A basic question we can ask is why bother with density matrices don't wave functions work just fine. The answer is wave functions work just fine as long as we deal with closed systems. The beauty of density matrices is they are better way to understand the dynamics of open systems. This point can be dramatized by noting that there are density matrices that have no wave function analog. An example of this is a completely mixed state i.e zero everywhere except on the diagonal or a scalar times the identity matrix. First we would like to derive the analog of schrodinger's equation for density matrices.
We start with what we know namely, Schrödinger’s equations
\begin{equation}
i \hbar \frac{\partial |\psi (t)\rangle}{\partial t} = H | \psi (t) \rangle \hspace{10mm} \text {eq.1}
\end{equation}
We can know imagine the wave function at t=0 and posit the existence of an operator whose sole job is to evolve the operator from time to another. We shall call this operator U. What this means is that
\begin{equation}
U(dt) | \psi (t) \rangle = | \psi (t +dt) \rangle \hspace{10mm} \text{eq.2}
\end{equation}
Thus \( \psi (t)\rangle = U(t) | \psi (0) \rangle \) and we place it in eq.1 to get
\begin{equation} i \hbar \frac{\partial U(t)}{\partial t} | \psi (0) \rangle = H U(t) \rangle | \psi (0) \rangle \hspace{10mm} \text {eq.3}
\end{equation}
The above equation holds for any initial wave function and at any time so we simply have
\begin{equation}
i \hbar \frac{\partial U(t)}{\partial t} = H U(t) \hspace{10mm} \text {eq.4}
\end{equation}
An analog equation can be got for the adjoint of U. This operator turns out to have the property that \( U U^{\dagger} = I \) .
Now recall,we introduced the density operator as being the outer product of the wave function \( \psi \rangle \) so \( \rho = |\psi \rangle \langle \psi | \). We therefore we can write
\begin{equation}
\rho(t) = U \rho_o U^{\dagger} \hspace{10mm} \text{eq.5}
\end{equation}
where \( \rho_o \) is the initial density matrix at initial time.
We can now take the derivative with respect to time on both sides of eq.5 and multiply by \( i \hbar \) on both sides then using eq.4 whe shall arrive at
\begin{equation}
\frac{d\rho(t)}{dt} = \frac{1}{i \hbar} [H(t), \rho(t)] \hspace{10mm} \text{eq.5}
\end{equation}
When we generalize to open quantum system and additional operator appears in eq.5 usually called the Lindblad operator
\begin{equation}
\frac{d\rho(t)}{dt} = \frac{1}{i \hbar} [H(t), \rho(t)] +\mathcal{L}(\rho) \hspace{10mm} \text{eq.5}
\end{equation}
The additonal operator will look different depending on the system under consideration.
We start with what we know namely, Schrödinger’s equations
\begin{equation}
i \hbar \frac{\partial |\psi (t)\rangle}{\partial t} = H | \psi (t) \rangle \hspace{10mm} \text {eq.1}
\end{equation}
We can know imagine the wave function at t=0 and posit the existence of an operator whose sole job is to evolve the operator from time to another. We shall call this operator U. What this means is that
\begin{equation}
U(dt) | \psi (t) \rangle = | \psi (t +dt) \rangle \hspace{10mm} \text{eq.2}
\end{equation}
Thus \( \psi (t)\rangle = U(t) | \psi (0) \rangle \) and we place it in eq.1 to get
\begin{equation} i \hbar \frac{\partial U(t)}{\partial t} | \psi (0) \rangle = H U(t) \rangle | \psi (0) \rangle \hspace{10mm} \text {eq.3}
\end{equation}
The above equation holds for any initial wave function and at any time so we simply have
\begin{equation}
i \hbar \frac{\partial U(t)}{\partial t} = H U(t) \hspace{10mm} \text {eq.4}
\end{equation}
An analog equation can be got for the adjoint of U. This operator turns out to have the property that \( U U^{\dagger} = I \) .
Now recall,we introduced the density operator as being the outer product of the wave function \( \psi \rangle \) so \( \rho = |\psi \rangle \langle \psi | \). We therefore we can write
\begin{equation}
\rho(t) = U \rho_o U^{\dagger} \hspace{10mm} \text{eq.5}
\end{equation}
where \( \rho_o \) is the initial density matrix at initial time.
We can now take the derivative with respect to time on both sides of eq.5 and multiply by \( i \hbar \) on both sides then using eq.4 whe shall arrive at
\begin{equation}
\frac{d\rho(t)}{dt} = \frac{1}{i \hbar} [H(t), \rho(t)] \hspace{10mm} \text{eq.5}
\end{equation}
When we generalize to open quantum system and additional operator appears in eq.5 usually called the Lindblad operator
\begin{equation}
\frac{d\rho(t)}{dt} = \frac{1}{i \hbar} [H(t), \rho(t)] +\mathcal{L}(\rho) \hspace{10mm} \text{eq.5}
\end{equation}
The additonal operator will look different depending on the system under consideration.
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