Can I treat a quantum process as a Markov process?
In general, no, but they are related. The main different is the first thing you ignored: the complex number $i$. It is indeed very important in the Schrodinger equation:
$$-i\frac{d}{dt}\psi = \hat{H} \psi \tag{1}$$
If there is no $i$ in the equation, then the solution would be
$$\psi(t) = e^{-\hat{H}t} \psi(0) \tag{2}$$
Since $\hat{H}$ is Hermitian matrix, its eigenvalues are real numbers. Components with positive eigenvalues decrease to zero and negative eigenvalues blow up to infinity. The only stable solutions are the one that corresponding to 0 eigenvalue. It implies that general states are not conserving probability. So, it could be neither a quantum system, nor a Markov chain.
You should noted that similar Mathematics is not necessary means they are the same. For example, in Wick rotation, we also use the replacement $it\to1/T$, but we cannot say time is temperature.
Also, the correct discrete time version in your question should be
$$\psi_{n+1} - \psi_n = \Delta t H \psi_n \tag{3}$$
Otherwise, the system could reach the equilibrium state for any finite time if we take $t\to 0$. The appropriate analogue should be the Continuous time Markov chain.
For the Markov process:
$$V_{n+1}-V_n=(M-1)V_n \tag{4}$$
the matrix $(M-1)$ is not symmetric, so it can have complex number with real part $\le 0$. That is when the real part of all eigenvalues are $<0$, it corresponds to damped oscillations of state. Equilibrium state will be eventually reach in long time.
A subtle thing happens when all eigenvalues of $(M-1)$ are pure imaginary number. The eigenstates are purely oscillate and certainly conserve probability. It actually give the same solution to Eq. (2) in the continuous case. Hence, in this case, there are direct relation between the probability and probability amplitude. I am not quite sure about its existence and the exact condition.
I want to know how to deal with a Markov process whose state space has infinite dimensions. I think maybe quantum mechanics could give me some help.
It is somehow helpful in concept. We can consider the energy level of infinite square well (or harmonic oscillator) which has discrete countable infinite number of state denoted by $|n\rangle$. If the probability of each of these state is given by
$$p_n = \frac{1}{\zeta(4)}\frac{1}{n^4} \tag{5}$$
where $\zeta(4)$ is Riemann zeta function, so the total probability is $\sum p_n = 1$.
Using this as an example, we can construct a infinite state of Markov chain: label each node by an index, and use the identity matrix. Well, it is trivial as there is no cross link.
A more general way is to find the infinite state Markov matrix $(M-1)$ with pure imaginary described above, which has a close connection to the quantum mechanics. A unitary transformation can therefore create states linked together.
A more sophisticated example is to construct a scale free network of $N$ nodes with power > 2 and out-link with probability $1/k$ of a node degree $k$. Since the equilibrium probability of a state (arXiv:cond-mat/0307719) is given by
$$p(k)=\frac{k}{N\langle k \rangle} \tag{6}$$
in the limit $N\to\infty$, we have a Markov chain with infinite state that probability won't diverge. There are more general way to construct these type of network, such as the construction of network from BEC (arXiv:cond-mat/0011224) when we add infinite boson:
These method only cover a part of way to construct those Markov chain.
Is quantum mechanics on a measurement level a deterministic theory or a probability theory?
Probability theory. Evidence: when physicists do quantum measurements they find the results of individual runs are unpredictable. Only frequencies of multiple runs are predictable and match the theoretical results of quantum mechanics.
How can this possibly be consistent with unitarity as described above?
During a quantum measurement (measuring a system S by an apparatus A) the complete system S+A viewed at the microscopic level undergoes unitary evolution. During that evolution the system S become entangled with the apparatus A. However, by experimental design, this entanglement when viewed as a macroscopic approximation is seen to have some simplifying features:
a. The apparatus is in a mixed state of pointer states
b. The possible eigenvectors of some observable of S have coupled to the pointer states
c. Off-diagonal "interference" terms have become suppressed by decoherence due to the many internal degrees of freedom of A.
Owing to the special nature of these pointer states of A (from OP "some many-particle systems may well be approximated as classical and can store the information of measurement outcomes") we now have an objective fact about our universe.
Only one of the pointer states has in fact actually occurred in our universe (we can make this statement whether on not a physicist actually reads the pointer and discovers which universe we are actually in).
We can then make the inference that for this particular run of the S+A interaction, the system S in fact belongs to the subensemble giving rise to the occurance of this pointer state. We can make this reduction of the original ensemble based on this objective information about our universe. Restricted to this subensemble, we still have unitary evolution when viewed at the exact microscopic level.
Disclaimer: I don't know whether this really makes any sense, but this is what the reference referred to by OP seems to be saying.
Follow up question: so can we say QM is a probability theory for practical purposes but deterministic in principle?
No I think not. Here is the confusion: having banished the need for explicit wave function collapse from the QM formalism it seems that all we are left with is deterministic unitary evolution of the wavefunction of our closed system. Hence surely QM is deterministic. But no. The indeterminism in the outcome of measurements is still present in the wavefunction.
In fact the QM formalism tells us precisely when it is able to be deterministic and when not: it is deterministic whenever the quantum state is an eigenvector of the operator related to the measurement in question. Remarkably from this one postulate it is possible to derive that quantum mechanics is probabilistic (i.e. we can derive the Born Rule).
Explicitly, we can show that it is deterministic that if the evolution of S+A is run $N$ times (with $N \rightarrow \infty$) then the frequencies of different results will follow precisely the Born rule probabilities. However for a single run there is no such determinism. For a single run it is only determined that there will be an outcome.
This approach to QM is described by Arkani-Hamed here.
Edit
For a more advanced discussion of these ideas I recommend Is the statistical interpretation of Quantum Mechanics dead?
Best Answer
I agree with much of what you write in your question. Whether quantum mechanics is considered to be deterministic is a matter of interpretation, summarised in this wiki comparison of interpretations. The wiki definition of determinism is this context, which I think is entirely satisfactory, is
In, for example, many-worlds interpretation, time evolution is unitary and is governed entirely by Schrödinger’s equation. There is nothing like the "collapse of the wave-function" or a Born rule for probabilities.
In other interpretations, for example, Copenhagen, there is a Born rule, which introduces a non-deterministic collapse along with the deterministic evolution of the wave-function by Schrödinger’s equation.
In your linked text, the author writes that quantum mechanics is non-deterministic. I assume the author rejects the many-worlds and other deterministic interpretations of quantum mechanics. Aspects of such interpretations remain somewhat unsatisfactory; for example, it is difficult to calculate probabilities correctly without the Born rule.