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9

I apologize, this is my third correction to my answer. This question is very subtle indeed. I hope this answer is the ultimate one! First of all, if you want to take advantage of Lie's theorem you mention (some time called third Lie theorem), the Lie algebra has to be real, as it must be the Lie algebra of a real Lie group. Then, if you are interested in ...


7

There is one more option. You can check that $aa$, $\{a,a^+\}$ and $a^+a^+$ form Lie algebra $sp(2)\sim sl(2)$. Then you can add $a^+$ and $a$ treating them as supergenerators. These are words that tell you to take anticommutators of $a$ and $a^+$ as I did in the first line. Then you get a $5$-dimensional superalgebra, which is $osp(1|2)$. There is a ...


4

Sure there was ! Just like we deduce the laws of Newton from relativity. There is a famous theorem in quantum mechanics named Ehrenfest's theorem, which states that quantum mechanical expectation-values follow classical laws. So after averaging out the quantum-behaviour you just get classical mechanics. For the correspondence with the classical mechanics ...


4

Reposting comment as an answer and expanding. The answer is yes. You can find an exposition in Condensed Matter Field Theory by Altland and Simons, starting on page 134 in the second edition. The troubles come from that spin can't be described with a Hamiltonian that is a function of $q$:s and their conjugate $p$:s. However the more general formulation of ...


3

if we have an electron in the valance band and excite it to the conduction band, and if there is no input of momentum during this process, then I always thought we would be left with an electron of wavevector $\vec k$ and a hole with wavevector $\vec k$ also. You wrote it yourself : there is no input of momentum during this process. Then why do you ...


3

Let's first consider a finite crystal with $M$ cells. By Bloch theorem we have $$\psi_n(\vec k)=e^{i\vec k\vec x}u_{n,\vec k}(\vec x),$$ where $u$ is periodic with crystal period. Now, for fixed $n$ we have exactly $M$ points in Brillouin zone, each corresponding to some $\vec k$. In some sense, $M$ can be understood as proportional to volume of Brillouin ...


3

The easiest thing for this exercise is to use Levi-Civita symbol for the vector product: $$\vec{a} \times \vec{b} = a_i b_j e_k \varepsilon_{ijk},$$ where I denote by $e_i$ the canonical basis of $\mathbb{R}^3$. Using this notation, we have: $$[L_j,p_i]=[r_k p_l \varepsilon_{klj},p_i]= i \hbar p_l \varepsilon_{ilj}.$$ and $$[L^2,\vec{p}]=e_i[L_j ...


3

You have catastrophically miscopied the formula you quote. The correct relation is $$ \bbox[5px,border:2px solid black]{E=h\times f.} $$ If the frequency is zero, the energy is also zero. There is no problem to begin with. It is possible you were thinking of $E=h/\lambda$ where $\lambda$ is the wavelength. In this case, a frequency of zero corresponds to ...


2

Suppose we have arbitrary vectors $|\alpha\rangle$ and $|\beta\rangle$ that are not necessarily aligned with one another. We can determine the component of $|\beta\rangle$ that lies along the direction of $|\alpha\rangle$ by defining an operator $$ \hat{P}=|\alpha\rangle\langle\alpha| $$ which we call the projection operator. Note that $\hat{P}$ is ...


2

Yes, e.g. all three Mandelstam variables $$ s~:=~(p_1+p_2)^2 ~\approx~ (m_1+m_2)^2 + \frac{m_1m_2}{2} ({\bf v}_1-{\bf v}_2)^2 ~>~0,$$ $$ t~:=~(p_1-p_3)^2~\approx~ (m_1-m_3)^2 - \frac{m_1m_3}{2} ({\bf v}_1-{\bf v}_3)^2 ~>~0,$$ $$ u~:=~(p_1-p_4)^2~\approx~ (m_1-m_4)^2 - \frac{m_1m_4}{2} ({\bf v}_1-{\bf v}_4)^2 ~>~0,$$ are strictly positive in ...


2

We interpret OP's question (v4) as: How do we recover the phase ambiguity from the generator of translation method in Ref. 1? Recall that an eigenvector for an operator can be rescaled with a non-zero multiplicative factor. The main point is that the position eigenket $| x \rangle$, which satisfies $$\tag{A} \hat{x}| x \rangle~=~ x| x \rangle, $$ ...


2

Usually, when you want to probe the lifetime $\tau$ of a particle (or a quasi-particle), what you basically do is looking for the way the associated wave-function $\psi$ is significantly decreasing : $$\psi\sim\psi_0\,e^{-\left(\frac{1}{\tau}+\,i\frac{E}{\hbar}\right)t}$$ Let us consider a free particle with a given energy $E$. As $\langle E\rangle=E$ is ...


2

Yes, you're right ! The uncertainty principle tells us that the thickness of the energy state $\Delta E$ is linked to the typical decay time of this energy level. If $\Delta E$ is large then $\Delta t \sim \tau$ (decay constant of the energy level) is small and then this energy state is very unstable.


2

When the system is a many-body system, it can't be solved analytically. Unlike classical mechanics, for $N = 3$ or $N = 4$ (few-body systems), there do exist Faddeev equations which can be solved by iteration. There should be more factors that I don't know. P.S. Here is a incomplete list of those systems with analytical solutions.


1

The most correct relation is the following general relation, that actually contains both terms. If you omit the inequality $(|z|^2\geq(Im(z))^2)$ from the derivation, the next steps toward the uncertainty relation would be: $$\sigma_A^2 \sigma_B^2\geq|\langle f\lvert g\rangle|^2$$ $$|\langle f\lvert g\rangle|^2=\langle f\lvert g\rangle\langle g\lvert ...


1

Spin1/2 particle Ususally, in this kind of Hamiltonian, people uses $s=s_z$, where $$s=s_z=\left[ \begin{array}{cc} 1 & 0 \\ 0 & -1\end{array} \right].$$ Then, your unperturbed hamiltonian $H_0$ is: $$H_0=-\mu s\cdot B_0 = -\mu \left[ \begin{array}{cc} 1 & 0 \\ 0 & -1\end{array} \right]B_{0,z}. $$ Then the eigen vectors of energy are: ...


1

It sounds like you're trying to find the shifts in the energy levels caused by $H_V$. To find the energy shift that goes as the first order in your small parameter you compute $$E_n^{(1)} = \langle \psi_n^{(0)}|H_V|\psi_n^{(0)} \rangle \quad (1)$$ as you noted. In this expression $|\psi_n^{(0)}\rangle$ means "the $n^{\textrm{th}}$ eigenstate of the ...


1

In the case of a particle decaying to two identical particles, this is true. There is no angle-dependence in the scattering amplitude, so the integration is indeed trivial. When calculating scattering cross sections, this is in general not true: the scattering amplitude can depend on the phase space angle.


1

I have never seen de Broglie's relation written with vector quantities. A quick search online reveals a lack of vectors as well. In the relation $$\lambda = \frac{h}{p}$$ it is implied that the quantity $p$ is the magnitude of the momentum $\left | \vec{p} \right |=p.$ Yes, the word momentum in a strict sense refers to a vector quantity, but often physicists ...


1

I am not sure that I get your question right, but let me try to answer according to my understanding. The spin part of the two electron wave function of the singlet state $|0,0\rangle=|l=0,m=0\rangle$ is $|0,0\rangle=(|\uparrow\downarrow\rangle-|\downarrow\uparrow\rangle)/\sqrt{2}$ The three triplet states look like this: ...


1

Well, I guess the answer is it becomes both! You're almost there! But then the third and fourth terms in your equation will vanish. So, since $\hat{n}$ is the number operator, a number state, such as $\vert m \rangle$ say will be an 'eigenfunction' or the $\hat{n}$ operator with 'eigenvalue' m. It seems that you probably already know this judging by how ...


1

Uncertainty principle is due to intrinsic uncertainty of nature. Yes, at any given time a particle exists both as a particle and a wave combined. Following part of your question is related to the complementarity principle (of the most accepted Copenhagen interpretation of quantum mechanics) claims that wave and particle are similar to two sides of a coin. ...


1

I'll provide an answer in non-relativistic quantum mechanics. The short answer is that momentum and mass commute, so a particle can have a well-defined momentum and mass simultaneously. But really, mass isn't considered an operator in quantum mechanics; it's a parameter, a number. So for some system, it is presumed that the mass is known always. There's no ...


1

Intuitively, if the potential energy is a function only of the position, if you measure the position precisely, you can just calculate the potential energy using that precise measurement. More formally, if $V(\hat x)$ is any function of $\hat x$, the position operator $$[V(\hat x), \hat x] = 0$$ which really is just that any operator commutes with itself.


1

First, there are effects from quantum mechanics that can be observed in macroscopic scale. I also have few words for "anomalies". Considering the following examples: Line in optical spectrum Floating metal with superconductivity Micro-Marco entanglement of millions photon Schrodinger cat state with hundreds For (1), when you use a prism, you can see ...


1

I think (but I'm not sure) that if you do it like if the photon has three physical polarization states then you get, as you said $S=0,1,2$. You can compute easily all of these states. But remember, you can express the states of the coupled basis in the uncoupled basis (via a Clebsch-Gordan decomposition). Then, you can delete the terms containing a non ...



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