# What causes continuous errors in a qubit?

I read that due to decoherence a qubit in a superpositon gets destroyed or put into one definite eigenstate. This kind of error seems to occur due to interactions with other stuff like the environment or other parts of the quantum computer.

However, I also read about "evil" continuous errors which can only happen to quantum computers. An error is supposed to look like this: $\lvert{\mathfrak{z}}\rangle\rightarrow \lvert{\tilde{\mathfrak{z}}}\rangle=a\lvert0\rangle+be^{i\cdot\theta}\lvert1\rangle$.

What would cause such an error?

• I can definitely help with this but first we need to clarify the question. Please define $|\mathfrak{z}\rangle$. Is it $a|0\rangle + b|1\rangle$? Also, why use such a bizarre and difficult to type symbol? – DanielSank Jul 5 '15 at 19:37
• I learned to write any general state vector with that old-school z. Most people write $\psi$ but then you got people thinking about wavefunctions and if you don't want to trigger those thoughts using a different letter can be very useful. The german word for state "Zustand" begins with a "z" and that is why it is a "z" instead of a "s" (I am german and study in Germany). – Thomas Elliot Jul 6 '15 at 7:36
• @DanielSank You assumption for z is correct. I am Sorry that I forgot to clarify it. – Thomas Elliot Jul 6 '15 at 7:37
• Despite the fact that there's an accepted answer here, I wonder why you say that continuous errors are "evil". – DanielSank Jul 6 '15 at 7:43
• One should also add that continuous errors can always be collapsed to Pauli errors, so they are no more evil than the latter. – Norbert Schuch Jul 6 '15 at 13:45

Suppose you have a standard spin qubit: spin "up" is $|1\rangle$ and spin "down" is $|0\rangle$ . Your qubit is in some storage location which does not couple the two spin directions to each other or have any different energy, you store $|\psi\rangle = a |0\rangle + b |1\rangle$ in this qubit. ` Now some proton flies by, or a minor fluctuation in the Earth's magnetic field occurs, or someone in the next door facility engages the magnets on the MRI machine. These will typically couple with the spin by some sort of tiny local Hamiltonian $\hat h /\hbar = \alpha |0\rangle\langle 0| + \beta |1\rangle\langle 1|$ for some time $\tau$. The state of the qubit is therefore changed to:$$|\psi'\rangle = a ~ e^{-i\alpha\tau} ~ |0\rangle + b ~e^{-i\beta\tau}~|1\rangle.$$A global phase factor is unobservable, but this system will behave as if $b\rightarrow b~e^{-i(\beta - \alpha)\tau}$, which could impact further experimentation.
These errors can also occur if the electronics for "switching on" or "switching off" an operational Hamiltonian are not 100% instantaneous and predictable; you will in general get either a little bit too little $e^{-i ~\hat h ~t/\hbar}$ or too much of it. So if the interaction Hamiltonian that was supposed to create $\psi$ worked by inducing $e^{-i ~ \omega ~ t ~ \hat x}$ on some previous state, where $\hat x$ is the usual $|0\rangle\langle 1| + |1\rangle\langle 0|$ quadrature, the shift from $t \rightarrow t + dt$ causes an approximate error:$$|\psi'\rangle = e^{-i~\omega~dt~\hat x} \left( a ~ |0\rangle + b ~|1\rangle \right) \approx |\psi\rangle - i ~\omega~dt~\left(b|0\rangle + a |1\rangle\right)$$That in turn could mess up not just the phases but the relative amplitudes of each term.