4
$\begingroup$

The Pauli Z gate inverts the phase of $\lvert1\rangle $ while leaving $\lvert0\rangle$ unaffected.

When I think about how $\lvert1\rangle $ and $\lvert0\rangle$ are physically realized, however, as in the Physical Implementations section here, there tends to be a physical symmetry between the physical realizations of the two.

It would seem, then, that a Z gate would need to be realized as some experiment which selectively delays the tuning of only spin-up electrons or nuclei, for example. Given the physical symmetry I expect between physical realizations of $\lvert1\rangle$ and $\lvert0\rangle$, this seems counterintuitive.

Is the fact that a physical $\lvert0\rangle$ cannot have phase delay, and a physical $\lvert1\rangle$ can, a matter of convention? If not, why can't I describe a single $\lvert0\rangle$ in a multi-qubit system as phase delayed relative to the others? E.g., $(I \otimes Z)\lvert10\rangle$ seems to be a perfectly reasonable experiment that delays my right qubit relative to my left, why is only the converse experiment allowed?

Is it, in fact, the case that the $Z$ gate is implemented as an experiment that only affects, e.g., spin-up nuclei but not spin-down nuclei? As background, I'd find an overview of how phase is physically implemented very helpful.

$\endgroup$
0
$\begingroup$

You're exactly right about the physical symmetry of $|0\rangle$ and $|1\rangle$. If I handed you a spin-1/2 particle in complete vacuum, the question of whether its in $|0\rangle$ or $|1\rangle$ is meaningless without a preferred coordinate system. So there needs to be some physically preferred direction to even set up computation.

Now say I have the ability to turn on an external magnetic field $\vec{B}$. When its turned on, its direction automatically sets a preferred coordinate system, and $|0\rangle$ and $|1\rangle$ are now defined as alignments of spins parallel or antiparallel to the field. Physically, this is because a spin-1/2 particle with some dipole moment $\vec{\mu}$ interacts with the field $\vec{B}$ according to the Hamiltonian

$$ H = - \vec{\mu} \cdot \vec{B} $$

which really just tells us how much energy our particles will have in the external field. Then, we'll choose a convention that $|0\rangle$ is the low-energy state (say $E_0=0$ for simplicity) and we're ready to compute!

In this case a Z-gate would just be flipping on the external field for some known time $\tau$, which corresponds to a time-evolution of $e^{-i H \tau/\hbar}$. This has the effect:

\begin{align} |0\rangle &\rightarrow |0\rangle \\ |1\rangle &\rightarrow -|1\rangle \end{align}

Not only does this describe spin qubits, but also simple superconducting qubits like charge qubits. Charge qubits were originally called "Cooper Pair Box" because they work by trapping a cooper pair (of charge $2e$) on a superconducting island separated from the grounded material by some distance. In a toy geometry, we'll say this distance is $x$, so that the trapped CP creates an electric dipole $|\vec{d}| = 2ex$. Now just add an external field to interact with this dipole and you're ready to apply quantum gates.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.