How do physical implementations of Z gate selectively affect $\lvert1\rangle $ basis vector?

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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.

Dragonsheep

Posted 2019-05-30T04:41:32.467

Reputation: 223

Answers

1

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.

forky40

Posted 2019-05-30T04:41:32.467

Reputation: 1 518