I've been reading about the standard phase estimation from the Qiskit tutorial and got stuck in interpreting the final state representation. What does the state $|2^t\theta\rangle$ mean? Is that the binary representation of the most probable state? Thanks for the help! enter image description here


2 Answers 2


There are "2" cases for the final result:

  1. The outcome imply that $2^t\theta=Integer$, in that case, yes, exactly, it will be a state with the binary encoding of $2^t\theta$
  2. In case $2^t\theta=NotInteger$, QPE will result (before measurement) to a state that has very high probability amplitude in the state nearest to the desired angle that you are looking for.

Schematically, before and after QFT (before measurement): enter image description here

Is mapped after measurement:

enter image description here

Notice that the term: enter image description here

will be close to 1 when $x$ is close to $2^t\theta$ , what will end and terms that do not (or just less) cancel each other after all the sum of the terms:

enter image description here


Given a unitary operator $U$ the algorithm estimates $\theta$ in $U|\psi \rangle = e^{2\pi i \theta} |\psi\rangle$. So $\theta$ is an angle that determines a complex eigenvalue $e^{2\pi i \theta}$ associated with an eigenvector $|\psi\rangle$ of $U$. If you know $\theta$ you know the eigenvalue of $U$.

QPE outputs a state $| 2^t \theta \rangle$ which is an integer (even if $\theta$ is not an integer). Of course, your integer will be represented in binary.

The notation $| 2^t \theta \rangle$ is a bit confusing, cause if $\theta$ was non-integer, say $\theta =0.2$, then we would be looking at $| 2^t 0.2\rangle$. One can infer that this state must be interpreted as a "rounded" to the nearest integer. For example, for $t=2$ and $\theta=0.2$ we would measure (with high probability) $|1\rangle$ because $| 2^2 0.2\rangle = |0.8\rangle \approx |1\rangle$ or in binary $|01\rangle$.

The conclusion is:
In the case when $\theta$ is an integer, the measurement on $t$ qubits yields the state $| 2^t \theta \rangle$ with certainty. Then you just solve for $\theta$.
If $\theta$ is not an integer then you will measure the closest integer to $2^t \theta$ with a probability $\geq 0.4$. Every measurement will be a binary encoding of some integer.


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