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0

The post measurement state is computed in the usual way: If you measure an observable with spectral decomposition $X = \sum_x x \ P_x$, $P$ being the projectors, the post measurement state if the outcome is $x$ is simply $\rho_x = \ P_x \ \rho \ P_x $ (up to normalization) To derive this in a neat way from the original postulate about pure states, you ...


1

It is easy to show that the total number of electrons in a 3D fermi sphere is : $$N(e)=\frac{V}{3\pi^2}*k_F^3$$ Where $k_F$ is the Fermi wave vector and $V$ is the real space volume of your sphere. Now if you rearrange for $k_F$ in terms of the total number of electrons you'll get a particular equation. It is know that ...


0

Honestly, the argument you're making here is a mess - the question is based on bad premises. So let me show you how to do it properly, and hopefully that will resolve your confusion. You're right that in order for a wavefunction to be normalized, it must satisfy $$\int_\text{all space} P(x)\mathrm{d}x = \int_\text{all space} \psi^*(x)\psi(x)\mathrm{d}x = ...


1

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.


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To quickly answer your question about the non-commutativity of matrices (and more generally operators): The uncertainty relation is entirely a consequence of non-commutativity. The uncertainly relation that you are talking about is a consequence of a much more general statement. It goes something like this... Let $A,B$ be operators (thinking of the as ...


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 ...


0

There is always a level of mathematical understanding. The trick is to be honest with yourself on where you can understand the articulation. The most basic introductory text that describes quantum fields with mathematics( and eventually matrix formulations) that I have encountered is Griffiths - Introduction to Quantum Mechanics. It assumes some knowledge of ...


0

You have omitted a complex conjugation, but the identity $$ \langle f \vert Og\rangle = \langle g | Of\rangle^* $$ holds because an observable by definition is Hermitian. An operator is Hermitian if $$(f , Og) = (Of ,g)$$ for all $f,g$. Since the inner product is conjugate-symmetric, $$(Of,g) = (g,Of)^*.$$


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 ...


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 ...


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 ...


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yes, good example will be solution of free particle.Where solution is like a plane wave solution hence such sols do not represent physically accepted states.this is the reason why any problem related to free particle should have a initial wave function which can be normalized other wise we cannot proceed further.


0

Even though you think of it as a single particle -- each of it's different properties like momentum, spin, etc (corresponding to each valid quantum number) sits in a Hilbert space of their own and the possible configurations of a particle sits in a tensor product of those Hilbert spaces. $$\mathcal{H_{particle}} = \mathcal{H_{momentum}} \otimes ...


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: ...


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

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 ...


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 ...


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 ...


2

Sure. In the limit $\hbar \to 0$ the Feynman path integral reduces to classical mechanics. You can also take the WKB approximation, which when $\hbar \to 0$ gives classical mechanics in the form of the Hamilton-Jacobi equation.


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.


0

I think that the problem has been stated in a way that begs the question. Any interpretation of QM that bases itself on the notion that the basis of the phenomena is wave interference by the particle itself, has to place some part of the "particle" in regions outside the box, so to speak. However, if one looks on the field that holds the particle as being ...


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 ...


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 ...


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 ...


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

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 ...


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 ...


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. ...


0

Well, the two are constrained by $$ p_{\mu}p^{\mu} = -m^2 $$ or $$ -\mathcal{E}_{\vec{p}}^2 + \vec{p}^2 = -m^2 $$ so I would say that yes they can. Let us consider the types of interactions people make in the LHC at CERN. I am not an experimentalist, but from my undergrad particle physics course, I seem to remember that linear-momentum can be measured (at ...


0

Let us take an electron's track in a bubble chamber where there is also a magnetic field. We can measure the momentum of the electron, the change due to ionisation, and its position as it goes through the spiral and finally know its final (x,y,z) at rest, and 0 momentum. Even though we are dealing with an elementary particle we are still, with our ...


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: ...


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 ...


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.


0

This is an excellent set of questions. The basic thing to realize here is that a wave function described by $Ae^{i(kx - \omega t)}$, where here $\omega \equiv E/\hbar = \hbar k^2 \ 2m$, extends with equal weight through all of the entire universe. These waves are called "plane waves". Because they are of infinite extent these wave functions are Not ...


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 ...


0

You may indeed have heard that an electron is antisymmetric, whereas a photon is symmetric. What does this mean? Suppose I have a system of several electrons. They could be orbiting a nucleus, for example. Their behaviour is described by a wave-function, $\psi$. If I swap the positions of two electrons labled $a$ and $b$ in the system, the wave-function ...


1

Well firstly, you're trying to put two concepts together that do not usually go. Lorentz invariance is a special property of relativistic quantum mechanics. You do not have Lorentz invariance in non-relativistic Quantum Mechanics. As regards your first expression, I do not think that this is correct. For one, it would give an infinite amplitude on-shell, ...


0

I think this is where Srednicki is being sloppy, a better way to do this is to take $$\langle f|i\rangle = \langle 0|T ~a_{1'}(+\infty)a_{2'}(+\infty)a_{1}^\dagger(-\infty)a_{2}^\dagger(-\infty)~|0\rangle$$ As the definition of the transition amplitude. Remember Srednicki does not try very hard to motivate his definition of the initial and final states ...


2

Recall $|E,l,m \rangle$ is the joint eigenvector of the Hamiltonian $H$, the total spin $L^2$, and a spin component (typically z) $L_z$, and the $E$, $l$ and $m$ label their respective eigenvalues. Notice all three of these operators act on the single particle that we're considering. Also recall $\langle k |$ is an element in the momentum basis, also in the ...


0

Okay, I think I got it after reading a lot of scattering-related papers. I will omit the wavefunctions normalization and the constants like $\hbar$, $m$, $\pi$, etc., where possible since the precise values are not so relevant. Consider a 2D rectangular pipe (for simplicity), infinite in $x$ direction and of finite height $H$ in $y$ direction. Under the ...


3

Seeing as $$\langle k|k_1k_2\rangle = \langle 0| a(\mathbf{k}) a^{\dagger}(\mathbf{k_1}) a^{\dagger}(\mathbf{k_2}) |0\rangle$$ and $$a(\mathbf{k})a^{\dagger}(\mathbf{k_1}) = a^{\dagger}(\mathbf{k_1})a(\mathbf{k}) + f(\omega)\delta(\mathbf{k}-\mathbf{k_1})$$ you'll get one term that vanishes because you cannot destroy the vacuum and one term that simply ...


-1

In nuclear fussion electrons and protons can fuse to form neutrons with the release of photons.


2

As you found, the two wavefunctions aren't orthogonal; what makes you think they should be? In general, degenerate eigenvectors of a symmetric matrix are not orthogonal. Note that since the two wavefunctions are linearly independent, the Gram–Schmidt process can be used to form an orthogonal basis for the eigenspace corresponding to the energy value $E$.


1

Short answer The reason why a physical quantity such as probability is given by $|\Psi|^2$ rather than some other function of $\Psi$ is geometry, namely Pythagoras' theorem. If you have a vector which points from the origin to the $(\hat x,\hat y,\hat z)$ coordinates $(x,y,z)$, then the length $\ell$ is given by $\ell^2=x^2+y^2+z^2$. Why is this the ...


1

Treat this factoring problem just like you would factor an algebraic expression. In this case, your expression is equivalent to $AX + AY + BX + BY$, so factor it similarly: $$\begin{aligned} |\Psi\rangle_{12}&=\frac12(|+_z\rangle_1|+_z\rangle_2+|+_z\rangle_1| -_z\rangle_2+|-_z\rangle_1|+_z\rangle_2+|-_z\rangle_1|-_z\rangle_2) \\ &= \frac12 ...


0

For the relation between the first and second equation, in the book you have : $$\frac{\partial H}{\partial t} + \nabla\cdot j_E = 0$$ and it's says also that : $$ j_E = \frac{1}{2}\int d^3r\left( R\frac{\partial\mathcal{H}}{\partial t} +\frac{\partial\mathcal{H}}{\partial t}R\right) $$ but $H$ and $\mathcal{H}$ are not the same objects, the first is the ...


3

Let us consider the famous double-slit experiment with photons. With the usual set-up, we denote the number of photons passing through by $N$ and we will denote the number of photons which hit the film between $y$ and $y + \Delta y$ by $n(y)$. The probability that a photon will be detected between $y$ and $y+ \Delta y$ at a time $t$ is given by: ...


3

I think the answer is "because it works". Early in the development of QM, that interpretation was given to the wave function, and over the decades it has proven to be a useful interpretation. It works. Additionally, the fact that it is possible to define an associated current, and that there is a quantum mechanical expression that guarantees that ...


2

When you have two operators $\hat{A}, \hat{B}$ satisfying equation $[\hat{A}, \hat{B}] = \imath \hat{C}$, you can prove with Schwarz inequality that $\sigma_{\hat{A}, \psi} \sigma_{\hat{B}, \psi} \geq \frac{1}{2} | \hat{C} |$. Unless there would be stronger inequality that can be used in calculations, it gives us the lower bound of uncertainty principle. ...



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