As a warmup, imagine the state ( |00> + |01> ) / √2 gets changed by a quantum operation into ( |00> + |11> ) / √2. The operation is “flip the first bit if the second is 1”. After that, if we have only the first qubit to play with, it’s statistically indistinguishable from a single qubit in the mixed state “|0> or |1> with probability 1⁄2 each”. Parts of a pure state, considered in isolation, are mixed states.
Mixed states involve probabilities, which makes us think of measurement. And indeed, the operation above can be seen as either “use the first qubit to measure the second” or “entangle the first qubit with the second”. When you’re inside a quantum system, it looks like measurement; when you’re outside and something inside is doing the measuring, it looks like entanglement. This lets us prove theorems using only linear operations, without worrying about measurement as a special case.
Moreover, every mixed state is a part of some pure state. Recall that every mixed state is an ellipsoid. We can rotate it to be axis-aligned, leading to a density matrix with nonnegative entries on the diagonal and zero everywhere else. These entries are the probabilities of getting each bit string, a.k.a. basis vector. Let’s say each basis vector |x_i> has probability p_i. Now we can make a bigger system with twice the dimensions, and build a superposition of basis vectors |x_i> ⊗ |x_i> with amplitudes √p_i, all other vectors get zero. That’s a pure state, and the first half of its qubits give our original mixed state, just like the first qubit of the pure state ( |00> + |11> ) / √2 gave us the mixed state “|0> or |1> with probability 1⁄2 each”. This lets us prove theorems using only pure states, without worrying about mixed states.
With the preliminaries out of the way...
No-cloning theorem: let’s say we have a qubit in unknown pure state and another qubit prepared as |0>. Then we can’t copy the first into the second. Proof: assume we can. Let’s say |x> is the rest of the universe. Then |00x> and |10x> should become |00y> and |11z> for some y and z. Then by linearity, (( |0> + |1> ) / √2) ⊗ |0x> becomes ( |00y> + |11z> ) / √2. But then measuring our qubits always gives either “00” or “11″, so they aren’t independent copies of ( |0> + |1> ) / √2 as we wanted. Done.
No-deleting theorem: let’s say we have two qubits in identical unknown pure states. Then we can’t erase one of them to |0> while leaving the other and the rest of the universe unchanged. Proof: if we could, we could do it in reverse and achieve cloning. Done.
I’m not sure these theorems and proofs are exactly right. Just looked up the rough theorem statements on Wikipedia and worked out the proofs here in the comment draft. If there are errors, I’m sure people will point them out.
(7/?)
No-cloning and no-deleting theorems.
As a warmup, imagine the state ( |00> + |01> ) / √2 gets changed by a quantum operation into ( |00> + |11> ) / √2. The operation is “flip the first bit if the second is 1”. After that, if we have only the first qubit to play with, it’s statistically indistinguishable from a single qubit in the mixed state “|0> or |1> with probability 1⁄2 each”. Parts of a pure state, considered in isolation, are mixed states.
Mixed states involve probabilities, which makes us think of measurement. And indeed, the operation above can be seen as either “use the first qubit to measure the second” or “entangle the first qubit with the second”. When you’re inside a quantum system, it looks like measurement; when you’re outside and something inside is doing the measuring, it looks like entanglement. This lets us prove theorems using only linear operations, without worrying about measurement as a special case.
Moreover, every mixed state is a part of some pure state. Recall that every mixed state is an ellipsoid. We can rotate it to be axis-aligned, leading to a density matrix with nonnegative entries on the diagonal and zero everywhere else. These entries are the probabilities of getting each bit string, a.k.a. basis vector. Let’s say each basis vector |x_i> has probability p_i. Now we can make a bigger system with twice the dimensions, and build a superposition of basis vectors |x_i> ⊗ |x_i> with amplitudes √p_i, all other vectors get zero. That’s a pure state, and the first half of its qubits give our original mixed state, just like the first qubit of the pure state ( |00> + |11> ) / √2 gave us the mixed state “|0> or |1> with probability 1⁄2 each”. This lets us prove theorems using only pure states, without worrying about mixed states.
With the preliminaries out of the way...
No-cloning theorem: let’s say we have a qubit in unknown pure state and another qubit prepared as |0>. Then we can’t copy the first into the second. Proof: assume we can. Let’s say |x> is the rest of the universe. Then |00x> and |10x> should become |00y> and |11z> for some y and z. Then by linearity, (( |0> + |1> ) / √2) ⊗ |0x> becomes ( |00y> + |11z> ) / √2. But then measuring our qubits always gives either “00” or “11″, so they aren’t independent copies of ( |0> + |1> ) / √2 as we wanted. Done.
No-deleting theorem: let’s say we have two qubits in identical unknown pure states. Then we can’t erase one of them to |0> while leaving the other and the rest of the universe unchanged. Proof: if we could, we could do it in reverse and achieve cloning. Done.
I’m not sure these theorems and proofs are exactly right. Just looked up the rough theorem statements on Wikipedia and worked out the proofs here in the comment draft. If there are errors, I’m sure people will point them out.