If we take a hydrogen atom as qubit, let

$\lvert0\rangle$ = unexcited state
$\lvert1\rangle$ = excited state

then what is the meaning of measuring the qubit value in the sign basis? If the atom may only be in excited or unexcited state, but $\lvert+\rangle$ and $\lvert-\rangle$ are superpositions of those states — then what would the outcome of the measurement be — also a superposition of $\lvert+\rangle$ and $\lvert-\rangle$? Can anyone please help to understand the idea behind the sign basis?


2 Answers 2


As the measurement postulate says, if you projectively measure a qubit, initially in a state $|\psi\rangle$, in the basis $\{|+\rangle,|-\rangle\}$, you will get the state $|+\rangle$ with probability $|\langle+|\psi\rangle|^2$, and similarly for $|-\rangle$.

For the particular implementation you mention, a two-level atom whose eigenstates are the logical $|0\rangle,|1\rangle$ states, there is no general, useful, real physical quantity${}^1$ represented by the operator $$X=|0\rangle\langle1|+|1\rangle\langle0|$$ whose eigenstates are $|+\rangle$ and $|-\rangle$ (check it!). To do a projective measurement on that basis, the standard (though not necessarily unique) procedure is to apply a $\pi/2$ Rabi pulse which will bring $|+\rangle$ to $|0\rangle$ and $|-\rangle$ to $|1\rangle$, and measure in the computational basis. One can then apply an inverse pulse if needed.

There are other implementations, however, where this basis has a more physical significance. For example, if your logical states are the up and down states of a spin-$\frac{1}{2}$ particle measured along the $z$ direction, then $X$ is the spin along the $x$ direction (which is no coincidence).

${}^1$ For any given atom, though, you can probably find detectable physical properties of interest. If, say, $|0\rangle$ is an $s$ state and $|1\rangle$ is a $p_z$ state, which may very well be the case, you'll find that the $|\pm\rangle$ states are localized towards either pole. A measurement of position above/below the $xy$ plane will closely approximate an $X$ measurement in most such circumstances. Similarly, a measurement of momentum going to positive or negative $z$ will approximate a measurement along $Y=i|0\rangle\langle1|-i|1\rangle\langle0|$, whose eigenstates $|\pm i\rangle=\frac{1}{\sqrt{2}}(|0\rangle\pm| i\rangle)$ look like $e^{\pm ikz}$ near the origin.


It is not true that the atom may be only in either unexcited or excited state; as everywhere in quantum mechanics, the atom may be in any superposition of those. However, those states are not eigenstates of the hydrogen atom, therefore the superposition will evolve (in the qubit picture, you'll get an intrinsic $z$-rotation of the qubit). Also note that the lifetime of your (evolving) superposition will not be longer than the lifetime of the excited state (and the emission of the photon on decay certainly counts as measurement in the eigenbasis, because the photon will only be emitted for the excited state).

However, superpositions of two longer-lived excited states of atoms are indeed used to implement qubits.

Note that, of course, you get the $\lvert+\rangle$ state in the atom as sum of the wave functions of the ground state and the excited state (times normalization, of course). Thus you can easily calculate what the state "looks like" in space (that of course depends on the specific excited state you've chosen).

  • 1
    $\begingroup$ The question was about the outcome of the measurement. Measurements collapse superpositions, so it's not helpful to inform that atoms can be in superpositions before measurement. What is being asked is what is being measured, physically. $\endgroup$
    – user1247
    Aug 26, 2012 at 20:51
  • $\begingroup$ @user1247: After an ideal measurement the atom is in the state you measured, that's how an ideal measurement is defined. Before the measurement it probably wasn't. $\endgroup$
    – celtschk
    Aug 26, 2012 at 20:53
  • $\begingroup$ Isn't the question how the measurement is made, not what the wave function might look like after it is made? $\endgroup$
    – user1247
    Aug 26, 2012 at 21:19
  • $\begingroup$ In other words, what physically corresponds to the operator |+->? $\endgroup$
    – user1247
    Aug 26, 2012 at 21:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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