Suppose, an election of a hydrogen atom has energy corresponding to $n=2$, where $n$ is the principal quantum number.

Is it possible to find the probability of the electron staying in $2s$, $2p_x$,$2p_y$ or $2p_z$ orbital?

Are these probabilities equal, as they are have equal energy?

  • $\begingroup$ Why don't you construct the most general wavefunction and then normalize it. The coefficient squared will give the probability. $\endgroup$
    – sbp
    Commented Jun 10, 2018 at 7:09
  • $\begingroup$ I know that. I was asking is it always possible to find the coefficients exactly, even if the sum of their squares add up to 1 $\endgroup$ Commented Jun 10, 2018 at 13:24
  • $\begingroup$ The sum of their squares will always add up to one. That's a necessity. By exactly do you mean exact analytical results? If yes, then I'm afraid no. $\endgroup$
    – sbp
    Commented Jun 10, 2018 at 14:37

2 Answers 2


If an atom is prepared in an $n=2$ atomic state it will decay via spontaneous emission into the ground, $n=1$ atomic state. You ask for the probability of the electron to stay in the excited state. A slightly more precise rephrasing would be if the atom is prepared in the excited state at time $t=0$ then what is the probability, $P(t)$, that the atom is in the excited state at time $t$? At $t=0$ we know $P(t=0) = 1$ since we know the atom was prepared in the excited state. As time goes on it becomes more and more likely that the atom has decayed. This means $P(t)$ decreases. Eventually, for very large $t$ we know that with almost certainty that the atom has decayed. This means $P(t\rightarrow \infty) \rightarrow 0$.

For typical atomic transitions $P(t)$ is an exponentially decaying function with decay rate $\Gamma$.

$$ P(t) = e^{-\Gamma t} $$

We call $\tau = \frac{1}{\Gamma}$ the lifetime of the atomic state because it is a timescale that roughly characterizes "how long" the atom stays in the excited state, thought it must be understood that there is non-zero probability for the atom to decay at a time earlier than $\tau$ and a non-zero probability for the atom to remain in the excited state after time $\tau$.

In any case, with this more sophisticated understanding of the atomic lifetime your main question can be rephrased.

You are essentially asking is it possible to calculate $P(t)$ for an atom starting in the $2s, 2p_x, 2p_y$, and $2p_z$ states. Or similarly, what are the lifetimes for the different atomic states. The answer is that yes it is possible to calculate the lifetimes for different atomic states.

Your next question is does the lifetime differ for the different atomic states. The answer is that in general yes, the lifetimes for different excited states is different. However, when there are certain symmetries it is possible for the lifetimes to be identical. For example, an atom in the $p_x$ state is the same as an atom in the $p_y$ state just with a rotation about the $z$ axis. Because the laws of physics are rotationally symmetric it would be surprising if the lifetimes of the $p_x$ and $p_y$ states were different. Of course, if a magnetic field is turned on or something breaks the symmetry this statement may no longer be true. However, the $s$ state and the $p$ states have different symmetries. This means that in general it is possible for the $s$ and $p$ states to have different lifetimes. It is possible to calculate and measure lifetimes for these different excited states in certain cases.

In general the lifetime, $\tau$ depends on how the wavefunction of the electron interacts with the electric field. This interaction depends on the shape of the electron wavefunction including the symmetry of the wavefunction. This means if different wavefunctions have different shapes ($2s$ and $2p$ for example) they can have different lifetimes.

edit: You ask if the probabilities for decay are equal because the energy of the two transitions (from $2s \rightarrow 1s$ or $2p \rightarrow 1s$) are equal. The answer here is no. Your intution is correct that the transition probability has a strong dependence on the energy difference between two states, however, there is also a factor related to the shape or symmetry of the two different states under consideration.

In fact, above I have failed to point out something very important. All of what I said above is very general for generic atomic states. What I will say below is specific to the specific transitions you asked about. When an atom spontaneously decays it typically decays into a quantum state which has the opposite parity. Notice that $2s$ and $1s$ have the same parity. This means that it is very very rare for a $2s$ atom to decay into a $1s$ atom. This means the lifetime $\tau_{2s}$ is very very long. However, it is easily possible for the $2p$ to decay into the $1s$ atom. This means that $\tau_{2p}$ is much shorter. Notice how the symmetry of the wavefunctions has an important role in the decay rate of the excited state. In the formulas for each there is a factor having to do with the energy difference and that factor is the same for each. It is in fact the symmetry factor which ends up mattering more for these particular transitions.

  • $\begingroup$ Your answer is very constructive. But still have one question. Is it possible to find the probability that the electron was in 2s, or the three individual 2p orbitals? We can take a linear combination of those orbitals and normalize it, but is it possible to evaluate the normalized coefficients? $\endgroup$ Commented Jun 10, 2018 at 13:29
  • $\begingroup$ If you know the linear combination of the orbitals then you just take the absolute magnitude of the coefficient of each state to get the probability the electron would be found in that state. Are you asking an experimental question about the determination of an unknown quantum state through measurement of the atom? $\endgroup$
    – Jagerber48
    Commented Jun 10, 2018 at 14:10
  • $\begingroup$ If possible, then I was asking about the analytic determination (which is not possible, according to the comments above). How can it be found theoretically? $\endgroup$ Commented Jun 10, 2018 at 17:59
  • $\begingroup$ Your question is not clear. Consider revising the question with more detail about what you are asking. What physical situation are you asking about? What are the initial and final conditions of the situation you are asking about? Right now it is not clear to me what you are asking. You are ask for "the probability that the electron was in $2s$ or the three individual $2p$ orbitals" but that is not a fully formed question. If you prepare the electron in the $2s$ state then the probability is obviously 100% that it is in the $2s$ state. The answer will vary depending on initial conditions. $\endgroup$
    – Jagerber48
    Commented Jun 11, 2018 at 1:01
  • $\begingroup$ The atom was in ground state and it is given enough energy for n=2. What are the probabilities that now the electron will be found in 2s, 2px, 2py, or 2pz orbital? Is it possible to measure this analytically? $\endgroup$ Commented Jun 11, 2018 at 8:49

The question asked in your comment is different from the question I answered so I'll put a new answer here to your question:

The atom was in ground state and it is given enough energy for n=2. What are the probabilities that now the electron will be found in 2s, 2px, 2py, or 2pz orbital? Is it possible to measure this analytically?

The answer is yes. This can be deteremined analytically. Often we have a cartoon picture in our heads where an atom "magically" absorbs a photon and moves into an excited state. This leaves out many of the important details.

The reason light interacts with atoms is because the electric field of the light pushes around the electron (and thus changes the shape of the electron wavefunction). This means that if you shine on light at a frequency which the electron will respond to you can push the electron from being in one spatial pattern (the ground state, for example) to being in another spatial pattern (an excited state, for example).

Consider the $p_x$ state. In this state the electron is kind of stretched in the $x$ direction. Well this means that you can get from the ground state to this state by shining on light which is linearly polarized along the $x$ direction. The electric field acts to stretch the atom along the $x$ direction. Similarly for the $y$ and $z$ directions.

So this means that which excited state the atom is driven to depends on the geometry of the electric field driving the atom as well as the geometry of the particular ground and excited states under consideration.

In the dipole approximation it is impossible to drive transitions from $1s$ to $2s$ because the states have the parity.

I haven't given you an analytic formula to calculated the different excited state fractions given a particular incident electric field so I haven't fully answered your question. I don't have time for that now but most atom/quantum optics textbooks will have a description of how to calculate the coupling strength between an optical field and an atomic transition, including clebsch-gordon coefficients and polarization vectors which capture the geometric effects I described earlier.

I like this textbook available online by Steck

  • $\begingroup$ The assumption that the energy is delivered by a (dipole) EM-induced excitation is reasonable, but it is not the only possibility. It would also be perfectly plausible to think of the energy being delivered via a collision (say, with an electron beam tuned to the energy difference), in which case a sizeable population in the $2s$ state would be quite a reasonable outcome. $\endgroup$ Commented Jun 12, 2018 at 0:18

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