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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.

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  • $\begingroup$ Do you know any methods for calculating the probabilities of measurement outcomes? $\endgroup$ – DaftWullie Sep 2 at 6:53
  • $\begingroup$ I know a general measurement probability if ket-psi = a|0> + b|1> then prob. of outcome 0 is (|a|)^2 $\endgroup$ – Chaitanya Reddy Sep 2 at 9:47
  • $\begingroup$ 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? $\endgroup$ – czwang Sep 2 at 15:17
  • $\begingroup$ yes they are same @czwang and you can assume N = 2^m $\endgroup$ – Chaitanya Reddy Sep 2 at 15:55
  • $\begingroup$ @czwang Is it a valid entangled state? if do sum of squares of coefficients of \left|j\right>_h\left|j\right>_t$$ we should get 1 right? $\endgroup$ – Chaitanya Reddy Sep 3 at 7:47
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Given your clarification in your comments (you might want to update your question to add these assumptions), I believe it is a valid entangled state of $2m$ qubits. If you are measuring the second set of $m$ qubits $|j\rangle_t$ in the computational basis, you will get the measurement result of $k$ (for $k \in \{0,\ldots,N-1\}$) with probability $$|\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.

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