# Is spontaneous excitation possible?

I know the $$|1\rangle$$ state can relax spontaneously to the $$|0\rangle$$ state, but can the opposite also happen?

• Welcome to the QCSE! You can use mathjax markup to make the math nicer, by using '$' signs before and after the math. For more info, please check math.meta.stackexchange.com/questions/5020/…. – JSdJ Aug 10 '20 at 16:02 • I think this is a good question, and I would defer to @JSdJ for more details, but note two things: 1. It is by convention that we equate the lowest energy to the ground state$\vert 0\rangle$, much as it is by convention that we equate the classical bit "$0$" to 0 volts. 2. Implicitly you assume that there is an energy difference between$\vert 0\rangle$and$\vert 1\rangle$. But not all encodes of qubits have such an asymmetry. For example qubits can be encoded in the polarization (horizontal=$\vert 0\rangle$and verticle=$\vert 1\rangle$) of photons of light. Aug 10 '20 at 18:46 • This question seems to be aimed at a particular physical implementation of the states$|0\rangle$and$|1\rangle\$ (since, as Mark S says, this is not true for all possible qubit implementations). Did you have a particular system in mind? Aug 11 '20 at 0:56

As a first note: the (uncontrolled) transition of $$|1\rangle$$ to $$|0\rangle$$ is generally not referred to as dephasing but as relaxation. The noise-process that involves (spontaneous) relaxation is also called the amplitude damping channel.

Now, if you have a system which has a finite temperature, meaning that there is some energy in the system, the reverse might spontaneously happen as well - the $$|0\rangle$$ state spontaneously transitions to the $$|1\rangle$$ state, something we call (spontaneous) excitation. So yeah, if there is any energy in the system (i.e. de facto every conceivable system), the reverse is also possible.

If you're familiar with the math, you might want to check out the generalized amplitude damping channel, for instance as introduced here. It formalizes the above idea that, in addition of a small relaxation probability, there is also a (small) probability of a spontaneous excitation.

Spontaneous emission from an excited state $$|1\rangle$$ to ground state $$|0\rangle$$ is a well known phenomenon, but spontaneous excitation is not discussed as often, although there has been discussion about spontaneous excitation for accelerating qubits, along with its relation to the Unruh effect which comes up when relativity is considered.

As JSdJ said in their answer, if the temperature is high, we can expect ground state systems to absorb photons or phonons or some form of energy which takes them from $$|0\rangle$$ to $$|1\rangle$$ without outside control of the situation (i.e. shining a laser on the system to force the excitation). In the absence of a laser, this excitation could be considered spontaneous excitation.

Keep in mind also that Wolfgang Ketterle's group has observed negative temperature at his lab, which means the opposite is true: $$|1\rangle$$ is the preferred state and you would need to provide external energy to force the system to go from $$|1\rangle$$ to $$|0\rangle$$ except in the case of spontaneous emission, which at negative temperature would be the equivalent of spontaneous excitation.

I tried to prepare a two simple circuits on IBM Q (1 qubit Armonk processor). First one was composed only of $$X$$ gate. So, when the gate was applied on state $$|0\rangle$$, it changed to $$|1\rangle$$. As you can see from the figure below, after measurement there is non-zero probability that the qubit is in state $$|0\rangle$$. This is a spontaneous relaxation.

When you replaced gate $$X$$ with $$I$$ (or there is no gate at all), the qubit remains in state $$|0\rangle$$. However, after measurement, there is also non-zero probability that the qubit is in state $$|1\rangle$$. This is a spontaneous excitation.

As you can see, the spontaneous excitation does not occur as often as the relaxation.

Note that these results can be also influenced by read-out errors and imperfect gates.