PART B (10 points possible)
Consider the operator S on an n-dimensional complex vector space so that
S(z1,z2,…,zn)=(0,z1,z2,…zn−1).
Assume the inner product is the standard one:
⟨(x1,…,xn),(y1,…,yn)⟩=x∗1y1+…+x∗nyn.
Calculate S†(w1,w2,…,wn) and fill in the boxes.
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To calculate the adjoint of an operator S, denoted as S†, we need to find the linear transformation that satisfies the equation ⟨S(x), y⟩ = ⟨x, S†(y)⟩ for all vectors x and y in the vector space.
In this case, we are given the operator S defined as:
S(z1, z2, ..., zn) = (0, z1, z2, ..., zn-1)
To calculate S†, we need to find the transformation that satisfies:
⟨S(x), y⟩ = ⟨x, S†(y)⟩
Let's start by substituting the definition of S and S† into this equation:
⟨(0, x1, x2, ..., xn-1), (y1, y2, ..., yn)⟩ = ⟨(w1, w2, ..., wn), (0, y1, y2, ..., yn-1)⟩
Expanding the left side using the standard inner product definition, we have:
0 * y1 + x1 * ¯y1 + x2 * ¯y2 + ... + xn-1 * ¯yn-1 = w1 * 0 + w2 * ¯y1 + w3 * ¯y2 + ... + wn * ¯yn-1
Notice that we only have non-zero terms on the diagonal and one term shifted to the right for each element. Therefore, we can simplify the equation to:
x1 * ¯y1 = w2 * ¯y1
x2 * ¯y2 = w3 * ¯y2
...
xn-1 * ¯yn-1 = wn * ¯yn-1
From these equations, we can see that for each element xi, we have:
xi = 0 if i ≤ 1
xi = wi+1 if i > 1
Now we can fill in the boxes for S†(w1, w2, ..., wn):
S†(w1, w2, ..., wn) = (0, w2, w3, ..., wn)
Therefore, the adjoint of the operator S is given by:
S†(w1, w2, ..., wn) = (0, w2, w3, ..., wn)