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I am looking into quantum computing, and have seen many say that a classical computer can consider 2n states (with n bits) at any single point in time.

My problem is that usually there is no explanation to how that is. I might be confusing myself, but to me it seems weird that in this case a classical computer can consider 128 states with 64 bits.

Can anybody explain the logic?

EDIT: See this video from IBM, https://youtu.be/WVv5OAR4Nik (skip to 1:37)

Moggel
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    2^n; with 64 bits you have quite a few more than 128 possible states, which would only require 7 bits... – jonrsharpe Jan 19 '20 at 17:28
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    It is not possible combinations, but the states representable by n bits at a single point in time. Try watching the video linked to. – Moggel Jan 20 '20 at 19:06
  • Quote from the video at the time where 2n is shown: *»[a classical computer] can represent and examine only one system state at a time«*. **¶** I also have no idea what the 2n is supposed to mean if it is neither the number of states processed at a time (1 for a classical computer) nor the number of states representable with n bits (2^n for a classical computer). – Socowi Jan 21 '20 at 22:03
  • https://youtu.be/lB_5pC1MkGg discusses this issue (from 1:38 - until 4:36). Also, related posts: https://physics.stackexchange.com/q/72765, https://cs.stackexchange.com/q/87870, and there is a discussion under the pinned comment here: https://youtu.be/aPCZcv-5qfA – Vladimir Fokow Oct 07 '22 at 01:55
  • https://youtu.be/zhQItO6_WoI?t=332 explains this from 5:32 until 6:20. As I understand, the n "classical/traditional bits" can take values only either 0 or 1, but the bits of "the quantum system" - are the amplitudes of each of 2^n possible states (so they are complex numbers whose squared norms must add up to 1). Am I correct? – Vladimir Fokow Oct 07 '22 at 02:26

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I'm guessing that something went wrong with the formatting of the article you read. n bits can be used to represent 2n ("two to the power of n") states - every bit has two states (0 and 1), and if you have n of them, you'd have 2n optional combinations - the first bit has two options, multiplied by the two options of the second bit, and so on, n times, giving 2n.

Mureinik
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    I don't hope so, it is probably my explanation of my problem then. It is based off of a video released by IBM about two years ago, see the video added to my question. – Moggel Jan 20 '20 at 19:09