# Superposition of quantum states

There has been many questions asked about superposition on this forum, but none of them provided me with a satisfactory answer for my following question.

I understand, quite well, the principle of superposition as a mathematical concept. I also understand quite well the superposition of wave functions of probability amplitudes.

However, there is a specific example where I have trouble figuring it out.

Dirac wrote: "A measurement always causes the system to jump into an eigenstate of the dynamical variable that is being measured, the eigenvalue this eigenstate belongs to being equal to the result of the measurement."

In the case of a spin-half particle in a magnetic field, the eigenstate of the Hamiltonian matrix for the z-spin corresponds to a given discrete spin value, so I have no problem imagining an electron in a state $|ψ>$ jumping to a state $|+>$.

Now, in the case of the simplified model of the 2 states of an ammonia molecule(nitrogen above or below the hydrogens), as presented by Feynman in chapters 8 and 9 of his lectures, the eigenstate correspond to a superposition of the up and down states (and if it starts in state $|1>$, its probability amplitude will oscillate in between 0 and 1 until it emits enough energy to settle in the lower energy state,$E_{0}-A$.

So my question is: How can the system jump (upon measurement) to an eigenstate which is a superposition of the up and down geometrical states, when in fact, it is possible to measure the angles of its tetrahedral conformation in experiments: geometrically, the nitrogen has to be above or below, it can't be in a flat plane,can it? Does it have anything to do with Heisenberg's uncertainty principle,especially the uncertainty associated to its position?

As per comments below, I'll provide more details on my question: Feynman shows the 2 geometrical states as |1> and |2>, and the 2 energy eigenstates as |I> and |II>. Each eigenstate's probability amplitude is a superposition of the 2 geometrical states' probability amplitudes, and conversely, each geometrical state is in a superposition of the energy eigenstates. The upper eigenstate has energy E+A, and the lower one E-A. If the molecule "starts" in the state |1>, then it is in a superposition of the 2 eigenstates, which have different frequencies ((E+A)/h and (E-A)/h respectively), so the molecule will switch between |1> and |2> at the frequency A/h. Since its eigenstate is not a geometrical state, I'm wondering where we would actually find it if we could observe it? From the answer below, I guess we can't differentiate between the 2 geometrical states because of Heisenberg's uncertainty principle?

Also, since we're on the subject: are the superposition of probability amplitudes and the superposition of quantum states 2 different concepts, or are they 2 ways to see the exact same thing?

• Hi Simon, have you read this post physics.stackexchange.com/questions/197601/… just in case it's a duplicate. – user108787 Oct 25 '16 at 18:37
• Thank you for editing, Mr Balaji. I haven't figured yet how to use TeX to write mathematical expression. However, the lower energy state is E0 - A, and the upper energy one is E0 + A. Would it be possible to edit your edit to change back that part, or does EoA means the same as E0 - A ? – Simon Oct 25 '16 at 18:38
• I think this provides a good explanation: colorado.edu/physics/phys5250/phys5250_fa14/…. Together with this graphic: hindawi.com/journals/apc/2012/164752/fig9 – Gert Oct 25 '16 at 18:50
• The superposition of probability amplitudes (i.e. wave functions) and of quantum states are, in essence, the same concepts. – freecharly Oct 26 '16 at 2:50
• Simon: Further on the part of your question “Since its eigenstate is not a geometrical state, I'm wondering where we would actually find it if we could observe it? ” and assuming by "it" you mean the geometrical state, please see new addition to my original Answer, below. – iSeeker Oct 26 '16 at 12:02

It is, indeed, possible that when the ammonia molecule is in an energy eigenstate, it is in a superposition of the "geometrical" nitrogen (N) up or down state. As you suspect correctly, this is a case of Heisenberg's uncertainty principle for the observables N position and energy. Energy and N position are observables that cannot be both accurately measured. If the molecule is in an energy eigenstate E1, its N up or down state is undetermined (probability 50% up and 50% down). Conversely, if the molecule is in an N up eigenstate, its energy is undetermined (50% probability E1 and 50% E2).

There are other simple quantum systems with similar behavior. For example the linear photon polarizations vertical and horizontal and the circular polarizations right circular and left circular. When a photon has e.g. vertical polarization, it is in a superposition of 50% probability right and 50% left circular polarization. Conversely, when it has a right circular polarization, it is in a superposition of 50% probability vertical and 50% horizontal polarizations.This is due to the fact that the observables linear and circular polarizations cannot be simultaneously measured accurately. This is also due to Heisenberg's uncertainty principle.

• OK. So Heisenberg's U.P. isn't only about momentum versus position? As long as we have 2 observable that are somewhat related to each other, then we will never get a proper determination of both? (I have to go now, but when I come back, if nobody disagreed with your answer, I'll accept it as the chosen answer) – Simon Oct 25 '16 at 22:16
• @Simon - Yes, you are right. There are many observables in quantum mechanics for which an uncertainty relation holds so that they cannot be determined accurately at the same time. There are other pairs of observables that can both be accurately measured because they are simultaneously in an eigenstate. – freecharly Oct 25 '16 at 22:43

The ammonia molecule’s inversion can be modelled, based on the energy change during its "umbrella" inversion. This involves a double potential well with a central barrier that corresponds to the energy required to get to the intermediate flat configuration.

This results in adjacent symmetric (+) and anti-symmetric (-) harmonic oscillator states with a very small energy difference, resulting in the two states being about equally populated at room temperature and accessible to each other by quantum mechanical tunnelling.

ADDITIONALLY (re your “wondering where we would actually find it if we could observe it”): From the csbsju.edu link below:

"If |NH3> represents the left-hand well and |H3N> the right-hand well, we can write the wave functions for these states symbolically as shown below.

|Psi>0 = 2^(-1/2) [|NH3> + |H3N>] and |Psi>1 = 2^(-1/2) [|NH3> - |H3N>]

[snip] However, |Psi>1 can be separated from |Psi>0 by electrostatic means and directed to a resonant cavity."

According to http://www.users.csbsju.edu/~frioux/super/MM-super.htm the states constitute "one-particle" superpositions in which a single particle or system is assumed to occupy a linear superposition of two states.[As Freecharly pointed out, it's not the whole molecule in a superposition.]

The bond angles are the same in both observable states (I think one can say that they won't be observed in the intermediate/transitional - flat configuration - that you're concerned about). See How are bond angles determined?

Two factors contribute to the rapidity of the inversion: a low energy barrier and a narrow width of the barrier itself.

Ammonia therefore exhibits a quantum tunnelling due to a narrow tunnelling barrier (rather than being the result of thermal excitation). As you seem to have realised, this is related to Heisenberg's Uncertainy Principle.

Superposition of the two states leads to energy level splitting, which is used in ammonia masers.

Hoping this helps.

• OK, so even though we can measure the angles, via X-ray crystallography or the VSEPR model, we have no way of telling whether the nitrogen was above or below the hydrogens, since the angle is the same in both configurations. So it exists in both geometrical configurations at the same time, and we can't tell the difference because of Heisenberg's Uncertainy Principle? Is my understanding correct? – Simon Oct 25 '16 at 19:03
• How do you know that at room temperature the ammonia molecule is in a "superposition of states"? And why only at room temperature? And how can you get the microwave absorption by transition from one eigenstate to the other when the molecule is in a "superposition of states"? – freecharly Oct 25 '16 at 20:30
• When, at room temperature, according to quantum statistics a certain number of ammonia molecules occupy a higher energy eigenstate and a certain number of molecules a lower energy eigenstate, does not mean that the ammonia molecule is in a superposition of states. – freecharly Oct 25 '16 at 20:46
• freecharly: Feynman shows the 2 geometrical states as |1> and |2>, and the 2 eigenstates as |I> and |II>. Each eigenstate is a superposition of the 2 geometrical states, and conversely, each geometrical state is a superposition of the eigenstates. The upper eigenstate has energy E+A, and the lower one E-A. If the molecule "starts" in the state |1>, then it is in a superposition of the 2 eigenstates, which have different frequencies ((E+A)/h and (E-A)/h respectively), so the molecule will switch between |1> and |2> at the frequency A/h. The energy difference becomes bigger under an elec. field – Simon Oct 25 '16 at 20:48
• @freecharly Please see www.users.csbsju.edu/~frioux/super/MM-super.htm for modelling superpositions of the N atoms on both sides of plane of the 3 H atoms of NH3. Energy difference of in-phase & out-of-phase superpositions is only 0.79 cm-1, so ~ equally populated at room temp: i.e. “dealing with “one-particle" superpositions in which a single particle or system is assumed to occupy a linear superposition of two states.” (Re comment above: Wiki <Nitrogen_inversion> is source of "ammonia interconversion is rapid at room temperature.") [If downvote was yours, maybe consider reviewing it?] Thanks – iSeeker Oct 25 '16 at 23:10