quantum mechanics - What does spontaneous symmetry breaking have to do with decoherence?

Background

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The question here by Prof. Wen, and the answers that follow point out that spontaneous symmetry breaking (SSB) has something to do with decoherence if I understand it crudely correctly.

But the usual reasoning why SSB does not occur in a quantum mechanical system (for example, a particle confined in a double-well potential) is that (due to the tunnelling effects) the ground state is a symmetric or antisymmetric linear superposition of the ground state wavefunctions localized around the classical minima of the potential which respects the symmetry of the Hamiltonian. It's only in field theory where one has infinite degrees of freedom and the tunnelling effects are shut down so that one can have SSB.

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Question

If decoherence were truly the reason of SSB then should one not expect SSB to happen even in quantum mechanics, and the system to go to a mixed state? But SSB doesn't happen in quantum mechanics.

I guess I wrongly understood the points explained there, and I would like to be clarified on this issue.

general relativity - Definition of derivative operator on a manifold

I'm hoping to understand the motivation for certain parts of the definition of a derivative operator $\nabla$ on a manifold $M$. In Wald's General Relativity, two clauses of the definition are:

1. 
   

   Commutativity with contraction: For all tensors $\mathit{A} \in \mathscr{T}(k,l)$, $\nabla_{d}(\mathcal{A}^{a_1...c...a_k}_{b_1...c...b_l}) = \nabla_{d}\mathcal{A}^{a_1...c...a_k}_{b_1...c...b_l}$. (the parenthesis indicates contraction)

   
   



2. 
   

   Consistency with the notion of tangent vectors as directional derivatives on scalar fields: for all functions $f \in \mathscr{F}:M \rightarrow \mathbb{R}$ and all tangent vectors $t^a \in V_p$, it is required that $t(f)=t^a\nabla_af$

   
   

What is the point of having derivative operators commute with contraction (where we sum over the vectors and dual vectors for some (i,j) slot in the tensor)? What theorems are not possible to prove if this commutativity isn't stipulated? The so-called "Leibnitz rule" for derivative operators on manifolds corresponds to the simple product rule in elementary calculus; does this commutativity-with-contraction rule correspond to something simple as well?

My second question is about the notation of clause 2. We know that the $t^a$ are tangent vectors in the tangent space $V_p$ at point $p$ on the manifold. Then what is $t$ supposed to be?

quantum mechanics - How can we represent uncertainty principle in phase space?

From what I understand I think if I construct the phase space by plotting $x$ component of momentum against $x$ for a quantum mechanical particle, I cannot specifically put a single point to mark the particle's state in that phase space. For that would mean I can state the position and momenta of the quantum particle with arbitrary precision, which is forbidden by the uncertainty principle. All I can do, at best, is to mark an area in the phase space.

Is this understanding of phase space of quantum system correct?

If it is, then my other question is:

Does the marked area of phase space (to specify the state of the particle) have an area of the order of $h$?

Answer

Your surmise is basically right.

Indeed, spikey δ-function distributions (perfectly localized) in phase space cannot describe quantum states. The narrowest possible quantum state has to extend to an area broader than h, as dictated by the uncertainty principle. Classical mechanics is precise and spikey, but quantum mechanics is uncertain and fuzzy.

In fact, for a phase-space quasi-probability distribution (Wigner function) describing a state, if it is normalized to one, $\int dx dp f(x,p)=1$, you may easily prove that is is of finite height, $|f|\leq 2/h$. It must extend to an h-base or broader to produce unit volume. This is in sharp contrast to classical localized states, and segues to them in the classical limit.

Much of this discussion is expanded and illustrated in our book, namely ISBN 978-981-4520-43-0, World Scientific Publishing 2014, A Concise Treatise on Quantum Mechanics in Phase Space, by Curtright, Fairlie, and myself.

Now, to be sure, δ(x)δ(p), a point/spike in phase space is meaningful in QM, but not as a specifier of a state: instead, it specifies the parity operator, in this Weyl-correspondence mapping of operators to phase-space functions; but this is highly technical and might not be of interest to you. There are also associated peculiarities of f actually going negative, but also in small phase-space regions of area smaller than h, so the uncertainty principle protects them from observation too--it works almost miraculously to fix things and eliminate paradoxes!

The takeaway is that quantum distributions are fluffy/fuzzy: think of them as marshmallows. They only look sharp and classical for systems with huge actions, on the scale of such huge actions themselves. (Scaling the phase-space variables down and f up to preserve the unit normalization ultimately collapses the base of the pillbox we considered above to an apparent "point" in phase space; and leads to a divergent height for f, so an apparently localized classical particle. However, the entropy has increased: several different quantum configurations reduce to this same limit, obliterating information.)

quantum mechanics - Gaussian integrals in Feynman and Hibbs

I was going through the calculation of the free-particle kernel in Feynman and Hibbs (pp 43). The book describes $$ \left(\frac{m}{2\pi i\hbar\epsilon}\right)\int_{-\infty}^{\infty}\exp\left(\frac{im}{2\hbar\epsilon}[(x_2-x_1)^2+(x_1-x_0)^2]\right)dx_1 \tag{3.4} $$ as a Gaussian integral. And the result of this integration is given as $$ \left(\frac{m}{2\pi i\hbar \cdot 2\epsilon}\right)^{1/2}\exp\left(\frac{im}{2\hbar\cdot 2\epsilon}(x_2-x_0)^2\right). \tag{3.4} $$ The only way I could do this integral was treating it as a Gaussian integral over $ix_1$, in which case I get the required solution but with a negative sign upfront. Is this the right way to do it, or is there something very obvious that I am not seeing?

On a related note, problem 3-8 (pp 63) also involves a similar "Gaussian" integral in order to obtain the kernel for the quantum harmonic oscillator. My technique of performing a Gaussian integral over $ix$ goes horribly wrong there and I am not even near the correct answer which should be of the form $$ \frac{1}{\sqrt{2\pi\sin\epsilon}}\exp\left(\frac{i}{2\sin\epsilon}[(x_0^2+x_2^2)\cos\epsilon-2x_2x_2]\right) $$ after performing the integration, $$ \exp\left(\frac{i\cot\epsilon}{2}(x_0^2+x_2^2)\right)\int\exp\left(\frac{i}{2\sin\epsilon}[2x_1^2\cos\epsilon-2x_1(x_0+x_2)]\right)dx_1. $$ Any help will be greatly appreciated.

Answer

1. 
   
   

   The variable $$\epsilon\equiv\Delta t^M~>~0$$ in eq. (3.4) is the change in Minkowski time.

   
   


2. 
   

   In order not to deal with purely oscillatory integrands, Feynman uses a prescription $$\Delta t^M\quad\longrightarrow\quad \Delta t^M-i0^+ $$ where an infinitesimal negative imaginary part is added to make the integrand exponentially decaying.

   
   


3. 
   

   In other words, under a Wick rotation $$\Delta t^E ~\equiv~ i \Delta t^M$$ to Euclidean time, $\Delta t^E$ should have a positive real part.

   
   


4. 
   

   Eq. (3.4) in Ref. 1 then follows from the Gaussian integral $$ \sqrt{\frac{m}{2\pi \hbar\Delta t^E_{21}}}\sqrt{\frac{m}{2\pi \hbar\Delta t^E_{10}}} \int_{\mathbb{R}}\!\mathrm{d}x_1~ \exp\left\{-\frac{m}{2\hbar}\left[\frac{\Delta x_{21}^2}{\Delta t^E_{21}} +\frac{\Delta x_{10}^2}{\Delta t^E_{10}}\right]\right\}$$ $$~=~\sqrt{\frac{m}{2\pi \hbar\Delta t^E_{20}}}\exp\left\{-\frac{m}{2\hbar}\frac{\Delta x_{20}^2}{\Delta t^E_{20}}\right\},$$ where $$ \Delta x_{ab}~:=~x_a-x_b, \qquad \Delta t_{ab}~:=~t_a-t_b, \qquad a,b~\in~\{0,1,2\}. $$

   
   
   


5. 
   

   The above square root factor is the famous Feynman's fudge factor, which can be understood in the Hamiltonian phase space path integral as arising from a Gaussian momentum integration, cf. e.g. my Phys.SE answer here.

   
   

References:

1. R.P. Feynman & A.R. Hibbs, Quantum Mechanics and Path Integrals, 1965.

mathematical physics - Does it make sense to speak in a total derivative of a functional? Part III

In this third part of the series, I will continue the deduction of Noether's theorem initiated in the previous post - Does it make sense to speak in a total derivative of a functional? Part II.

Situation 1

Here, I will consider the validity of the total derivative \begin{equation} \frac{d\mathcal{L}}{dx^{μ}} =\frac{\partial\mathcal{L}}{\partialφ_{r}}\partial_{\mu}φ_{r}+\frac{\partial\mathcal{L}}{\partial\big(\partial_{ν}φ_{r}\big)}\partial_{\mu}\big(\partial_{ν}φ_{r}\big)+∂_{μ}\mathcal{L}.\tag{III.1}\label{eq1} \end{equation}

We have expressed in Eq. (\ref{eq24}) of the previous post (Does it make sense to speak in a total derivative of a functional? Part II) that \begin{multline} \dfrac{S^{\prime}-S}{\varepsilon} \approx \int_{\mathbb{\Omega}}d^{D}x~\left\{ \dfrac{\partial\mathcal{L}}{\partial\phi_{r}}\zeta_{r} + \dfrac{\partial \mathcal{L}}{\partial\partial_{\nu}\phi_{r}}\partial_{\nu}\zeta_{r}\right. \\ \left. + \xi^{\mu }\left( \dfrac{\partial\mathcal{L}}{\partial\phi_{r}}\partial_{\mu}\phi _{r}+\dfrac{\partial\mathcal{L}}{\partial\partial_{\nu}\phi_{r}}\partial_{\mu }\partial_{\nu}\phi_{r}+\partial_{\mu}\mathcal{L}\right) +\partial_{\mu}% \xi^{\mu}\mathcal{L}\right\}, \tag{II. 24}\label{eq24}% \end{multline} where I'd like to remember that $\zeta_r\equiv\zeta_r(x)$ and $\xi^{\mu}\equiv\xi^{\mu}(x)$.

If what we ask about Eq. (I.$9$) in the first post of this Series (Does it make sense to speak in a total derivative of a functional? Part I) has a yes as an answer, then the following identifications must be valid: \begin{equation} \frac{d\zeta_{r}}{dx^{\mu}}=\partial_{\mu}\zeta_{r} \quad\text{and}\quad \frac{d\xi^{\mu}}{dx^{\mu}}=\partial_{\mu}\xi^{\mu}.\tag{III.2} \end{equation} Thus, the Eq. (\ref{eq24}) becomes \begin{equation} \dfrac{S^{\prime}-S}{\varepsilon}\approx\int_{\mathbb{\Omega}}d^{D}x~\left\{ \dfrac{\partial\mathcal{L}}{\partial\phi_{r}}\zeta_{r}+\dfrac{\partial \mathcal{L}}{\partial\partial_{\nu}\phi_{r}}\dfrac{d\zeta_{r}}{dx^{\nu}} +\xi^{\mu}\dfrac{d\mathcal{L}}{dx^{\mu}}+\mathcal{L}\dfrac{d\xi^{\mu}} {dx^{\mu}}\right\}.\tag{III.3}\label{eq3} \end{equation} Now, we do use of identity \begin{equation} \dfrac{\partial\mathcal{L}}{\partial\partial_{\mu}\phi_{r}} \dfrac{d\zeta_{r}}{dx^{\mu}}=\frac{d}{dx^{\mu}}\left(\zeta_{r}\frac{\partial\mathcal{L} }{\partial\partial_{\mu}\phi_{r}}\right)-\zeta_{r}\frac{d}{dx^{\mu}}\frac{\partial\mathcal{L}}{\partial\partial_{\mu}\phi_{r}},\tag{III.4}\label{eq4} \end{equation} such that \begin{equation} \dfrac{S^{\prime}-S}{\varepsilon}\approx\int_{\mathbb{\Omega}}d^{D}x~\left\{ \left(\dfrac{\partial\mathcal{L}}{\partial\phi_{r}}-\dfrac{d}{dx^{\nu} }\dfrac{\partial\mathcal{L}}{\partial\partial_{\nu}\phi_{r}}\right) \zeta _{r}+\dfrac{d}{dx^{\nu}}\left( \zeta_{r}\dfrac{\partial\mathcal{L}}{\partial\partial_{\nu}\phi_{r}}+\xi^{\mu}\mathcal{L}\right) \right\},\tag{III.5}\label{eq5} \end{equation} where we have used \begin{equation} \xi^{\mu}\dfrac{d\mathcal{L}}{dx^{\mu}}+\mathcal{L}\dfrac{d\xi^{\mu}}{dx^{\mu }}=\dfrac{d}{dx^{\mu}}\left( \xi^{\mu}\mathcal{L}\right). \end{equation}

We have to said in Does it make sense to speak in a total derivative of a functional? Part II, Eq.(\ref{II19}), that \begin{equation} \zeta_{r}\left( x\right) +\xi^{\mu }\left( x\right) \partial_{\mu}\phi_{r}\left( x\right) =\dfrac{\tilde{\delta}\phi_{r}}{\varepsilon}=\chi_{r}\left( x\right) ,\tag{II.19}\label{II19} \end{equation} so that (\ref{eq5}) becomes \begin{multline} \dfrac{S^{\prime}-S}{\varepsilon}\approx\int_{\mathbb{\Omega}}d^{D}x~\left( \dfrac{\partial\mathcal{L}}{\partial\phi_{r}}-\dfrac{d}{dx^{\nu}} \dfrac{\partial\mathcal{L}}{\partial\partial_{\nu}\phi_{r}}\right) \zeta _{r}\\ +\int_{\mathbb{\Omega}}d^{D}x~\dfrac{d}{dx^{\mu}}\left[ \dfrac {\partial\mathcal{L}}{\partial\partial_{\mu}\phi_{r}}\chi_{r}-\left( \dfrac{\partial\mathcal{L}}{\partial\partial_{\mu}\phi_{r}}\partial_{\nu} \phi_{r}-\delta_{\nu}^{\mu}\mathcal{L}\right) \xi^{\nu}\right].\tag{III.6}\label{eq6} \end{multline}

And now comes the question: how can we apply the generalized divergence theorem in the second integral on the right side-hand if instead of a partial derivative we have a total derivative?

Situation 2

Before asking the question, let's see what happens if we do not use Eq. (\ref{eq1}). In this case, we can rewrite the Eq. (\ref{eq24}) as: \begin{equation} \dfrac{S^{\prime}-S}{\varepsilon}\approx\int_{\mathbb{\Omega}}d^{D}x~\left\{ \dfrac{\partial\mathcal{L}}{\partial\phi_{r}}\chi_{r}+\dfrac{\partial \mathcal{L}}{\partial\partial_{\nu}\phi_{r}}\left( \partial_{\nu}\zeta _{r}+\xi^{\mu}\partial_{\mu}\partial_{\nu}\phi_{r}\right) +\partial_{\nu }\left( \xi^{\nu}\mathcal{L}\right) \right\},\tag{III.7}\label{eq7} \end{equation} where we have used (\ref{II19}).

If we add and subtract the term $\partial_{\mu} \phi_{r}\partial_{\nu}\xi^{\mu}$ in the expression in parentheses of the second term, that last equation becomes \begin{equation} \dfrac{S^{\prime}-S}{\varepsilon}\approx\int_{\mathbb{\Omega}}d^{D}x~\left\{ \dfrac{\partial\mathcal{L}}{\partial\phi_{r}}\chi_{r}+\dfrac{\partial \mathcal{L}}{\partial\partial_{\nu}\phi_{r}}\partial_{\nu}\chi_{r} -\dfrac{\partial\mathcal{L}}{\partial\partial_{\nu}\phi_{r}}\partial_{\mu} \phi_{r}\partial_{\nu}\xi^{\mu}+\partial_{\nu}\left( \xi^{\nu}\mathcal{L} \right) \right\}.\tag{III.8} \end{equation} Now, using the identities \begin{align} \dfrac{\partial\mathcal{L}}{\partial\partial_{\nu}\phi_{r}}\partial_{\nu} \chi_{r}&=\partial_{\nu}\left( \chi_{r}\dfrac{\partial\mathcal{L}} {\partial\partial_{\nu}\phi_{r}}\right) -\chi_{r}\partial_{\nu} \dfrac{\partial\mathcal{L}}{\partial\partial_{\nu}\phi_{r}},\tag{III.9}\label{eq9}\\ -\dfrac{\partial\mathcal{L}}{\partial\partial_{\nu}\phi_{r}}\partial_{\mu} \phi_{r}\partial_{\nu}\xi^{\mu}&=-\partial_{\nu}\left( \dfrac{\partial \mathcal{L}}{\partial\partial_{\nu}\phi_{r}}\partial_{\mu}\phi_{r}\xi^{\mu }\right) +\xi^{\mu}\partial_{\nu}\left( \dfrac{\partial\mathcal{L}} {\partial\partial_{\nu}\phi_{r}}\partial_{\mu}\phi_{r}\right),\tag{III.10}\label{eq10} \end{align} we obtain \begin{multline} \dfrac{S^{\prime}-S}{\varepsilon}\approx\int_{\mathbb{\Omega} }d^{D}x~\xi^{\mu}\partial_{\nu}\left( \dfrac{\partial\mathcal{L}} {\partial\partial_{\nu}\phi_{r}}\partial_{\mu}\phi_{r}\right)+\int_{\mathbb{\Omega}}d^{D}x~\left( \dfrac{\partial\mathcal{L}}{\partial\phi_{r}}-\partial_{\nu}\dfrac {\partial\mathcal{L}}{\partial\partial_{\nu}\phi_{r}}\right) \chi_{r} \\+ \int_{\mathbb{\Omega}}d^{D}x~\partial_{\nu}\left[ \dfrac{\partial \mathcal{L}}{\partial\partial_{\nu}\phi_{r}}\chi_{r}-\left( \dfrac {\partial\mathcal{L}}{\partial\partial_{\nu}\phi_{r}}\partial_{\mu}\phi _{r}-\xi^{\nu}\mathcal{L}\right) \xi^{\mu}\right].\tag{III.11}\label{eq11} \end{multline}

Here, considering the validity of Euler-Lagrange's equation \begin{equation} \dfrac{\partial\mathcal{L}}{\partial\phi_{r}}-\partial_{\nu}\dfrac {\partial\mathcal{L}}{\partial\partial_{\nu}\phi_{r}}=0, \tag{III.12}\label{eq12} \end{equation} and the applicability of divergence theorem over to third integral (Which now seems to be quite reasonable!) \begin{equation} \int_{\mathbb{\Omega}}d^{D}x~\partial_{\nu}J^{\nu}=\oint_{\partial \mathbb{\Omega}}dS_{\nu}~J^{\nu}=0,\tag{III.13}\label{eq13} \end{equation} with \begin{equation} J^{\nu}=\dfrac{\partial\mathcal{L}}{\partial\partial_{\nu}\phi_{r}}\chi _{r}-\left( \dfrac{\partial\mathcal{L}}{\partial\partial_{\nu}\phi_{r} }\partial_{\mu}\phi_{r}-\delta_{\mu}^{\nu}\mathcal{L}\right) \xi^{\mu},\tag{III.14}\label{eq14} \end{equation} when $\varepsilon\rightarrow 0$, we have found \begin{equation} \lim_{\varepsilon\rightarrow 0} \dfrac{S^{\prime}-S}{\varepsilon}\approx\int_{\mathbb{\Omega}}d^{D}x~\xi^{\mu }\partial_{\nu}\left( \dfrac{\partial\mathcal{L}}{\partial\partial_{\nu} \phi_{r}}\partial_{\mu}\phi_{r}\right),\tag{III.15}\label{eq15} \end{equation} which at first seems to be non-zero.

As we know, it is hoped that \begin{equation} \lim_{\varepsilon\rightarrow 0} \dfrac{S^{\prime}-S}{\varepsilon}\approx 0.\tag{III.16}\label{eq16} \end{equation}

Questions

We have, therefore, two questions:

1. 
   

   In the situation (1), when we use the total derivative (\ref{eq1}), the divergence theorem seems nonapplicable over the second integral of the Eq. (\ref{eq6}), so the question is: Is it still possible to apply the divergence theorem the second integral (Eq.(\ref{eq6}))?

   
   


2. 
   
   

   In the situation (2), when we do not use the total derivative, we have a remaining term that is apparently is not null. The question is: Could this term become null? What does it really represent?

   
   

Of course, I am considering a possibility of that I have committed some mistake in all the way follow at here, but, at the point of view mathematical, all my calculations seem to be correct. I would be very grateful if anyone could see something besides what I have seen.

Answer

Concerning situation 1, the main point seems to be that the generalized divergence theorem works with total derivatives, not partial derivatives.

thermodynamics - Why does earth cool?

This question is boggling me for some time. We know that heat can be transferred from matter to matter and heat is nothing more than tiny atoms vibration intensity (correct me if I'm wrong). But space is a vacuum, and so arises the question: Can you heat vacuum? :)) How can it be that earth loses it's heat to space itself?

Thanks

Answer

Earth can lose heat to space through radiation.

The earth behaves roughly as a blackbody and so radiates electromagnetic radiation into space at a rate of roughly 120 PW.

cosmology - What should I consider as an observer to measure the speed of cosmic objects?

I mean for example if earth is the observer, then there might be entire galaxies travelling faster than the speed of light relative to earth. So according to Einstein relativity this shouldn't be possible so I want to know what I should consider as an observer to measure cosmic objects' speeds.

Answer

First about reference frames of value for cosmological observations.

So, the preferred reference frame for cosmological measurements is the comoving one, meaning it is moving along with the average flow of galaxies due to the expansion. In that reference frame one is said to have zero peculiar velocity. We have to vectorially subtract our peculiar velocity. It is mostly due to the galaxy and local cluster moving wrt to the cosmic flow. The total is about 360 Kms/sec. When we do that adjustment vectorially right,, in the observations, we see the universe as on the average homogeneous and isotropic, and it is the reference frame where the CMB is also isotropic.

The comments and answer already explained that in fact galaxies can are are moving away from us faster than light. For galaxies at distances of about 14 billion light years, at a redshift of about z = 1.5, they are super-luminal (faster than light), but we still see them. We can see now out to the so called particle horizon, about 45 billion light years.

The fact that they are faster than light does not break relativity. The galaxies are not moving that fast, just the space between us is expanding, GROWING, and at distances greater than about 14 billion light years the space is expanding faster than light. There is no problem with that, general relativity explains it and has a pretty good model of it, with many cosmological values measured. As @Cort Ammon said in his comment, this whole thing should conceptually bother you. Nevertheless it is true, and it is fun to read and understand it and try to get convinced it's real. See the wiki articles on the expansion of the universe, and maybe some of the math. Try https://en.m.wikipedia.org/wiki/Metric_expansion_of_space For a good intro w/o the math.