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It is fairly simple to show that the Schrodinger equation is of the form, $$\frac{d|\psi(t)\rangle}{dt}=-\frac{i}{\hbar}\hat{A}|\psi(t)\rangle$$ for some Hermitian $\hat{A}$, from the assumption that time evolution is the result of an operator on the Hilbert space acting on the initial state, $$|\psi(t')\rangle=\hat{U}(t,t')|\psi(t)\rangle.$$ By conservation of probability, we can infer that $\hat{U}(t,t')$ must be unitary, and we can deduce that it must obey composition and that it tends to the identity at $t'$ approaches $t$. Since unitary operators can be written as the exponential of some Hermitian operator ($\hat{B}$), we can express the infinitesimal time evolution operator $$\hat{U}(t+dt,t)=e^{i\hat{B}},$$ we find that all necessary properties are satisfied $\hat{B}=-\frac{1}{\hbar}\hat{A}dt$. Applying the infinitesimal time evolution operator, we have $$|\psi(t+dt)\rangle=(\mathbb{1}-\frac{i}{\hbar}\hat{A}dt +\mathcal{O}(dt^{2}))|\psi(t)\rangle$$ which can be rearranged to give the first equation.

In this sense, I can clearly understand why the Schrodinger equation takes the form it does, based on the fact that we are representing states as vectors on a Hilbert space, and the properties of states and operators on Hilbert spaces.

On the other hand, general non unitary evolution can be described by a quantum channel, $\mathcal{E}$, a completely positive, linear, trace preserving superoperator acting on density operators in a Hilbert space. More specifically, we expect a family of channels, $\{\mathcal{V}(t)\}$ which will describe the whole future evolution of the system. $$\hat{\rho}(t)=\mathcal{V}(t)\left(\hat{\rho}(0)\right)$$ Can we use similar properties of superoperators and density operators (and any necessary simplifying approximations) to infer the form of the master equation $$\frac{d}{dt}\hat{\rho}=\mathcal{L}[\hat{\rho}(t)]$$ or, equivalently, can we explain why the family of operators will evolve according t0: $$\mathcal{V}(t)=\exp(\mathcal{L}t)$$

Thanks in advance for any help

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2 Answers 2

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The result you refer to in the first part of your question is called Stone's Theorem. A crucial ingredient here is that the unitary propagators $U(t)$ form a strongly continuous semigroup, i.e., $U(t+s) = U(t) U(s)$ and $U(t) |\psi\rangle \to U(t_0) |\psi\rangle$ as $t \to t_0$ for any vector $|\psi\rangle$. Stone's Theorem states that then there is a self-adjoint operator $A$ with $U(t) = e^{iAt}$.

In the more general setting where we talk about quantum channels, the propagators $\mathcal V(t)$ are not unitary. However, they still form a strongly continuous semigroup, i.e. $\mathcal V(t+s) = \mathcal V(t) \mathcal V(s)$ and $\mathcal V(t) \rho \to \mathcal V(t_0) \rho$ as $t \to t_0$ for any self-adjoint $\rho$. This property is all that is needed to define a generator $\mathcal L$ as follows: $$ \mathcal L \rho \equiv \lim_{t \to 0} \frac{V_t \rho - \rho}{t} . $$ The domain of $\mathcal L$ consists of all self-adjoint $\rho$ where the limit exists. Then $\mathcal L$ is closed and densely defined, and $\mathcal V(t) = e^{\mathcal L t}$. A proof can be found for example at the beginning of chapter X.8 of Reed & Simon, Vol. II. This result is a part of the Hille-Yosida-Phillips Theorem, which additionally provides some conditions on the spectrum and the resolvent of $\mathcal L$.

When we add the fact that $\mathcal V(t)$ are completely positive trace-preserving operators, we get the familiar "Lindblad" structure of $\mathcal L$. That was proven by Gorini, Kossakowski, Sudarshan and Lindblad.

Edit in response to comments:

  • Why do we assume that the propagators form a continuous semigroup?
    The semigroup property is physically quite intuitive. It says that evolving the system for a time $s$ and then for a time $t$ is the same as evolving it for the time $s+t$. You are correct that there is an assumption of Markovianity in here. In addition, there is an assumption that there is no time-dependent driving, see below.
    I don't have such an intuitive motivation for the continuity, except that, well, usually we expect stuff to be continuous in Physics.
  • Why is the semigroup property so important?
    First of all note that if we have $\mathcal V(t) = e^{\mathcal L t}$, then the semigroup property is obviously satisfied. The theorems above tell us that the semigroup property alone (+strong continuity) is enough to be able to write the function in this exponential form.
  • For the sake of completeness, a comment on time-dependent driving:
    If there are system parameters that are varied externally, the propagators depend on start and end time and the semigroup property must be replaced by the composition law $\mathcal V(t_3, t_1) = \mathcal V(t_3, t_2) \mathcal V(t_2, t_1)$ which is again physically intuitive. The solution is then given by a time-ordered exponential $$\mathcal V(t_2, t_1) = \mathcal T e^{\int_{t_1}^{t_2} \mathcal L(t) dt} \tag{*}$$ with a time-dependent generator. Again, if (*) is given, the composition property is obvious, and there are theorems guaranteeing that the composition property alone (+continuity) is enough to guarantee the form (*).
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  • $\begingroup$ Thanks for your response. I didnt quite understand the significance of the fact that our channels form a semigroup. Is it the fact that they are continuous which demands that they are a semigroup? $\endgroup$ Apr 21, 2023 at 13:00
  • $\begingroup$ Further, is it because we are assuming Markovian evolution that we can say the evolution under the channels obey the composition relation/ are continuous? $\endgroup$ Apr 21, 2023 at 13:02
  • $\begingroup$ @AdrienAmour I have edited the answer. $\endgroup$
    – Noiralef
    Apr 21, 2023 at 15:35
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The answer by @Noiralef is well-informed. I just would like to add a pedestrian view, which hopefully helps to see the actual physical forest behind the mathematical trees.

The reason why master equations are written for the density matrix and not for the wave function is that density matrix is more general - it does not assume that we are in a pure state (i.e., a state describable by a wave function.) And this is very likely the case, when we deal with a system coupled to the bath/reservoir/thermostat/environment. Indeed, the goal of Lindbad equations is to describe relaxation towards thermodynamic equilibrium (or dynamic behavior in presence of such relaxation) and quantum statistical physics describes this equilibrium in terms of density matrix.

Density matrix obeys the von Neumann equation, which is the quantum version of the classical Liouville equation, but which is also the generalization of the Schrödinger equation to the case, where a system is described by the density matrix: $$i\hbar\frac{\partial \rho}{\partial t} = [H,\rho]=H\rho-\rho H $$ This is a matrix equation, which is not very convenient to work with, so we rewrite the density matrix as a column vector and do similar transformation with the right-hand side, obtaining a simple equation for a vector (which is routinely studied in linear algebra): $$ i\hbar\frac{\partial \rho}{\partial t} = \mathcal{L}\rho. $$ It is an easy excercise to show how $\mathcal{L}$ is expressed in terms of $H$.

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