# Could we use varying voltage with programmable gates?

One of the benefits I'm reading about qubits is that they can be in an infinite number of states. I'm aware of Holevo's bound (even though I don't fully understand it). However, it made me think of why we haven't tried varying voltage on classical computers and have programmable gates to control what passes in terms of a voltage. In that way, we could simulate quantum computing more closely.

This reminds me of another question we had here: What's the difference between a set of qubits and a capacitor with a subdivided plate?

One of the benefits I'm reading about qubits is that they can be in an infinite number of states.

Yes, qubits don't have to be in state 0 or state 1, but can be in an infinite number of states represented by something called the Bloch sphere. However, while it sounds impressive that you can be in an infinite number of states, this alone is not what gives quantum computers their full power!

Indeed, an analog classical computer can be in an infinite number of states too.

In order to really get the full power of a quantum computer, you need at least 2 qubits. The pair of qubits can exist in a state that no digital or analog classical computer can be in, which is a mixture of being in (0,0) and (1,1) at the same time. I explained this in my answer to that previous question: What's the difference between a set of qubits and a capacitor with a subdivided plate?

I'm aware of Holevo's bound (even though I don't fully understand it).

If you have a specific question on Holevo's theorem, or Ashwin Nayak's generalization of it, I'm sure you'll get an answer if you ask that as a separate question here :)

However, it made me think of why we haven't tried varying voltage on classical computers and have programmable gates to control what passes in terms of a voltage.

I suppose you are suggesting this because voltage can be in any state between, for example, 0V and 20V (i.e. can be 0, or 1, or anything in between, like maybe 0.5 would be 10V). Qubits can also be in an infinite number of states other than 0 or 1, but that is not what gives them their full power. The power comes from how two qubits can interact.

If you have a pair of voltage-based bits, can the pair be in a state where the two bits are 0 and the two bits are also 1, at the same time?

You can have the two bits being in the states:
(0V, 0V), or
(0V, 20V), or
(20V, 0V), or
(20V, 20V), or, since you want to allow an infinite number of voltages,
(15V, 12.3V),

but you cannot have: $$\frac{1}{\sqrt{2}}\left[(0\rm{V},0\rm{V}) + (20\rm{V},20\rm{V})\right]$$

which means you're both in the (0V,0V) state and the (20V,20V) state at the same time (like Schrodinger's cat is alive and dead at the same time).

In conclusion: Even the ability to be in an infinite number of different states (like in analog classical computing), is not enough to do what a quantum computer can do!

• Could 2 qubits be simulated w/ 2 pairs of voltage-based bits? – user820789 Oct 30 '18 at 17:00
• Unfortunately not. To simulate quantum mechanical things (qubits) with classical hardware (voltage-based bits), you need to simulate a $2^n$ dimensional vector for $n$ qubits. The number of voltage-based bits you need is therefore larger than 2, except if n=1. This is also why 1 qubit (as suggested in this question) is not enough to demonstrate the full power of quantum computing! Great question! I had not thought of that before! – user1271772 Oct 30 '18 at 22:49
• If simulating 1 qubit requires 2 v-based bits, wouldn't simulating 2 qubits require 4 v-based bits (phrased differently, 2 pairs) [$2^2=4$]? Would a qutrit be simulated w/ $3^n$ v-based bits? Qudit w/ $k^n$ v-based bits? – user820789 Oct 31 '18 at 0:21
• @meowzz: You are right! 1 qubit = 2 bits, 2 qubits = 4 bits. For a qutrit we need 3 bits. For 2 qutrits we need 9, and for 3 qutrits we need 27 bits. For qudits we need $k^n$. You are correct on all 4 accounts. – user1271772 Oct 31 '18 at 0:42