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21

Not all atoms are the same. Uranium is very different to Hydrogen and behaves very differently, because it has many more protons/neutrons in its nucleus and many more electrons. It is a much more complex atom. Different atoms form different molecules, with different properties. When you compare the behavior of physical materials, you are really comparing ...

9

I wanted to complement the answer by @Time4Tea with a little more specifics. In your question you mention colour and hardness, so I am going to expand a little about those two. Colour: Most matter we know is made up of chemical compounds (the only monoatomic compound you will normally find in stable form is Helium). The colours we observe come from 1) the ...

8

If we ignore molecules for a moment and only look at reasonably pure samples of every element, you'll see that many of them look very similar. It's a sea of slight variations of 'gray', with only a few solids that have different colors (copper, gold), and a few liquids and gases that add some color. The portraits in this image are elements that have been ...

4

Here is a way to generalize most of the answers given here by others. What we experience of atoms in our everyday lives is determined mostly by the outermost electrons that those atoms possess. It is those electrons that determine how the atoms bind themselves to other atoms, and it is the specific nature of those bonds which in turn determine if the ...

3

The other answer did not go into the different colors (I am assuming you are only asking about visible light). Now to our knowledge today, atoms are made up of quarks (and gluons) and electrons, and these are the elementary building blocks. Now these elementary building blocks can be combined in different ways, building up different atoms. These ...

3

It is important to note that there are two different boundary conditions, the first is the boundary condition for real spin model, the second is for fermion model. In fact, for real spin model with a certain boundary condition, there may be different corresponding boundary conditions for fermion operator. When we make a simulation in reality, we need to deal ...

3

A mapping from the Brillouin zone to a sphere cannot be continuous and invertible (i.e. a homeomorphism) because continuous, invertible maps preserve topological properties, and the torus and the sphere are topologically distinct. More specifically, the 2-sphere is simply connected, which means that any curve drawn on the surface can be shrunk to a point ...

3

The exponentials are a common factor of both sides of your penultimate equation, so he just cancels them. At the last step the ${\mathcal E}_\lambda$'s cancel, then he multiplies in the bra from the left so that he gets a number rather than a ket. The $|\Psi_\lambda\rangle$ are normalized so that the $\partial \phi/\partial t$ is just multiplied by unity.

3

The mean field approximation doesn't refer to some external field, but to the assumption that we can replace interactions between the particles with some effective average behavior. Under this assumption, when we have an interaction of the sort you gave, we treat the average densities as numbers, and then we indeed have $$H_I^{MF} = u\sum_{i,j}(n_j c^{\... 3 To arrive at the exchange correlation potential starting from the exchange correlation energy, one indeed does not take a total derivative with respect to the density. The exchange correlation energy is a functional of the density, and the exchange correlation potential is given by the functional derivative of the exchange correlation energy$$E_{\rm xc} = \...

2

Interesting question! These are just some thoughts based on general knowledge of physics. I don't have any detailed knowledge of this topic. I could be wrong. If the resistance is going to depend on the variables $\rho$, $H$, and $F$, then the exponents follow from dimensional analysis. Given the units associated with the Meyer hardness, plus this ...

1

The "breathing kagome lattice" refers to a kagome Heisenberg model with a "breathing" anisotropy, where the Heisenberg interaction strength on inequivalent (up-/down-triangles) is different, see e.g. Eq. (1) in https://arxiv.org/abs/1706.10105.

1

In the first expression after the equals sign on the first line you have $c_{i\alpha}c^\dagger_{j\beta}$. It should be $c^\dagger_{j\beta}c_{i\alpha}$ since $(AB)^\dagger = B^\dagger A^\dagger$.

1

I have not tried to do the integral in this case. But usually these topological things boil down to writing the hamiltonian in the form $d_x\sigma_x+d_y\sigma_y+d_z\sigma_z$ and then computing (by simply looking at the formula) the winding number of the map $\hat {\bf d}:{\rm Brillouin{\phantom i}zone} \to S^2$ where $\hat {\bf d}$ is the unit vector in the ...

1

The Hamiltonian of a your $H_2^+$ molecule is given by $$\hat{H} = \hat{T}_\text{electron} + \hat{T}_\text{nuclei} + \hat{V}_\text{electron-nucleus} + \hat{V}_\text{nucleus-nucleus}.$$ If you want to get an energy for the electron from this, it is kind of arbitrary how you divide the contribution $\hat{V}_\text{electron-nucleus}$ between the nuclei and the ...

1

The half comes from the fract that the gaussian integral over the bosonic fields gives you the reciprocal of the square root of the Fredholm determinant: In $A$ is an $n$-by-$n$ matrix $$\int d^nx\, e^{ -x^TAx}= \pi^{n/2} [{\rm det}(A)]^{-1/2}.$$

1

One straight forward use of homotopy theory is the physics of Solitons where you try to find static,finite energy solutions to classical field theories. You might want to read this book: Manton, N., & Sutcliffe, P. (2004). Topological Solitons (Cambridge Monographs on Mathematical Physics). Cambridge: Cambridge University Press. doi:10.1017/...

1

Try to do a "parsimonious" QR decomposition, where R is a square matrix and Q an isometry. (Differently speaking, remove the zero rows of R and the corresponding columns of Q.) Then Q has dimension $d\chi\times\chi$.

1

In the scaling/continuum limit, the microscopic constants should really be taken to $J \rightarrow \infty$, $a \rightarrow 0$, and $g \rightarrow 1$, so one shouldn't refer to the "scaling dimensions" of $a$ and $J$. Once one puts the theory into its scaling form (Eq. (10.25) in Sachdev, which crucially must be independent of the microscopic constants), the ...

1

In principle, yes, the energy of the nuclear motion can be anything, but for the configurations being described here, you're not interested in the full arbitrary range of values a given variable can hold ─ you're interested in the range of values that it actually does hold in practice. This is why the text emphasizes that this is the typical range for the ...

1

A lot of work has been done since the 1984 review by Moriya and Takahashi, both theory and experiments. On the theoretical side, calculations have been refined a lot. The effect of the on-site $dd$ correlation energy $U$ was taken into account first by LSDA+U, then by LSD+DMFT (a dynamic mean field theory), for example this paper from 2001 about high ...

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