When exactly is “God playing dice”? (Question on Hamiltonian and ground state, actually)

There is something I don't get:

• So I read that by applying an operator to the wavefunction (aka. measuring stuff), it is as if the wavefunction collapses onto one defined state which is an eigenstate. Which specific state it actually is seems to be "chosen" completely at random (i.e. "God playing dice").
• At the very same time, we are told that the eigenstate in question is not random at all for the Hamiltonian (lower energies are favoured and at 0K, we are in the ground state).

What did I misunderstand?

• Hum, that's not what I read: which eigenvalue is chosen depends on the particular eigenvector/eigenstate the wavefunction collapses into. The probability of obtaining any given eigenvector/eigenstate is given by the eigenvector's coefficient squared. So should I assume that these coeeficients are temperature dependent for the Hamiltonian and that they are hugely favouring the lowest energy states? – Jacomo23 Feb 18 '17 at 20:59
• For me, randomness is just opposed to determinism. In particular, it is not equivalent to having a process with equiprobable outcomes (i.e. rectangular pdf). As it is explained, it seems that the "choice" of eigenvalues to collapse into is deterministic for the Hamiltonian while it is said to be random for the others. (Again, random as "we can't know for sure even though some may be more likely"). – Jacomo23 Feb 19 '17 at 9:20
• +1 Jacomo I upvoted your post and I hope you get a proper answer. I deleted my comments as I don't feel they help clarify your question. I realize now I just don't have the experience TBH, best of luck with your question. I look forward to reading an answer from someone with more knowledge than I. – user140606 Feb 19 '17 at 10:21

Random does not necessarily mean uniformly random. That is, while you can construct states in quantum mechanics that are equally likely - forcing the observer to see one of two equally likely outcomes, this is not the general case.

You can have three states that are equallly likely, but you could also have a quantum mechanical state in which the states are 50%, 25% and 25% likely. These likelihood's are determined by the wavefunction of the state. While the wavefunction choosing a random outcome when you "observe" it, it evolves deterministically when not "observed." For example, the state could go from (50%, 25% and 25%) to (10%, 40% and 50%), and in fact we could manipulate the states surroundings such that it changes into such a state (as long as we don't explicitly measure which state it's in).

The easiest way to make sure there's no randomness in your state is simply to measure it. Once you measure it, it's in an eigenstate of the system (it's in a fixed energy) and it'll remain like that forever!

• @StevenSagona-After a measurement, the eigenstate will not remain the same forever. No eigenstate has a fixed energy, due to the uncertainty principle (the eigenstates of an electron in a hydrogen atom have a limited spatial extent, and so the momentum has too, which is to say, an uncertainty in energy). It's because of the constant interaction between the electron and the proton that the eigenstate doesn't develop but stays constant. For a free particle with spin 1/2, after a measurement, the spin will be in an eigenstate, but after that, a component of the opposite spin develops. – descheleschilder Feb 19 '17 at 11:07

I think you have to make a distinction between bound states and free particles. If you make a spin measurement on a free electron, then the two outcomes, $+\frac 1 2 \hbar$ and $-\frac 1 2\hbar$ (with respect to a certain axis usually called the z-axis), have an equal probability (you can see this process as inherently random, which I find difficult to imagine, or as deterministic, presuming the existence of non-local hidden variables). The wave function after the measurement of the spin (after applying the spin operator to the electron's wavefunction), will in time develop two different spin directions unless you measure the spin again directly after the first measurement, which would yield the same spin direction.

The electron in a hydrogen atom is bound to the proton and the normalised energy eigenstates of the electron form a complete set in Hilbert space (just like the two spin eigenstates in the case of the measurement of spin, ignoring the spatial part). If the electron is in a superposition of eigenstates (which is a normalised wave function) and we make a measurement of the energy, one of these eigenstates is "chosen" randomly (or, if you prefer, is determined by non-local variables). This state isn't gonna last forever, but neither will it develop into a new superposition of eigenstates. The interaction between the electron and the proton (and some say this is due to interactions with the vacuum fluctuations) will force the newly formed eigenstate into the lowest energy state.

So I read that by applying an operator to the wavefunction (aka. measuring stuff),

It is one of the postulates of quantum mechanics:

With every physical observable q there is associated an operator Q, which when operating upon the wavefunction associated with a definite value of that observable will yield that value times the wavefunction.

together with

Associated with any particle moving in a conservative field of force is a wave function which determines everything that can be known about the system.

you continue:

it is as if the wavefunction collapses onto one defined state which is an eigenstate. Which specific state it actually is seems to be "chosen" completely at random (i.e. "God playing dice").

What waves in the wavefunction? In postulate number 5

. For a system described by a given wavefunction, the expectation value of any property q can be found by performing the expectation value integral with respect to that wavefunction.

To relate a quantum mechanical calculation to something you can observe in the laboratory, the "expectation value" of the measurable parameter is calculated. For the position x, the expectation value is defined as

where x is the x position operator.

This integral can be interpreted as the average value of x that we would expect to obtain from a large number of measurements. Alternatively it could be viewed as the average value of position for a large number of particles which are described by the same wavefunction. For example, the expectation value of the radius of the electron in the ground state of the hydrogen atom is the average value you expect to obtain from making the measurement for a large number of hydrogen atoms. etc in the link

What waves is the probability of getting a specific measurement. That is why measuring the electron scattering off two slits gives an interference pattern. The individual measurement is a point on the screen, the distribution shows the wave solution, i.e. te probability of finding the electron at that (x,y) on the screen.

Measuring stuff, one individual measurement is the throw of the dice, has a probability of displaying the value. The dice has a probability of 1/6 of throwing a five, the measurement (throw) of a single electron in the double slit experiment gives a throw from the probability distribution.

Now the expectation value of the energy operator, will in a similar manner give a single value for the energy, which will be one instance in the probability distribution for that particular potential problem. It is a fact that energy levels are concentrated in value around a central value because there exists a width to the energy lines.

you state:

At the very same time, we are told that the eigenstate in question is not random at all for the Hamiltonian (lower energies are favoured and at 0K, we are in the ground state).

Systems relax to the lowest energy state is a principle through all physics frameworks, in quantum mechanics it appears as

The ground state of a quantum mechanical system is its lowest-energy state; the energy of the ground state is known as the zero-point energy of the system. An excited state is any state with energy greater than the ground state. In the quantum field theory, the ground state is usually called the vacuum state or the vacuum.

The solutions of the Hamiltonian are mathematical. A particular physical system is modeled by it. A single hydrogen atom will be in the ground state of the solution. To measure this, one has to scatter with a photon with the energy ( within the width) of the difference between the lowest state and the next one, excite the atom, i.e. a different wave function represents it, and verify the measurement by the relaxation and emission of a photon.

By the way, collapse is an unfortunate term to use for a probability distribution. A specific measurement described by a probability distribution classical or quantum mechanical does not collapse anything. It gives a probability of one for that single measurement to have happened. The accumulation of measurements will give the probability distribution, i.e. the width of the ground state in the hydrogen example.