# If superposition is possible in QM, why do we often assume systems are already in their eigenstates?

My understanding is that an arbitrary quantum-mechanical wavefunction can be written as a linear combination of eigenfunctions of some Hermitian operator, most commonly the Hamiltonian; when a measurement corresponding to that operator is made on this superposition state, the wavefunction collapses and only specific values are observed -- namely, the eigenvalues of the particular eigenstates that comprised the wavefunction. (Moreover, the probability of measuring the eigenvalue $E_i$ is proportional to ${\|c_i\|}^2$, the square of the coefficient of that eigenstate in the linear combination, etc.)

And yet, in many situations, it seems to be assumed that the system is already in an eigenstate and that superposition is not possible. For example:

• the electron in the hydrogen atom is said to be in, e.g., the 1s $^{2}S_{1/2}$ state or the 2s $^{2}S_{1/2}$ state, but never a superposition of the two.

• the possible angular momentum vectors for a QM rigid rotor with fixed $l$ are sometimes drawn as discrete "cones"... but couldn't the average $\bf{L}$ point in any direction, since a rigid rotor might be in a superposition of states?

• when deriving Boltzmann statistics, we consider how to place $N_i$ particles into the level with energy $\epsilon_i$, but there is no consideration that a particle might occupy two (or more) energy levels simultaneously.

Why do we ignore superposition in these and similar cases?

• If you understand the basics of density matrix, then I think you can find a satisfactory answer here link Commented May 1, 2018 at 16:00

Often, we don't observe single quantum systems. Often, we observe ensembles of identical or nearly identical systems, often in thermal equilibrium.

In some cases, it can be utterly indistinguishable whether each individual system is in an eigenstate, or each individual system is in a superposition state with random relative phase. See this question for one such case: Are these two quantum systems distinguishable?

When a substantial fraction of an ensemble is in a superposition state and this population shares a common relative phase between the superposed states, we can observe the superposition much more easily.

As a concrete example, I've used a femtosecond laser to line up a bunch of nitrogen molecules. This changes the index of refraction of nitrogen gas, and you can watch how the index changes with time due to molecular rotation. It turns out that quantum superposition effects explain observed results beautifully. However, collisions between molecules disrupt the synchronous rotation, and the nitrogens drift out of relative alignment with one another. They might be driven towards rotational eigenstates, or they might stay in superposition states but be driven out of sync. The interesting thing is, you can't tell the difference between these two cases by watching the index of refraction - and I think you actually can't even tell the difference in principle, which is kinda neat.

• The link provided is very helpful, as is your concrete example. I think my fundamental misconception, as I said in response to Mark, was that some observable difference had to exist between a set of systems in superposed states and a set of systems in constituent eigenstates. I still have some qualms about the Boltzmann distribution and superposition, and have a much more detailed question... I'm not sure if I should post it as a comment, edit it in the above question, or post a new question.
– WP-
Commented Nov 14, 2012 at 6:00
• New question seems like the right approach to me. Commented Nov 15, 2012 at 14:01

We don't ignore superposition. I'm not sure who told you an electron can't be in a superposition of energy states, but of course it can.

You probably mean "expectation value of angular momentum" for "average angular momentum". Yes, it could point in between those discrete cones. Again I doubt you find anyone denying this.

In the last case, you're just talking about the solution to a counting problem. Without being more specific about the derivation you're discussing, I don't know what else to tell you.

Basically, I reject the premise that people ignore superposition. It may be that they just don't explicitly state "and there can be a superposition of these states" in the same way that every statement has context that you are assumed to understand.

• No one told me an electron can't be in a superposition of states, or anything like that. I just assumed that a superposed state would have properties different than those of its constituent eigenstates -- and therefore the physical behavior of the superposed state would differ somehow. At the very least something would be lost by not considering superposition. That potential omission bothered me. But Andrew's answer points out my misconception: an ensemble of systems in superposed states is essentially indistinguishable from a properly constructed ensemble of systems in eigenstates.
– WP-
Commented Nov 14, 2012 at 6:01
• A small nitpick: there are ensembles of systems in superposed states that can't be replaced with an equivalent ensemble of systems in energy eigenstates. For example, any ensemble that was all in the same superposition of ground and first excited state and all having the same relative phase between these two states. However, any ensemble of systems in energy eigenstates can be indistinguishably replaced with an ensemble of systems in superposition states, but with random relative phases. Commented Nov 15, 2012 at 14:05

In the case of the Hydrogen atom at room temperature, we expect the superposed state to rapidly decohere and collapse to the ground state emitting a photon. We'd only expect to find Hydrogen atoms in a superposition of the ground and excited states if some mechanism was continually injecting energy into them.

If you heated Hydrogen towards the temperature where it forms a plasma you would have to start treating the atoms as superposed states, and indeed if you performed some suitable spectroscopy on them you'd find that a proportion of the atoms collapsed into an excited state.

• You expect an H atom at room temperature to be in the electronic ground state, but you would not expect to find an H_2 molecule in the rotational ground state unless it was much colder. However, it turns out that high temperatures don't require you to treat atoms or molecules as superposed states - unless they're coherent with one another, which they won't be, in thermal equilibrium. Commented Nov 13, 2012 at 13:22

This question is addressed in another post https://physics.stackexchange.com/posts/402366/edit

Depending on specific situations, I can think of two independent answers to the question "why we mostly only care about stationary states":

1. decoherence by "slow environment" (I won't focus on this aspect);

2. all (effectively nondegenerate) systems becomes (improper) ensembles of stationary states IF we have uniform ignorance of time (I will focus on explaining this idea).

First, I assume you have basic knowledge about density matrix. (I think I have to do so, since otherwise my answer would be too long.)

Given an arbitrary initial density matrix $$\rho_0$$, we construct a time ensemble based on uniform distribution of time. Why do we want such time ensemble? Say we set the initial state, and then we wait while we don't record the time. Not recording time roughly means we have an uniform ignorance of time. Given that uniform ignorance of time, the density matrix describing the system is the time ensemble $$\rho_{\text{time}}$$. Say $$U_t$$ is the time evolution operator.

$$\rho_{\text{time}} =\lim_{t \rightarrow \infty} \frac{1}{t} \int^t_0 \mathrm{d}t \ U_t\rho_0 U_t^\dagger$$

Say the initial state is in a pure state superposition $$\rho_0=\left(\sum_m C_m \left|m \right>\right) \left( \sum_n C_n^\dagger \left< n \right|\right)$$

$$\rho_{\text{time}} =\lim_{t \rightarrow \infty} \frac{1}{t} \int^t_0 \mathrm{d}t \ U_t\rho_0 U_t^\dagger = \lim_{t \rightarrow \infty} \frac{1}{t} \int^t_0 \mathrm{d}t \sum_{m,n} e^{-i (E_m-E_n)t} C_m C_n^\dagger \left| m \right> \, \left< n \right|$$

Taking the limit first gives the delta function $$\delta_{m,n}$$

$$\rho_{\text{time}}=\sum_{m,n} \delta_{m,n} C_m C_n^\dagger \left| m \right> \left< n \right|=\sum_{n} \left|C_n \right|^2 \left|n \right> \left< n \right|$$

Here we go! If we have an initial state and an uniform ignorance of time, all phase information is lost and the density matrix is diagonal in energy so it's an (improper) ensemble of energy states. For thermodynamic properties which are independent of time, such time ensemble capture all information about the system. Therefore, only energy states and the occupation probabilities of the states matter. The ideas of superpositon of energy states and interference between energy states become non important roughly speaking. (The above derivation requires non-degeneracy. The reason why it's often reasonable is that there is rarely perfect symmetry. I won't elaborate this more.)