How many bits of information can be stored in a qubit? On making a measurement on a qubit, one can catch it in either $|0\rangle$ or $| 1\rangle$. Seems that measurement on a qubit can not reveal more than a single bit. But does it mean the a qubit can't store more than 1 bit of information?
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2 Answers
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No more than one. It follows from Holevo's theorem that $n$ qubits cannot be used to store more than $n$ bits of information. See for example these notes for an explanation of Holevo's theorem (interestingly, the main question answered by these notes is exactly the title of this question).
Very roughly speaking, the reason for this is that if you try to encode $n$ bits of information in a system of dimension $d<2^n$ (so, for example, in less than $n$ qubits), then there is no reliable way to retrieve such information. This is because if $d<2^n$, then there are too few orthogonal states on which to store the information (if two states are not orthogonal, they cannot be deterministically distinguished).
Note that this is very different than saying that a single qubit requires a single bit to be described. In fact, in general, you need an infinite amount of bits to describe the state of a single qubit, because a single qubit is characterised by continuous numbers. This means that, in some sense, a single qubit can encode an arbitrarily large amount of information. The problem is that you can't deterministically decode such information afterwards, using a single copy of said qubit. You could use many copies of the same qubits to completely reconstruct its state, and therefore recover all of the information encoded in it, but this turns out to be a pretty inefficient information storage scheme.
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$\begingroup$ Great answer. There should probably be some mention superdense coding, too. Superdense coding does not strictly pack more than n bits of information into n qubits, but it is a predictable source of confusion on the subject. en.wikipedia.org/wiki/Superdense_coding physics.stackexchange.com/questions/412558/… $\endgroup$ Dec 14, 2019 at 18:00
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$\begingroup$ Your answer has really been quite helpful. I was struggling in getting this. I just want to know one thing. Why do we need infinite bits to store the state of a qubit? And on measuring the qubit, we get one bit of information ( either 0 or 1 depending on the collapsed state). How can we construct the state of the qubit by repeated measurement when in effect we would be getting 0 or 1 again and again? $\endgroup$ Mar 1, 2021 at 15:05
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$\begingroup$ What do you mean by qubit sare characterised by continuous numbers? $\endgroup$ Mar 1, 2021 at 15:13
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$\begingroup$ @Shashaank you can open new threads to ask separate questions $\endgroup$– glSMar 1, 2021 at 19:34
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$\begingroup$ @Shashaank We need infinite bits to store the state of a qubit because the state of a qubit is a superposition of the state 0 and 1, which can be described as a weighted sum of the two states. The weights determine the phases and amplitudes of the qubit which determine the probability of observing the outcome 0 and 1 states. Because the weights have continuous values, we need infinite bits to encode them exactly, e.g., 0.14159265... $\endgroup$– THNJun 16, 2021 at 11:46
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What is really contained in a single qubit are 2 real numbers. This is the amplitude of the $|1\rangle$ state (the other one is given by the normalization), and the relative phase of the two states. When measured, we get out a single bit of information, as you correctly stated, but with which probability we get this specific bit is determined by these real numbers. Measuring equivilantly prepared states multiple times allows us to infer something about the 2 variables contained in the state.
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$\begingroup$ What do you mean by equivalently prepared states? It is impossible to have two copies of a qubit by the no-cloning theorem right? $\endgroup$– JamālOct 25, 2020 at 7:13
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1$\begingroup$ The no cloning theorem only states there cannot be a device which takes an arbitrary state as input and outputs two copies of that state. Nothing prevents us from just running a quantum circuit more than once. $\endgroup$– noahOct 25, 2020 at 10:47
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$\begingroup$ Hmm, i don't get it fully. How would you prepare equivalent states? Btw i should add that I'm not from a physics background so i only know some elementary stuff mathematically. $\endgroup$– JamālOct 25, 2020 at 11:47
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1$\begingroup$ Then have a look at how we prepare quantum states in the first place. I am talking about just doing that twice. Say, you have a very easy circuit: initialize to $|0\rangle$, apply a y-rotation gate by some angle and measure in the z-basis. Record your findings and do the experiment again. By the statistical distribution of many outcomes you can determine the amplitude of the 1 and the 0 state after that circuit. This is exaclty what the other answer is talking about. You need to run your whole setup multiple times, so you cannot (in one shot) extract more than one bit of information. $\endgroup$– noahOct 25, 2020 at 11:54
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$\begingroup$ Ok i guess i have to do a little more research on that. It seems to me that someone already has to know what the amplitudes will be in order to make the machine you described in which case they already have complete information about the quantum state but another person will have to measure the state and find out the amplitudes statistically. I guess the first person could make a ton of those states and the second person will measure them. Anyway thanks a lot for giving me a direction to keep thinking. $\endgroup$– JamālOct 25, 2020 at 13:02