Physical meaning of projection operator acting on a quantum state

Question

Let's say I have a quantum state such as $$ | \psi \rangle = \alpha |00\rangle + \beta |11\rangle $$ for some pair of qubits. I am wondering how to interpret an operator like $$ P_1 = | 0 0 \rangle \langle 0 0 | + | 1 1 \rangle \langle 1 1 | , $$ or $$ P_2 = | 1 0 \rangle \langle 1 0 | + | 0 1 \rangle \langle 0 1 | , $$ which might be applied to such a state. (States and operators similar to these occur in discussions of quantum bit-flip error correction which is why I am thinking about this).

Is it physically possible to perform the operation corresponding to, e.g., $P_1$? It looks like this means, "Use a device to measure the state of the qubits. If the device finds $|00\rangle$ or $|11\rangle$ then it must react the same way physically, giving no indication of which of these two it found. It might display a "1" (for measurement result 1) on a read-out screen in this case. If it finds one of $|01\rangle$ or $|10\rangle$ instead, it should react in some other way, displaying maybe a "2" (measurement result 2) on the screen."

One example I've thought up that might be an illustration of this is a system where the total energy depends on the relative spin of two spin-1/2 states. If they are parallel, meaning $|00\rangle$ or $|11\rangle$, then there is an energy, say $+J$. If instead, they are antiparallel, i.e. $|10\rangle$ or $|01\rangle$ then it has an energy $-J$. Then the measurement device I am trying to construct above would simply measure the total energy of the pair and read "1" if it measures $+J$ and "2" if it measures "-J".

Is this device physically plausible or is it just a fantasy?

Answer 1

If these are qubits in a general-purpose quantum computer, you can do it. For example, starting with

$$α_{00}|00\rangle+α_{01}|01\rangle+α_{10}|10\rangle+α_{11}|11\rangle$$

apply a CNOT gate controlled by the first qubit to the second qubit to get

$$α_{00}|00\rangle+α_{01}|01\rangle+α_{10}|11\rangle+α_{11}|10\rangle$$

then measure the second qubit, then apply the CNOT gate again, to get either $α_{00}|00\rangle+α_{11}|11\rangle$ or $α_{01}|01\rangle+α_{10}|10\rangle$ (times a normalization factor). Of course, there is no way to choose which one you will get.

More generally, you can project to any family of orthogonal subspaces by adding some ancillary qubits, computing some sort of "subspace number" in them, then measuring and discarding them. For example, add a third qubit initialized to $|0\rangle$, apply a Toffoli gate controlled by the other two qubits, then measure and discard the third qubit, and you'll end up with either $α_{00}|00\rangle+α_{01}|01\rangle+α_{10}|10\rangle$ renormalized, or $|11\rangle$.

If you don't have a working quantum computer, but just some physical spin systems, then it may not be easy to do these manipulations.

Answer 2

Every measurement is an irreversible energy transfer.

That should be the first sentence students get to hear in a QM 101 class. Unfortunately it's not and it is probably not for a reason. The problem with von Neumann's "state" description of quantum systems is that these systems don't consist of "states". They consist of quantum fields. An irreversible energy exchange is therefor the exchange of a quantum of energy, momentum, angular momentum and charges between the relevant quantum field and the environment. What that quantum "is" depends on the actual physics of the system. In high energy physics these quanta are the "particles" that form the standard model. In a quantum optical system it's photons. In a solid state implementation of a "qbit" it could be a phonon or a magnon or any one of a dozen other quantized lattice excitations, in a superconducting system it's a Cooper pair etc..

By writing down state only you have completely eliminated the actual physics from the description. The only way to bring it back is by working with an actual interaction Hamiltonian that spells out the properties of the interaction (an electromagnetic coupling with dipole symmetry gives us a photon, for instance). You have to decide whether you are actually interested in the details of the physical interaction. Most theorists in the quantum mechanics space are not. They are only interested in how the free system evolves and how it "projects" to a hypothetical (perfect) physical measurement result. That is abstract math and it's perfectly fine to stay on that level of description. There are plenty of interesting things happening at that level, e.g. for the application of quantum computing. If you want to get down and dirty with "energy", however, then you have to throw an actual physical Hamiltonian in, which opens another can of worms entirely. If you want to see how complex that gets, just look at an advanced textbook on atomic and molecular physics.