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3 votes

How to take derivative of density operator?

This is just a product rule. Take $\rho(t) = e^{-iHt/\hbar} \rho(0) e^{iHt/\hbar}$ and use the product rule: $$\frac{d}{dt}\rho(t) = \frac{d}{dt}[e^{-iHt/\hbar} \rho(0) e^{iHt/\hbar}]$$ This is just:...
MrDBrane's user avatar
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3 votes

How to take derivative of density operator?

The product rule works essentially as you think it might. You just must be careful about the commutation of of operators. For example, $$\frac{d}{dt}\big(AB \big) = \left(\frac{d}{dt}A\right) B + A \...
J. Murray's user avatar
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1 vote
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Sum of two state functions is not path independent

I would venture to offer a simple formal explanation of why the arguments leading to the paradoxical inequality of mixed partial derivatives of the state function are incorrect. There were adequate ...
Gec's user avatar
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1 vote

What's the difference between $\nabla\cdot(\rho v)$ and $\rho(\nabla\cdot v)$ as a physical intuition?

In general, using the product rule for the divergence operator $\nabla \cdot$, we have $\nabla \cdot (\rho \vec v) = \rho \nabla \cdot \vec v + (\nabla \rho) \cdot \vec v$ We can therefore see that a ...
gandalf61's user avatar
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1 vote
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What's the difference between $\nabla\cdot(\rho v)$ and $\rho(\nabla\cdot v)$ as a physical intuition?

What's the difference between $\nabla\bullet(\rho v)$ and $\rho(\nabla\bullet v)$ as a physical intuition? Physically, we are often interested in $\rho \vec v$, where $\vec v$ is the fluid velocity ...
hft's user avatar
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1 vote
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From where does the expression of the tangential accerelation come from?

The portion of the acceleration in spherical coordinates you have shown is the portion that is in the direction of the $\hat{\theta}$ unit vector. This vector is in the direction of a small change in ...
toomanyfeet's user avatar
1 vote
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Is the derivative of the adjoint the adjoint of the derivative?

The derivative of a bra function is indeed the negative of the adjoint of the corresponding ket acted on by the "derivative operator". This statement is kind of confusing because maybe its ...
Er Jio's user avatar
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1 vote

Is the derivative of the adjoint the adjoint of the derivative?

The derivative of a bra is equal to the derivative of the adjoint of the corresponding ket with changed sign.
Franc's user avatar
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1 vote

Is the derivative of the adjoint the adjoint of the derivative?

Your question is hard to answer, as it is not quite well defined. In any case, here is how you get out of the woods. Recall $$ \langle x| \hat p / \hbar = -i\partial_x \langle x| \Leftrightarrow \\ (...
Cosmas Zachos's user avatar
0 votes

Why does $\delta \vec{r} = \delta \vec{ \theta} \times \vec{r}$?

$\delta \vec{r} = \delta \vec{ \theta} \times \vec{r} \tag{1}\;,$ ...but I don't quite catch where that particular equation (1) It is easiest to start in two dimensions. Recall the equation for the ...
hft's user avatar
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0 votes

Why is the regular Taylor expansion so similar to the construction of the Hamiltonian from a Lagrangian/Legendre transform?

No, Legendre transforms are not the same as Taylor expansions. Well ok, so derivatives are by definition linear approximations, and by definition of Taylor polynomials, the first-order term is by ...
peek-a-boo's user avatar
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1 vote

Why does $\delta \vec{r} = \delta \vec{ \theta} \times \vec{r}$?

starting with $$\mathbf r=\mathbf R\,\mathbf r_0\tag 1$$ the rotation matrix $~\mathbf R~$ is described with three Euler angles $~\theta_i(t)~$ for a small angles $$\mathbf R\mapsto I+\left[ \begin {...
Eli's user avatar
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5 votes

Why does $\delta \vec{r} = \delta \vec{ \theta} \times \vec{r}$?

Let us actually discuss more generally, because your picture and argument are very limited to 3D (especially with the invocation of cross products), and make too much appeal to vague intuition (which ...
peek-a-boo's user avatar
  • 5,670
1 vote

Why does $\delta \vec{r} = \delta \vec{ \theta} \times \vec{r}$?

Suppose you have a vector $\vec{r}_0$ that is rotating counter clockwise with constant angular velocity $\vec{\omega}$. Our task is to determine the tangential velocity $\vec{v}$. We can do this in ...
Silly Goose's user avatar
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2 votes

Why does $\delta \vec{r} = \delta \vec{ \theta} \times \vec{r}$?

The length of an arc that subtends an angle $\delta\theta$ is just $\text{arc length} =(\text{radius})(\delta\theta)$. This radius is the horizontal distance from the vertical axis in your picture ...
Er Jio's user avatar
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1 vote
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"Why is $n$ held constant when taking the time derivative in the course of the Van Kampen's system size expansion?"

When you have a master equation for $P(n,t)$ is describes how the probability of a certain state (in this case the state corresponds to the particle number n) changes over time. It is important to ...
emir sezik's user avatar
  • 1,758
2 votes

Show that $dE/dt = -bv^2$ (Help with differentiation)

It's not like we're having to use the chain rule here, like we do during the differentiation of 1/2 m(dx/dt)^2. What am I missing? That's precisely what we have to do, so that's what you're missing.
J. Murray's user avatar
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