Understand Decoherence in Quantum Computers: How Heat Affects Qubits

[michael879]

ok I just read something that doesn't make much sense to me. It said that quantum computers can only use reversible logic gates because irreversible gates generate heat/lost information. It makes sense to me that everything in physics is reversible and that NAND gates produce heat because there is a loss of information but what I don't get is how the heat generated by a NAND gate in a quantum computer would cause the computer to stop working.

This article said that the heat would cause decoherence in the qubits which makes sense since the heat gives information about their state. What I don't get is how that would effect a quantum circuit. The qubits are still randomly 1 or 0 with whatever probability you made them so wouldn't whatever algorithm your running still come out right? would using irreversible gates in a quantum computer than can factor 15 really cause it to not give 5 and 3 as an answer?

 

[Hurkyl]

If I have the state

|0>

and do a 45° rotation in the state space, I now have (a multiple of)

|0>+|1>

If I do another 45° rotation, I now have

|1>

And when I measure the result, I'm guaranteed to get |1>.

-----------------------------

If I have the state

|0>

and do a 45° rotation in the state space, I now have (a multiple of)

|0>+|1>

If decoherence happens, I now have

50% chance of |0>
50% chance of |1>

If I do another 45° rotation, I now have

50% chance of |0> + |1>
50% chance of -|0> + |1>

And when I measure the result, I have a 50% chance of getting a |1>, and a 50% chance of getting a |0>.


An efficient quantum algorithm relies on doing similar sorts of things, and as we see, decoherence will change the behavior of your program.

 

[michael879]

is a 45 degree rotation the same as a square root not gate? If its not I still get what you mean cause the same thing would happen with those. Thanks a lot for the help, I just started seriously reading about quantum computers the other day.

 

[michael879]

and wait if you do 2 45 degree rotations on a |0> your guaranteed to get a |1>? That sounds kinda weird. The description of a sqrt-not gate I read was to send a photon at an electron with half of the energy required to make it change states (a not gate would have twice the energy). If you do this twice to an electron in the ground state it will always be excited in the end?

 

[Hurkyl]

By a 45° rotation, I mean the operation that sends

a |0> + b |1>

to

[(a - b) |0> + (a + b) |1>] / sqrt(2)


Appling this transformation twice will leave you with

b |0> + a |1>

which is, indeed, a not gate.


If what you described is a physical realization of that transformation, then yes, that's what I mean.