Understanding Qubits: How Superposition Powers Quantum Computers

[T-Oreilly]

Hi,

I'm having trouble understanding the power of qubits relating to quantum computers. I've read a number of times that the power comes from the fact that instead of simply holding an on or off state (1/0), they can hold both at the same time (superposition). However, when we measure them they 'decide' on a state.

My question is, how can this third state of superposition provide a huge benefit over the 1 and 0 states of transistors, since once we use them (observe them) the states available to us is still only a 1 or a 0?

Many thanks in advance http://www.thephysicsforum.com/images/smilies/smile.png

 

[Heinera]

It's because the system of qubits can do a lot of computations before you measure the final result, and during this time, it will be in a superposition of states.

See more here: https://en.wikipedia.org/wiki/Quantum_computing

And if you have time to spare, these lecture notes by Scott Aaronson:
http://www.scottaaronson.com/democritus/

 

[Hornbein]

And if you have time to spare, these lecture notes by Scott Aaronson:
http://www.scottaaronson.com/democritus/

I'm looking at lecture 9 and there are many gaps. Maybe he's giving a slide show and the slides are not included.

 

[Strilanc]

Quantum computers are not ternary computers. Qubits are not just bits with a third state.

A qubit is a 2-level quantum system that can store states like $a \left|0\right\rangle + b \left|1\right\rangle$ where $a^2 + b^2 = 1$.

Put $n$ qubits together, and you get a $2^n$-level quantum system. For example, 3 qubits can store states like $a \left|000\right\rangle + b \left|001\right\rangle + c \left|010\right\rangle + d \left|011\right\rangle + e \left|100\right\rangle + f \left|101\right\rangle + g \left|110\right\rangle + h \left|111\right\rangle$ where $a^2 + b^2 + c^2 + d^2 + e^2 + f^2 + g^2 + h^2 = 1$.

In other words, quantum computers can store a linear combination of the classical states. But the weights of the linear combination are not probabilities, which would have to satisfy $a + b + ... + h = 1$, they are the square roots of probabilities and must satisfy $a^2 + b^2 + ... + h^2 = 1$.

Everything else flows from that square-root-of-probability thing. Operations correspond to complex orthonormal matrices. Destructive interference is possible. Copying doesn't quite work. Everything is reversible. Entanglement is a thing. Etc.

 

[Heinera]

I'm looking at lecture 9 and there are many gaps. Maybe he's giving a slide show and the slides are not included.

Do you mean that you don't see any figures or formulas?

 

[Hornbein]

Do you mean that you don't see any figures or formulas?

I see

ask me to
exp
the Bell inequality to them.

BUT when I copy this from the lecture notes and paste here, it comes out correctly!

ask me to explain the Bell inequality to them.

So I can paste the entire lecture to a Physics Forums reply box and read it that way. Golly. Well, whatever works, works.