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  Entangled or unentangled?

+ 7 like - 0 dislike
536 views

I got a little puzzled when thinking about two entangled fermions.

Say that we have a Hilbert space in which we have two fermionic orbitals $a$ and $b$. Then the Hilbert space $H$'s dimension is just $4$, since it is spanned by \begin{align} \{ |0\rangle, c_a^\dagger |0\rangle, c_b^\dagger |0\rangle, c_a^\dagger c_b^\dagger |0\rangle\}, \end{align} where $c_i^\dagger$ are the fermionic operators that create a fermion in orbital $i$.

Say we have a state $c_a^\dagger c_b^\dagger |0\rangle$. Then if I partition my Hilbert space into two by looking at the tensor product of the Hilbert spaces of each orbital, i.e. $H = H_a \otimes H_b$, then my state can be written as $c_a^\dagger |0\rangle_a \otimes c_b^\dagger |0\rangle_b$, from which it is obvious that this state is unentangled ($|0\rangle = |0\rangle_a \otimes |0\rangle_b$).

Now I was thinking about writing the state in first quantized i.e. a wavefunction. Let $\phi_a(r), \phi_b(r)$ be the wavefunctions of the orbitals $a$ and $b$. Then \begin{align} \psi(x_1,x_2) = \langle x_1 x_2 | c_a^\dagger c_b^\dagger |0 \rangle = \phi_a(x_1) \phi_b(x_2) - \phi_a(x_1)\phi_b(x_2). \end{align} This is where I got confused. What object is $\psi(x_1,x_2)$, i.e. what Hilbert space does it belong to? What exactly are we doing when we do $\langle x_1 x_2 | c_a^\dagger c_b^\dagger |0 \rangle$? We seem to be changing/expanding our Hilbert space by taking the position representation?

Written in this way, and assuming the same partition $H_a \otimes H_b$, the unentangled nature of the original state is no longer manifest. I'm not sure what the partition $H_a \otimes H_b$ even means in this context. Would that be saying $\psi(x_1, x_2) = \psi_a(x_1, x_2) \times \psi_b(x_1,x_2)$ where $\psi_i(x_1,x_2)$ is a linear combination of $\phi_i(x_1), \phi_i(x_2)$? This does not seem right to me.

Regardless, now I have a state written in two different but supposedly equivalent ways, with the same partition of the Hilbert space, yet it is unentangled in one way and entangled in the other.

Help?

This post imported from StackExchange Physics at 2014-04-04 16:42 (UCT), posted by SE-user nervxxx
asked Apr 25, 2013 in Theoretical Physics by nervxxx (210 points) [ no revision ]
You cant partition a two-fermion sector of the Fock space as tensor product - underlying states are always anti-symmetriezed

This post imported from StackExchange Physics at 2014-04-04 16:42 (UCT), posted by SE-user Slaviks
@ Slaviks : sorry, I didn't quite get what you meant. Are you saying $H = H_a \otimes H_b$ is wrong? but that's how you build up multi-fermion Fock states. each operator $c_i^\dagger$ acts in a Hilbert space which is just the vacuum and an excitation, and so for $N$ fermions $H = \bigotimes_{i=1}^N H_i$ ($2^N$ possible states). Here i'm just considering $N = 2$.

This post imported from StackExchange Physics at 2014-04-04 16:42 (UCT), posted by SE-user nervxxx
@Slaviks - if you apply a Jordan-Wigner transformation you immediately get a tensor-product-decomposition into 2-dimensional Hilbert spaces. I think this question is an ongoing debate in the literature. There are different approaches, since it appears not to be particularly easy to deal with the (anti-)symmetrization. In my view, the answer must be found by considering the result of measurements and their correlations and therefore (probably) factorization properties of one- and two-body density matrices in the simplest cases. See e.g. arXiv:0902.1684 and PRA 67, 024301 (2003) and their refs.

This post imported from StackExchange Physics at 2014-04-04 16:42 (UCT), posted by SE-user S. Gammelmark
@nervxxx I see what you mean. I'm saying that occupation number decomposition $H_1 \bigotimes H_2$ is not the same as (the impossible) decomposition into the tensor product of single-particle Hilbert spaces. But need to think more, especially in light of S. Gammelmark's comment.

This post imported from StackExchange Physics at 2014-04-04 16:42 (UCT), posted by SE-user Slaviks
Indeed, there is much literature around discussing the fact that the antisymmetrization that you automatically get in multi fermion states shouldn't be considered as operationally useful "entanglement" e.g. Zanardi et al state that it's not useful to discuss entanglement without specifying the manner in which one can manipulate and probe its constituent physical degrees of freedom. In this sense entanglement is always relative to a particular set of experimental capabilities.

This post imported from StackExchange Physics at 2014-04-04 16:42 (UCT), posted by SE-user twistor59
@ Slaviks : still confused. My $H_i$ is the Hilbert space of single orbitals, not a single-particle Hilbert space. The particle number is undetermined (but constrained to be $ 0 \leq 2 $ in this case), until I give a state. i.e. i am not looking at the 2-particle Fock space.

This post imported from StackExchange Physics at 2014-04-04 16:42 (UCT), posted by SE-user nervxxx
To expand on twistor59's comment: In this case, an important restriction is that the fermions are indistinguishable. In a traditional Bell state we can make any measurement on either system individually. But in a 1st quantized picture of 2 fermions occupying 2 modes, there's no way to make a measurement on just 1 of the fermions. It is a virtue of the 2nd quantized picture that this apparent restriction is shown be a confusion: you can't distinguish between the two fermions because they aren't two different objects. They just count the excitation of the more fundamental objects, the modes.

This post imported from StackExchange Physics at 2014-04-04 16:42 (UCT), posted by SE-user Jess Riedel

3 Answers

+ 2 like - 0 dislike

Let me remind you that the Fock space of multiple fermions is defined to be the antisymmetric (fermionic) subspace of the full Fock space

$$ \Gamma_a=\bigoplus_{n=0}^{\infty}H^{\wedge n}, $$

where $\wedge$ stands for the antisymmetric tensor product

$$ v_1\wedge\ldots\wedge v_n=\frac{1}{n!}\sum_{p\in P_n}\sigma_p v_{p(1)}\wedge\ldots\wedge v_{p(n)}. $$

Here $\sigma_p$ is the sign of the permutation $p$ in the group of permutations $P_n$.

Thus the confusion here comes from the fact that $c_a^{\dagger}c_b^{\dagger}|0\rangle\neq|ab\rangle$ as you seem to state.

Recall that the creation and annihilation operators are defined within the occupation number representation, i.e. $c_a^{\dagger}c_b^{\dagger}|0\rangle=|11\rangle$, where the first number denotes the number of fermions in state $a$ while the second denotes the number of fermions in state $b$. On the other hand, states written in the occupation number representation are defined to be properly antisymmetrized (for fermions) many-body basis states, as forced upon us by the particle indistinguishability. Therefore they are defined within the fermionic Fock space. Any textbook will show so, take a look at the first chapter of Many-Body Quantum Theory in Condensed Matter Physics: An Introduction by Bruus and Flensberg for example. For two fermions described via a single particle basis $\{|a\rangle,|b\rangle\}$ one possible choice is:

$$ |11\rangle=\frac{1}{\sqrt{2}}\left(|ab\rangle-|ba\rangle\right). $$ Therefore

$$ c_a^{\dagger}c_b^{\dagger}|0\rangle=\frac{1}{\sqrt{2}}\left(c_a^{\dagger}|0\rangle_1\otimes c_b^{\dagger}|0\rangle_2-c_b^{\dagger}|0\rangle_1\otimes c_a^{\dagger}|0\rangle_2\right) $$

The familiar anticommutativity of these operators is now obvious from this from

$$ c_b^{\dagger}c_a^{\dagger}|0\rangle=\frac{1}{\sqrt{2}}\left(c_b^{\dagger}|0\rangle_1\otimes c_a^{\dagger}|0\rangle_2-c_a^{\dagger}|0\rangle_1\otimes c_b^{\dagger}|0\rangle_2\right)=-c_a^{\dagger}c_b^{\dagger}|0\rangle $$

In fact one of the great advantages of creation and annihilation operators is that they include the antisymmetry (for fermions) of the wave function implicitly.

Dotting this with $\langle x_1x_2|$ we obtain:

$$ \langle x_1x_2|c_a^{\dagger}c_b^{\dagger}|0\rangle=\frac{1}{\sqrt{2}}\left(\phi_a(x_1)\phi_b(x_2)-\phi_b(x_1)\phi_a(x_2)\right). $$

Thus there is no inconsistency, both representations show that the particles are entangled.

On the other hand dotting $\langle x_1x_2|$ with $c_a^{\dagger}|0\rangle_1\otimes c_b^{\dagger}|0\rangle_2$ would simply produce

$$ \psi(x_1,x_2)=\phi_a(x_1)\phi_b(x_2) $$

Therefore there is no inconsistency here also, however, as I said, the important thing to remember is that

$$ c_a^{\dagger}c_b^{\dagger}|0\rangle\neq c_a^{\dagger}|0\rangle_1\otimes c_b^{\dagger}|0\rangle_2 $$

This post imported from StackExchange Physics at 2014-04-04 16:42 (UCT), posted by SE-user mgphys
answered Jun 3, 2013 by mgphys (20 points) [ no revision ]
No, that is not correct. With the choice of partitioning of the Hilbert space I have specified in the problem, the two subsystems are unequivocally unentangled. I cite, for example, the use of the orbital cut in the entanglement spectrum of QH states (go look it up) - in there, IQH states are unentangled because they have the form $|\psi\rangle = \prod_i c_i^\dagger |0\rangle$.

This post imported from StackExchange Physics at 2014-04-04 16:42 (UCT), posted by SE-user nervxxx
Next, $c_a^{\dagger}c_b^{\dagger}|0\rangle=\frac{1}{\sqrt{2}}\left(c_a^{\dagger}|0 \rangle_1 \otimes c_b^{\dagger}|0\rangle_2-c_b^{\dagger}|0\rangle_1\otimes c_a^{\dagger}|0\rangle_2\right)$ is wrong. The multiparticle Fock state vacuum is built up from the tensor product of individual vacuums $\prod_i |0\rangle_i$. Then a Fock state representation arises from acting the creation operators on this: $c_a^\dagger c_b^\dagger |0\rangle = |0\rangle_i \cdots c_a^\dagger |0\rangle_a \otimes \cdots c_b^\dagger |0\rangle_b \cdots|0\rangle_j$

This post imported from StackExchange Physics at 2014-04-04 16:42 (UCT), posted by SE-user nervxxx
The anticommutativty arises because of the canonical anticommutation relations $\{ c_i, c_j^\dagger \} = \delta_{ij}$. This then results in $c_a^\dagger c_b^\dagger |0\rangle = -c_b^\dagger c_a^\dagger |0\rangle$. the whole point of the Fock state number representation is that it treats the particles immediately as indistinguishable and does not have an unphysical label (the positions, $x_1, x_2, \cdots$) for each particle like a first quantized notation (wavefunction) would. The first quantized notation is really cumbersome and awkward, because what it does

This post imported from StackExchange Physics at 2014-04-04 16:42 (UCT), posted by SE-user nervxxx
is to say that each particle is distinguishable with position labels $x_1, x_2, \cdots$, but then it enforces the rule that they're actually indistinguishable and should be odd under a swap of labels, leading to the Slater determinant / antisymmetrization scheme. So, -1.

This post imported from StackExchange Physics at 2014-04-04 16:42 (UCT), posted by SE-user nervxxx
@nervxxx, I have edited my answer to further explain my point. However I would also like you to provide me with concrete references, since "go look it up" is not one.

This post imported from StackExchange Physics at 2014-04-04 16:42 (UCT), posted by SE-user mgphys
@nervxxx, as far as multiparticle Fock space is concerned, I explained in my answer that for fermions one uses the antisymmetric tensor product. Therefore for two fermions $c_a^{\dagger}c_b^{\dagger}|0\rangle=\frac{1}{\sqrt{2}}\left(c_a^{\dagger} |0\rangle_1 \otimes c_b^{\dagger}|0\rangle_2-c_b^{\dagger}|0\rangle_1\otimes c_a^{\dagger}|0\rangle_2\right)$ is correct.

This post imported from StackExchange Physics at 2014-04-04 16:42 (UCT), posted by SE-user mgphys
I agree with the rest of your comments, and point out that my whole answer is based on that.

This post imported from StackExchange Physics at 2014-04-04 16:42 (UCT), posted by SE-user mgphys
you might need some background in QH physics but the $\nu = 1$ IQH state is a pure product state princeton.edu/~jmaciejk/entanglement2012/slides/… slide 8, and arxiv.org/pdf/1111.2810.pdf page 2 "An easy inspection reveals that the ground state $\rho$ is a product state in the orbital basis and, hence, $\rho_A$ is a pure state for any orbital bi-partition (where all $N_A$ orbitals in subsystem $A$ are filled) of this system. Consequently, the ES (and, hence, the entanglement entropy) is trivial, consisting of exactly one level"

This post imported from StackExchange Physics at 2014-04-04 16:42 (UCT), posted by SE-user nervxxx
the gist is that for a pure product state, with the kind of partitioning I described in my post, there is no entanglement, i.e. it has only one Schmidt value.

This post imported from StackExchange Physics at 2014-04-04 16:42 (UCT), posted by SE-user nervxxx
Dear @nervxxx, thank you for the resources. I understand and agree with these arguments. The only point I was trying to make is that if you want a $H=H_a\otimes H_b$ partition you can't use the $c_a^{\dagger}c_b^{\dagger}|0\rangle$ vector, since it implies a antisymmetric partition. This, I believe, is the source of confusion.

This post imported from StackExchange Physics at 2014-04-04 16:42 (UCT), posted by SE-user mgphys
that statement is completely baseless. Why can't you use the vector? You choose a vector simply from the full Hilbert space $H$. The partitioning you impose on the Hilbert space, is a completely independent choice. For the resources I gave you, look: in the lowest Landau level, the Hilbert space is given by the tensor product of orbital Hilbert spaces $H = \otimes_{i=1}^{l} H_i$. The orbital cut is given by the choice of partiotning $H = H_a \otimes H_b \equiv (\otimes_{i=1}^{m} H_i) \otimes (\otimes_{i={m+1}}^{l} H_i)$. The IQH state, which is given by

This post imported from StackExchange Physics at 2014-04-04 16:42 (UCT), posted by SE-user nervxxx
$(\prod_{i=1}^l c_i^\dagger ) |0\rangle$, under this partitioning, becomes $\prod_{i=1}^m c_i^\dagger |0\rangle \otimes \prod_{i=m+1}^l c_i^\dagger |0\rangle$. So there is no entanglement. This is what the authors mean by the orbital cut having only one Schmidt level. I believe instead that you are quite confused with your understanding of the antisymmetrized Fock space.

This post imported from StackExchange Physics at 2014-04-04 16:42 (UCT), posted by SE-user nervxxx
I would appreciate it if you could specify what you think is wrong in my answer. Did you mean to say that $c_a^{\dagger}c_b^{\dagger}|0\rangle$ does not imply antisymmetric tensor product?

This post imported from StackExchange Physics at 2014-04-04 16:42 (UCT), posted by SE-user mgphys
Note that in the way you posed the question, it has very little to do with IQH. It also does not go into the nature of entanglement of indistinguishable particles, which is discussed in e.g. physik.rwth-aachen.de/fileadmin/user_upload/www_physik/… and sqig.math.ist.utl.pt/pub/PaunkovicN/04-P-phdthesis.pdf

This post imported from StackExchange Physics at 2014-04-04 16:42 (UCT), posted by SE-user mgphys
Also it seems to me that the question you asked is not the one that you meant to ask, and the answer to that latter one might be here physics.stackexchange.com/questions/35185/…

This post imported from StackExchange Physics at 2014-04-04 16:42 (UCT), posted by SE-user mgphys
I wouldn't trust the bachelor's thesis you gave me - it has errors. the phd thesis is ok. my question has everything to do with the IQH - I'm just restricting the number of orbitals to 2 for illustrative purposes, while in the IQHE, on a compact manifold the number of orbitals is $N_{orb}$. Ok what's wrong with your comment is specified in my second comment in the chain. Your last assertion, that $c_a^\dagger c_b^\dagger |0\rangle \neq \cdots $ is not correct. You are completely mixing up second quantized notation and first!

This post imported from StackExchange Physics at 2014-04-04 16:42 (UCT), posted by SE-user nervxxx
But anyway I see the trouble now, sorry it took so long. My $H$ is not the same as your $H$. Your $H$ is the single-particle Hilbert space. My $H_i$ is the Hilbert space of the $i$-th fermionic mode. I shall illustrate this difference by counting arguments. Say that we have $M$ orbitals (with different quantum numbers). Each orbital has its associated raising and lowering operators $c_i, c_i^\dagger$. So the Hilbert space of the orbital (which I have been using) is dimension 2. The full Hilbert space is the tensor product of the $M$ orbitals, $\otimes_i H_i$, which has dim $2^M$.

This post imported from StackExchange Physics at 2014-04-04 16:42 (UCT), posted by SE-user nervxxx
This tensor product is neither symmetrized nor anti-symmetrized. It is a plain tensor product. Now let's look at YOUR $H$, which we call $H_1$, the single particle Hilbert space. It is spanned by $M$ possible states $\phi_1(x), \cdots \phi_M(x)$. We can have at most $M$ electrons in this system, so the dimension of the full Hilbert space must be finite. Let's count it explicitly. what's the dimension of the 2 particle Hilbert space? Is it $H_1 \otimes H_1$? No, because we are using first quantized notation, and to obey spin-statistics theorem we need to antisymmetrize this space.

This post imported from StackExchange Physics at 2014-04-04 16:42 (UCT), posted by SE-user nervxxx
This gives us $H_2 = S_\nu H_1 \otimes H_1 \subset H_1 \otimes H_1$, where $S_\nu$ is the antisymmterization operator on a tensor. What's the dimension of $H_2$? It's ${M \choose 2}$. Then similarly, dim $H_3 = {M \choose 3}$, and so on, until $H_M$. Lastly we note that there is $H_0$, the vacuum, which has NO wavefunction associated with it (no arguments!), and has dimension 1. This defines the Fock space as you have stated, $H = H_0 \oplus H_1 \oplus \cdots H_M$, which has dim ${M \choose 0} + {M \choose 1} + \cdots + {M \choose M} = 2^M$, the same as before!

This post imported from StackExchange Physics at 2014-04-04 16:42 (UCT), posted by SE-user nervxxx
The key difference, though, is in what $H$ actually means. Don't get confused. Remember, my $H$ is the orbital Hilbert space, which has dim $2$. Your single-particle Hilbert space has dim $M$, which can possibly be infinite. The creation and annihilation operators give the so called second quantization formulation of QM, which totally avoids (or rather, hides) the issue of antisymmeterization in its commutation relations. That's where you are wrong because you are using $c_a^\dagger, c_b^\dagger$ like in first quantized notation.

This post imported from StackExchange Physics at 2014-04-04 16:42 (UCT), posted by SE-user nervxxx
I don't have the patience to read the whole exchange, but I wonder if the confusion hinges on what kind of partitioning you're doing. In that case I'm going to take a stab at commenting. The $\nu=1$ QH state is pure under orbital partitioning, but not under "particle partitioning". Maybe this will help -- arxiv.org/abs/0905.4024 @nervxxx, Your 2-particle state might bepure under orbital partitioning, but it is entangled under particle partitioning. Entanglement is completely dependent on how you choose to partition your system.

This post imported from StackExchange Physics at 2014-04-04 16:42 (UCT), posted by SE-user Siva
I've elaborated that comment in my answer.

This post imported from StackExchange Physics at 2014-04-04 16:42 (UCT), posted by SE-user Siva
+ 0 like - 0 dislike

I'm posting a modified version of my comment as an answer, as more people will see it this way.

I think the confusion hinges crucially on what kind of partitioning you're doing. The $\nu=1$ QH state is pure under orbital partitioning, but not under "particle partitioning". Maybe arXiv:0905.4204 will help. IIRC, they work out a simple example about this detail, in the 2nd section.

@nervxxx, Your 2-particle state might be pure under orbital partitioning, but it is entangled under particle partitioning. Due to the antisymmetrization, it looks like a singlet Bell state.

So the bottomline is that entanglement is completely dependent on how you choose to partition your system. The subtlety is not widely appreciated. For a nuanced discussion, see this article http://rspa.royalsocietypublishing.org/content/463/2085/2277.full

This post imported from StackExchange Physics at 2014-04-04 16:42 (UCT), posted by SE-user Siva
answered Sep 20, 2013 by Siva (720 points) [ no revision ]
Yes, thanks. That's what I concluded eventually. But what irks me (and the reason for the whole long exchange above) is that almost everyone else doesn't understand the Fock space (with single particle Hilbert space of dim $M$) is isomorphic to the tensor product of $M$ orbital spaces. One can see this from looking at the dimension of the Hilbert spaces - both are $2^M$, and one has a 1-1 between the two spaces, so they are isomorphic! That implies that one can actually decompose the Fock space into a tensor product - except that because this isomorphism only works for the full spaces,

This post imported from StackExchange Physics at 2014-04-04 16:42 (UCT), posted by SE-user nervxxx
one cannot decompose the singlet wavefunction into a tensor product as $c_a^\dagger | 0 \rangle \otimes c_b^\dagger | 0 \rangle$ and represent it as some $f(x_1) \otimes g(x_2)$.

This post imported from StackExchange Physics at 2014-04-04 16:42 (UCT), posted by SE-user nervxxx
+ 0 like - 2 dislike

We have to be careful with the bra-ket formalism and its meaning. Unlike $|x_1\rangle$, I am not sure that the notation $|x_1 x_2\rangle$ where $x_1$ and $x_2$ are positional coordinates makes any sense. In literature [1] the notation $|ab\rangle$ designates the Slater determinant or Hartree-Fock state, i.e.:

$$|ab\rangle=c_a^{\dagger}c_b^{\dagger}|0\rangle=\phi_a(x_1) \phi_b(x_2)-\phi_a(x_2) \phi_b(x_1)$$

My feeling is that your confusion is related to mixing of the occupation numbers formalism and real space representation.

[1] Szabo, Ostlund, "Modern quantum chemistry: introduction to advanced electronic structure theory"

This post imported from StackExchange Physics at 2014-04-04 16:42 (UCT), posted by SE-user freude
answered Apr 25, 2013 by freude (-20 points) [ no revision ]
If somebody will explain meaning standing behind |r1 r2> notation, I will be very grateful.

This post imported from StackExchange Physics at 2014-04-04 16:42 (UCT), posted by SE-user freude
$\newcommand\ket[1]{\left|#1\right>} \ket{x_1x_2}$ is a common shorthand for $\ket{x_1}\otimes\ket{x_2}$. In general, what is written inside a ket is just a label: its meaning is heavily context dependent. Mixing the two representations is not uncommon, as long as there is no confusion.

This post imported from StackExchange Physics at 2014-04-04 16:42 (UCT), posted by SE-user Frédéric Grosshans
ok, I totally agree with that. But let us concern a particular case from the question above, i.e. the case when $x_1$ and $x_2$ are quantum numbers labeling the eigenstates of the positional operator $\hat{x}$. What is the meaning of decomposing $\ket{x_1}\otimes\ket{x_2}$ ? Also, I am curious if there is any references containing such notation for coordinate space. It would be interesting to read more about.

This post imported from StackExchange Physics at 2014-04-04 16:42 (UCT), posted by SE-user freude
I untastood $x_1$ and $x_2$ as labelling the position of the two fermions. But then, of course the ket $\ket{x_1x_2}$ is not antisymmetrized and does not correspond to an allowed state. Only $\ket{x_1x_2} -\ket{x_2x_1}$ is a valid state.

This post imported from StackExchange Physics at 2014-04-04 16:42 (UCT), posted by SE-user Frédéric Grosshans
Does labeling means quantum numbers? I think symbols in bra and ket vectors in the Dirac notation can not be just labels. They should be proper quantum numbers. That is why I do not see clear sense in $|x_1 x_2>$ unless somebody explains me what is the operator states $|x_1 x_2>$ belong to?

This post imported from StackExchange Physics at 2014-04-04 16:42 (UCT), posted by SE-user freude
In general, they are labels to proper vectors in the Hilbert space. If "proper quantum numbers" means " labels to the eigenbasis of a given operator", then they don't have to be "proper quantum number". If the basis is free, then I don't exactly know what "proper" means

This post imported from StackExchange Physics at 2014-04-04 16:42 (UCT), posted by SE-user Frédéric Grosshans
With Fermions, the situation is complicated by the fact that you impose a global symmetry. $\newcommand\ket[1]{\left|#1\right>}$ $\ket{x1x2}\ket{\uparrow\uparrow}$ is not a valid state, but $\ket{x1x2}(\ket{\uparrow\downarrow}-\ket{\downarrow\uparrow}$ is

This post imported from StackExchange Physics at 2014-04-04 16:42 (UCT), posted by SE-user Frédéric Grosshans

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