As a lead-in question, why do we use only reversible logic? Why can’t we erase a qubit? We can, but it’s not pretty. Let’s say we have the state ( |00> + |11> ) / √2 and want to erase the first qubit, ideally ending up with ( |0> + |1> ) / √2. Here’s a couple things we could try:
Pry away the first qubit and send it off in a spaceship
Measure the first qubit and then forget the classical bit
The two methods aren’t mutually exclusive, because the guy in the spaceship could decide to measure the qubit as well. We shouldn’t be able to instantly learn that, so both methods should give the same result. But it’s easy to compute what happens in method 2, because we start with a superposition of bit strings “00” and “11″ with amplitude 1/√2 each. Either the measurement yields 0 and the remaining state is |0>, or the measurement yields 1 and the remaining state is |1>, with probability 1⁄2 each. Forgetting the bit, we end up with a classical probability distribution over quantum amplitude distributions: |0> or |1> with probability 1⁄2 each.
That’s a new kind of object, a “mixed” state. It’s different from our desired ( |0> + |1> ) / √2, because rotating the latter by 45 degrees and then measuring it leads to 1 with certainty, while for the former it’s not true. Moreover, it’s different from every “pure” state! Because for every pure state there’s some rotation that makes it 1 with certainty, but rotating our mixed state by any angle and then measuring it always gives 0 or 1 with probability 1⁄2 each.
Okay, that’s weird thing #1. Erasing a qubit doesn’t work on individual bit strings as you’d expect. Instead it gives us something new—a mixed state with both classical probabilities and quantum amplitudes.
Weird thing #2: our mixed state “|0> or |1> with probability 1⁄2 each” can also be described in other ways, like “( |0> + |1> ) / √2 or ( |0> - |1> ) / √2 with probability 1⁄2 each”. Don’t believe me? With both these descriptions, you can check that rotating by any angle, possibly negating, and then measuring always leads to 0 or 1 with probability 1⁄2 each. So they are indistinguishable by all our allowed tools.
In fact, since our mixed state is invariant under rotation, it has infinitely many descriptions—just rotate it by any angle. For a mixed state, there’s no fact about which pure states (“worlds”) live inside it. This mystery is kind of similar to the Born probabilities, in that you can spend lots of effort trying to figure it out, but as a student it’s better to move on. There’s more to learn.
(There is a unique way to describe mixed states, just not in terms of pure states. It’s called density matrices. I might come up with a simple explanation someday, but for now just check Wikipedia.)
(2/?)
Mixed states.
As a lead-in question, why do we use only reversible logic? Why can’t we erase a qubit? We can, but it’s not pretty. Let’s say we have the state ( |00> + |11> ) / √2 and want to erase the first qubit, ideally ending up with ( |0> + |1> ) / √2. Here’s a couple things we could try:
Pry away the first qubit and send it off in a spaceship
Measure the first qubit and then forget the classical bit
The two methods aren’t mutually exclusive, because the guy in the spaceship could decide to measure the qubit as well. We shouldn’t be able to instantly learn that, so both methods should give the same result. But it’s easy to compute what happens in method 2, because we start with a superposition of bit strings “00” and “11″ with amplitude 1/√2 each. Either the measurement yields 0 and the remaining state is |0>, or the measurement yields 1 and the remaining state is |1>, with probability 1⁄2 each. Forgetting the bit, we end up with a classical probability distribution over quantum amplitude distributions: |0> or |1> with probability 1⁄2 each.
That’s a new kind of object, a “mixed” state. It’s different from our desired ( |0> + |1> ) / √2, because rotating the latter by 45 degrees and then measuring it leads to 1 with certainty, while for the former it’s not true. Moreover, it’s different from every “pure” state! Because for every pure state there’s some rotation that makes it 1 with certainty, but rotating our mixed state by any angle and then measuring it always gives 0 or 1 with probability 1⁄2 each.
Okay, that’s weird thing #1. Erasing a qubit doesn’t work on individual bit strings as you’d expect. Instead it gives us something new—a mixed state with both classical probabilities and quantum amplitudes.
Weird thing #2: our mixed state “|0> or |1> with probability 1⁄2 each” can also be described in other ways, like “( |0> + |1> ) / √2 or ( |0> - |1> ) / √2 with probability 1⁄2 each”. Don’t believe me? With both these descriptions, you can check that rotating by any angle, possibly negating, and then measuring always leads to 0 or 1 with probability 1⁄2 each. So they are indistinguishable by all our allowed tools.
In fact, since our mixed state is invariant under rotation, it has infinitely many descriptions—just rotate it by any angle. For a mixed state, there’s no fact about which pure states (“worlds”) live inside it. This mystery is kind of similar to the Born probabilities, in that you can spend lots of effort trying to figure it out, but as a student it’s better to move on. There’s more to learn.
(There is a unique way to describe mixed states, just not in terms of pure states. It’s called density matrices. I might come up with a simple explanation someday, but for now just check Wikipedia.)