What is the difference between implicit, explicit, and total time dependence, e.g. $\frac{\partial \rho}{\partial t}$ and $\frac{d \rho} {dt}$? What is the difference between implicit, explicit, and total time dependence, e.g. $\frac{\partial \rho}{\partial t}$ and $\frac{d \rho} {dt}$?
I know one is a partial derivative and the other is a total derivative. But physically I cannot distinguish between them. I have a clue that my doubt really might be understanding the difference between implicit, explicit and  total time dependence. 
 A: I usually explain it this way:
$$\rho = \rho(t,x(t),p(t))$$
$$\frac{\partial\rho}{\partial t} = \lim_{\Delta t \to 0} \frac{\rho(t+\Delta t,x(t),p(t))-\rho(t,x(t),p(t))}{\Delta t}$$
$$\frac{d\rho}{d t} = \lim_{\Delta t \to 0} \frac{\rho(t+\Delta t,x(t+\Delta t),p(t+\Delta t))-\rho(t,x(t),p(t))}{\Delta t}$$
A: You are essentially asking about the material derivative when discussing a total derivative with respect to time.
Let's say you are looking at the velocity of the air in your room. There is a different velocity everywhere, and it changes with time, so
$$v = v(x,y,z,t)$$
When you take a derivative like 
$$\frac{\partial v}{\partial t}$$
you are saying "I will keep sampling the wind velocity at the exact same point in my room, and find how quickly that velocity changes."
If, on the other hand, you take
$$\frac{\textrm{d}v}{\textrm{d}t}$$
you are now saying, "keep following one particular little bit of air, and see how quickly its velocity changes (i.e. find its acceleration)."
(note: Marek has made a nice clarification about the difference between these two uses of $t$ in the comments to this answer.)
They are related by the chain rule
$$\frac{\textrm{d}v}{\textrm{d}t} = \frac{\partial v}{\partial t} + \frac{\partial v}{\partial x}\frac{\textrm{d}x}{\textrm{d}t} + \frac{\partial v}{\partial y}\frac{ \textrm{d}y}{\textrm{d}t} + \frac{\partial v}{\partial z}\frac{\textrm{d}z}{\textrm{d}t}$$
This says that the if you look at one particular little air particle, its velocity is changing partially because the entire velocity field is changing. But even if the entire velocity field weren't changing, the particle's velocity would still change because it moves to a new spot, and the velocity is different at that spot, too.
As another example, say there is an ant crawling over a hill. It has a height that is a function of two-dimensional position
$$h = h(x,y)$$
If we look at $\partial h/\partial x$, we're looking at the slope in the x-direction. You find it by moving a little bit in the x-direction while keeping y the same, finding the change in z, and dividing by how far you moved.
On the other hand, since we're tracking the ant, we might want to know how much its height changes when it moves a little bit in the x-direction. But the ant is traveling along its own convoluted path, and when it moves on the x-direction, it winds up changing its y-coordinate as well.
The total change in the ant's height is the change in its height due to moving in the x-direction plus the change due to moving in the y-direction. The distance the ant moves in the y-direction in turn depends on the x-direction movement. So now we have
$$\frac{\textrm{d}h}{\textrm{d}x} = \frac{\partial h}{\partial x} + \frac{\partial h}{\partial y}\frac{\textrm{d}y}{\textrm{d}x}$$
On the right hand side of that equation, the first term corresponds to the change in height due to moving in the x-direction. The second term is the change in height due to moving in the y-direction. The first part of that, $\partial h/\partial y$ is the change in height due to changing y, while the second part, $\textrm{d}y/\textrm{d}x$ describes how much y itself actually changes as you change x, and depends on the particulars of the ant's movement.
Edit
I now see that you're specifically concerned with the quantum mechanics equation
$$\frac{\textrm{d}}{\textrm{d}t}\langle A \rangle = -\frac{\imath}{\hbar}\langle[A,H]\rangle + \langle \partial A/\partial t \rangle$$
Here, $\langle \partial A/\partial t\rangle$ is the expectation value of the partial derivative of the operator $A$ with respect to time.  For example, if $A$ is the Hamiltonian for a particle in a time-dependent electric field, that operator would contain time explicitly.  We begin by formally differentiating the operator itself, then taking the expectation value.
On the other hand $\langle A \rangle$ is simply a real-valued function of time (if $A$ is Hermitian), so $\textrm{d} \langle A \rangle / \textrm{d} t$ is the usual derivative of a real function of a single variable.
A: Maybe an intuitive answer is best given in terms of classical physics. Suppose you are looking at the movement of a classical particle. The relevant variables here are position and momentum. If you solve the motion of your system, you are presented with functions $x(t)$ and $p(t)$. 
Now, there are a lot of derived quantities you can build from these trajectories. For example, angular momentum $\vec{L} = \vec{x} \times \vec{p}$. Since $x$ and $p$ depend on the time, $L$ also depends on time, but in this case it does so only because $x$ and $p$ depend on time. You have basically a function $L = L(x,p)$ which then becomes $L(x(t), p(t))$. This is because in the definition of $L$, the time does not play a role. Therefore, we say that this quantity has only an implicit time dependence. In particular, 
$\frac{\partial L}{\partial t} = 0$.
If, however, your derived quantity $f$ is defined for some reason such that the time occurs explicitly in the definition, then $\frac{\partial f}{\partial t} \not= 0$. For example, you might want to add a time-dependent phase factor to your quantitiy, e.g.
$f = \vec{x} \cdot \vec{p} \cdot e^{i\omega t}$.
Then we have $f = f(x,p,t) = f(x(t), p(t), t)$, and now $\frac{\partial f}{\partial t}$ is not zero.
