# What is the probability of measuring $|j\rangle$ with $j\in\{0, 1, 2, … N-1\}$?

Consider the state $$\frac{1}{\sqrt{N}}\sum_{j=0}^{N-1}e^{2\pi i\frac{x_1}{N}j}\left|j\right>_h\left|j\right>_t.$$

In the above entangled state, what is the probability of getting $$|j\rangle_t$$ where $$j$$ belongs to $$\left\lbrace 0, 1, 2, \ldots, N-1\right\rbrace$$? The 1st $$m$$ qubits are $$|j\rangle_h$$ and 2nd m qubits $$|j\rangle_t$$. We are measuring 2nd $$m$$ qubits here.

• Do you know any methods for calculating the probabilities of measurement outcomes? – DaftWullie Sep 2 '19 at 6:53
• I know a general measurement probability if ket-psi = a|0> + b|1> then prob. of outcome 0 is (|a|)^2 – Chaitanya Reddy Sep 2 '19 at 9:47
• The notation here seems to be a bit confusing. First I assume $N=2^m$. Second, you have a single index j, but two set of qubits using that index, are these two sets of qubits always the same? – czwang Sep 2 '19 at 15:17
• yes they are same @czwang and you can assume N = 2^m – Chaitanya Reddy Sep 2 '19 at 15:55
• @czwang Is it a valid entangled state? if do sum of squares of coefficients of \left|j\right>_h\left|j\right>_t we should get 1 right? – Chaitanya Reddy Sep 3 '19 at 7:47

Given your clarification in your comments (you might want to update your question to add these assumptions), I believe it is a valid entangled state of $$2m$$ qubits. If you are measuring the second set of $$m$$ qubits $$|j\rangle_t$$ in the computational basis, you will get the measurement result of $$k$$ (for $$k \in \{0,\ldots,N-1\}$$) with probability $$|\frac{1}{\sqrt{N}} e^{2\pi i \frac{x_1}{N} k}|^2=\frac{1}{N} e^{2\pi i \frac{x_1}{N} k} e^{-2\pi i \frac{x_1}{N} k} =\frac{1}{N},$$ and the first $$m$$ qubits $$|j\rangle_h$$ also in state $$|k\rangle$$, which gives $$k$$ upon measurement.