Suppose we ran m steps of Grover's algorithm on some function f (which has one marked element y) and the resulting superposition was exactly |y⟩.
If we run for m+1 additional steps (i.e. total of 2m+1 steps from the initial state), what is the resulting superposition? Note that you can describe the superposition by specifying two numbers, αy and αx for x≠y. You may use K to denote the total number of elements. (K should be uppercase.) HINT: You may want to review Problems 2, 3, and 5 of Assignment 6.
Now if we run for another m steps, what is the resulting superposition?
What about after yet another m+1 steps?
This is fairly simple. You just need to remember that Grover's algorithm is cyclical
You start out with an equal superposition for both the y x answers
0 steps: ay = 1/sqrt(K), ax = 1/sqrt(k)
then after m steps, you solve it
@m steps: ay - 1, ax = o
Since you started from 0 instead of 1, it takes M+1 steps to get it back to the original position, but it's inverted
@2m+1 steps: ay = -1/sqrt(K), ax = -1/sqrt(k)
Then it takes another m steps to find the answer (but inverted)
@3m+1 steps: ay = -1, ax = 0
and then m+1 steps later, you're back to the initial position.
@4m+2 steps: ay = 1/sqrt(K), ax = 1/sqrt(k)
and the cycle repeats.
@5m+2 steps: ay = 1, ax = 0
@6m+3 steps: ay = -1/sqrt(K), ax = -1/sqrt(k)
To determine the resulting superposition after running m+1 additional steps of Grover's algorithm, let's analyze the algorithm step by step. Here's the step-by-step explanation:
1. Initial State: The initial state of Grover's algorithm is a uniform superposition of all elements. For simplicity, let's assume there are K total elements, so the initial state can be represented as |ψ⟩ = (1/√K)Σ|x⟩.
2. Oracle Application: In the first m steps, we apply the Grover oracle function to the initial state |ψ⟩. This oracle function marks the desired element y. After m steps, the state becomes |ψ'⟩ = -|y⟩ + (√(K-1)/√K)Σ|x⟩ (negative sign on the marked element).
3. Inversion about the Average: After applying the oracle, we perform the inversion about the average operation. This step involves reflecting the state |ψ'⟩ about the average amplitude of all elements.
a. The average amplitude is calculated as 2(1/K)Σ|x⟩, which is 2 times the initial state |ψ⟩.
b. We subtract the average from |ψ'⟩ and multiply by the reflection matrix.
c. Applying the inversion about the average operation, the state becomes |ψ''⟩ = 2(1/K)|y⟩ - (2√(K-1)/K)Σ|x⟩.
4. Result after m+1 Steps: To find the superposition after m+1 steps, we repeat steps 2 and 3 one more time.
a. After applying the oracle for the second time, the state becomes |ψ'⟩ = -2(1/K)|y⟩ - [(2√(K-1)/K) - (2(1/K))]|x⟩.
b. Simplifying the expression, we have |ψ'⟩ = -2(1/K)|y⟩ - 2√(K-1)/K)|x⟩.
c. This superposition can be represented by the amplitudes αy = -2(1/K) and αx = -2√(K-1)/K.
5. Result after m+2 Steps: If we continue running for another m steps, the process is repeated.
a. After applying the oracle for the third time, the state becomes |ψ'⟩ = 2(1/K)|y⟩ - [(2√(K-1)/K) - (2(1/K))]|x⟩.
b. Simplifying the expression, we have |ψ'⟩ = 2(1/K)|y⟩ - 2√(K-1)/K)|x⟩.
c. This superposition can be represented by the amplitudes αy = 2(1/K) and αx = -2√(K-1)/K.
6. Result after m+3 Steps: Following a similar pattern, after each additional m steps, the superposition repeats with alternating signs for αy and αx.
To summarize, the resulting superposition after running m+1 steps is characterized by αy = -2(1/K) and αx = -2√(K-1)/K. After running m+2 steps, it becomes αy = 2(1/K) and αx = -2√(K-1)/K. The pattern continues with alternating signs for αy and αx after each additional m steps.