2(x)=85(3x2−1). Note that the new basis ∣ej⟩ are constructed step by step in a way that they are all mutually orthogonal to each other and they are normalized. Therefore we have an orthonormal basis now!
Since the set of 3 basis vectors ∣ej⟩ span the space, the dimension of the vector space P(3) is d=3. This means that linear operators in this space can be represented as 3×3 matrices. We know the derivative operator D^=∂x∂ is a linear operator in this space. Use the relation
Dij=⟨ei∣D^∣ej⟩ to derive the matrix representation of the D^ operator using this basis vectors
D=⎝⎛D00D10D20D01D11D21D02D12D22⎠⎞. Discuss whether this matrix is Hermitian (D†=D) or skew-Hermitian (D†=−D) or neither.
Solution
First, we calculate the derivative of each basis function
D^∣e0⟩D^∣e1⟩D^∣e2⟩=0=23=325x Now, the systematic way to calculate Dij elements is to calculate the inner product of these new functions with each of the basis states. This will be 9 integrations in total. But because we know the representation of any function in the basis is unique, it suffices if we can manually write these new functions as a linear combination of the basis functions.
D^∣e0⟩D^∣e1⟩D^∣e2⟩=0=0∣e0⟩+0∣e1⟩+0∣e2⟩=23=3∣e0⟩+0∣e1⟩+0∣e2⟩=85x=0∣e0⟩+15∣e1⟩+0∣e2⟩ Which means the matrix D will be
D=⎝⎛0003000150⎠⎞. As we can see, this matrix is neither Hermitian nor skew-Hermitian.