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Thanks to comment by @gIS, I realized that I was mixing up the order. If I write $j$ as $|j1\cdots j_t\rangle$$|j_1\dots j_t\rangle$, of course it will be equal to $j_12^{t-1} \dots j_12^0$$j_12^{t-1} \cdots j_t2^0$. I was confused about the numbering of the qubits.

Thanks to comment by @gIS, I realized that I was mixing up the order. If I write $j$ as $|j1\cdots j_t\rangle$, of course it will be equal to $j_12^{t-1} \dots j_12^0$. I was confused about the numbering of the qubits.

Thanks to comment by @gIS, I realized that I was mixing up the order. If I write $j$ as $|j_1\dots j_t\rangle$, of course it will be equal to $j_12^{t-1} \cdots j_t2^0$. I was confused about the numbering of the qubits.

Source Link
usercs
usercs
  • 471
  • 2
  • 9

Thanks to comment by @gIS, I realized that I was mixing up the order. If I write $j$ as $|j1\cdots j_t\rangle$, of course it will be equal to $j_12^{t-1} \dots j_12^0$. I was confused about the numbering of the qubits.