Hamiltonians may ``make no sense'', but not for the reasons mentioned. The Hamiltonian should be bounded from below, else the correlation functions, that determine the observables can't be defined, since the partition function won't be normalizable. However there's no problem with a Hamiltonian that has the properties mentioned in the question. An even simpler example is provided by the kinetic and the potential energy of a particle. The former is diagonal in the momentum basis, the latter in the position basis. These properties don't imply any  problem for studying the harmonic oscillator or the anharmonic oscillator. They simply mean that energy eigenstates are neither position eigenstates, nor momentum eigenstates, but superpositions thereof, that's all. So the eigenstates of H(b1) will be superpositions of the eigenstates of H(b2) and vice versa. The example with the magnetic field is misleading, as stated, and requires more precision. But, of course, it can be treated the same way. 

Hamiltonians may ``make no sense'', but not for the reasons mentioned. The Hamiltonian should be bounded from below, else the correlation functions, that determine the observables can't be defined, since the partition function won't be normalizable. However there's no problem with a Hamiltonian that has the properties mentioned in the question. An even simpler example is provided by the kinetic and the potential energy of a particle. The former is diagonal in the momentum basis, the latter in the position basis. These properties don't imply any  problem for studying the harmonic oscillator or the anharmonic oscillator. They simply mean that energy eigenstates are neither position eigenstates, nor momentum eigenstates, but superpositions thereof, that's all. So the eigenstates of H(b1) will be superpositions of the eigenstates of H(b2) and vice versa. The example with the magnetic field is misleading, as stated, and requires more precision. But, of course, it can be treated the same way. 

There is a large field of research concerning time-dependent Hamiltonians H(t). For example, we may assume H→H_1 as t→-∞ and H→H_2 as t→+∞. If the change is slow enough (adiabatic change) the eigenvectors of H_1 go to eigenvectors of H_2, if eigenvalues are separated at any time. If eigenvalues cross at some moments of time, interesting phenomena may occur.

Alternatively, if the change is abrupt (`quantum quench'), i.e. H=H_1 for t0, eigenvectors of H_1 are necessarily disintegrate into complicated combinations of eigenvectors of H_2, so that a stationary state for t0.

I was thinking more about the eigenvalues

Paraphrasing Landau and Lifschitz, Ch 1, if two operators have simultaneous

eigenvalues, then they commute.

If A implies B, then not B implies not A, so if they do not commute they cannot

have simultaneous eigenvalues. But separately they do,

Suppose its the free Dirac eq. Then you could not define the derivative

of H with respect to p?

Thanks for the answer, will think., both of you.

Νο, if they have the same eigenvectors, then they commute.

But you do believe that the origin of the uncertainty principle lies in

non commutivity?

That if AB-BA=C

Then delta(A) delta (B) larger or equal to hbar ?

To me that is what gives the simultaneous uncertainty of the two eigenvaloues.

It's not a question of belief-but a mathematical statement. The uncertainty principle does not have anything to do with any uncertainty of the eigenvalues of the operators-which is absent-but with the variance of their values in a state. That's why doesn't make sense, without specifying in which state the average value is taken, same with the variances.  And some care is required: If two operators do not commute, then they don't have common eigenvectors. But the eigenvectors of any one are a basis in the space in which both act-otherwise the calculations don't make sense-therefore the eigenvectors of one can be expanded as superpositions of the eigenvectors of any other-if they commute the superposition is just one term. The average value of A in an eigenstate, |a>, is the corresponding eigenvalue, λ: A|a>=λ|a> => =λ  and ΔA=-()^2=0-this value is certain. In the same state, B|a>=sum c_a|a>, i.e. a superposition of eigenstates of A. Therefore =sum |c_a|^2 and = ΔB isn't zero. And so on. 

These calculations, however, rely on a crucial assumption: that the eigenstate, |a>,  is normalizable, is finite (and may, thus,  be normalized to 1).  For unbounded operators, such as the position operator and the momentum operator on unbounded domains this isn't the case-only wavepackets are normalizable to finite norm.  This is what leads to the uncertainty principle, since the only states are superpositions of eigenstates of A or B, not individual eigenstates. Therefore, ΔΑ and ΔΒ both are non-zero. If they are conjugate, in the sense of mechanics, then the product of these variances has a lower bound. It is possible to say quite a bit, in general, however, cf. http://arxiv.org/abs/1211.4169

Say you have H(t)= H0, t=t1

So until t1, you have a set of eigenstates and a particle in one of them.  Since then H changes, you should NOT expect to find the particle in the same eigenstate. It will, in general be in some sort of linear combinations of the original eigenstates. It might  even be in the same eigenstate and stay there, if that eigenstate also happens to be an eigenstate of H1. So I am not sure what the problem is: The fact that Ho and H1 may not commute does not mean you cannot measure time. No matter what H(t) is, H0 and H1 work in the same space and hence have the same number of dimensions and  base states.  A state is merely a vector.  Just because it is not in the x-direction in a given coordinae system does not mean we cannot measure its 'x-content', i.e. projection along the x-axis; it  merely means that it may have other contents, e.g. 'y-content' and that these contents may not be constant in time.  Quantum mechanis in that is a simple vector space: You can have ANY basis you need, but  one basis (the basis of eigenvectors) allows a simpler treatment of a given problem (i.e. Hamiltonian), just like Cartesian coordinates are simpler for some problems and spherical or cylindrical for others.

Ok, based on a premise that non commuting operators cannot be measured

simultaneously(almost everyone says that) you might have some restrictions

in defining a derivative dH/dt at the point t1 because H(t1-)does not commute with

H(t1+) and you need the two eigenvalues at the same time for the limit def of derivative.

Also Dirac gets c for an answer when evalueating the speed of the electron

using dH/dp. Maybe there is no actual problem, but you have to watch out.

When you square and use HH, it does become self commuting, and then you can evaluate dH/dp

P.D.

cannot even answer if the premise is denied, maybe physics has changed since

i retired.

1) Noncommuting operators(like position and velocity in the same direction) cannot be measured simultaneously: True, but irrelevant

2)A moot point: You can imagine a sharp but continuous change in H(t) if it bothers you and this is what you could atually achieve experimentally, e.g  using a switch.

3) Why do you need two eigenvalues at the same time? You have prepared a state that has maybe a definite momentum (say H(t vector in Cartesian coordinates will have to be decomposed in whatever eigestates the new H(t) has, say cylindrical coordinates, |x>=a|r>+b|phi>+c|z> and the new time evolution will be |psi(t)>=|x>exp(-iomega_x t), t[aexp(-iomega_r  (t-t_1))|r>+bexp(-iomega_phi (t-t1))|phi>+c exp(-iomega_zt()-t1)|z>]exp(-iomeg_x t_1)

If what bothers you is 'what eigenvalue does the particle actually have at t=t1', i.e. exactly  at the turn of the interaction, the answer is it does not have any, because it is NOT in a definite state of the new H. Experimentally  if you want to measure the value of an operator, say momentum and repeated the experiment with identical systems a large number of times, yoy would observe all three possible results,  with probabilities that depend on the amplitudes , and , i.e. a, b and c

If you want the energy of the particle, again it does not have a definite energy, as it is a linear combination of states with different energies in the new H(t). Experimentally you esentially need to observe for a long time (so that you have at least complete cycles of the imaginary exponential) and again when you do that to get the energy, you will get  different  results with the said probabilities.

Try thinking  about it this way(although this is a rather simplistic example): A particle is moving with constant speed in a straight line , say the x-axis, until t1. Then an attractive center is introduced, which forces the particle to do a circular orbit with the tangential velocity the particle had at the time the center was created. What you are asking is "what is the velocity(a vector) of the particle ?" and the answer is, it is not constant, so when you measure it you may get different results .

I am not sure why you refer to  Dirac, dH/dp, the speed of light etc

Physics has not changed on that since,well about 1926. See if you understand this simplest example before  adding more complex hamiltonians, as the essence of what you are asking seems to be answerable very simply.

``Ok, based on a premise that non commuting operators cannot be measured simultaneously (almost everyone says that) you might have some restrictions in defining a derivative dH/dt at the point t1 because H(t1-)does not commute with H(t1+) and you need the two eigenvalues at the same time for the limit def of derivative.''

There are some things that are correct, and a few that are misunderstandings. ``non commuting operators cannot be measured simultaneously'': correct: we cannot define a unique procedure for measuring a common value of A and B, when these do not commute.

However, there is no problem in *adding* two non-commuting operators. Thus, for traditional Hamiltonians, p^2 + V(x) is the sum of two non-commuting parts. That just means the total Hamiltonian cannot be reduced to being a combination of kinetic and potential energy: the two are not simultaneously measurable, but the sum is well-defined. The problem is that the eigenvalues of the sum have no relation to the sums of the eigenvalues, and the eigenvectors of the sum are equally unrelated to the summands' eigenvectors.

So we may define a derivative. Indeed, it is correct to say that H(t+) will in general not commute with H(t-), but we may subtract them and take the limit t+->t and t-->t.

However, the non-commutativity issue you raise does have important consequences. Thus the derivative of exp[A(t)] is in fact neither

dot A(t) exp[A(t)]

nor

exp[A(t)] dot A(t) 

nor anything else of that nature. It is, in fact, hard to compute, as pointed out in several other posts on that thread. 

I can see you getting some operator, as a result of this derivative; but i dont see

how one can define based on the difference of eigenvalues, the derivative.

So there is some doubt in my mind of wheather the derivative operator has the

right eigenvalues.

As I mentioned above, the free Dirac eq., taking dH/dp, leads to a supposed

velocity operator with eigenvalues +c or -c, obviously incorrect.

Because the dirac matrices do not commute, I believe H(p1) does not commute with

H(p2), a non self commuting operator, unless p1 is parallel to P2. It becommes self commuting, squaring H., and then you get the right velocity.

About a simple harmonic oscilator, I do not think this is a good example.

In this case H(x1,p1) does commute with H(x2, p2) because the derivative with

respect to x1 will not affect x2, which acts as a constant and so on.

You are right: the eigenvalues of the derivative are definitely not the derivative of the eigenvalues. Thus, if you look at p^2/2 + lambda V(x), then the derivative with respect to lambda is V(x), and the eigenvalues of V(x) are the values of V(x), with the eigenvectors being dellta functions. But the derivatives of the eigenvalues are something entirely different. In particular, if V(x) is a potential that leads to bound states, all the eigenvalues are discrete and depend on lambda in a quite non-obvious manner.