In quantum mechanics, the unitary time translation operator $\hat{U}(t_1,t_2)$ is defined by $\hat{U}(t_1,t_2)|ψ(t_1)\rangle = |ψ(t_2)\rangle$, and the Hamiltonian operator $\hat{H}(t)$ is defined as the limit of $i\hbar\frac{\hat{U}(t,t+\Delta t)-1}{\Delta t}$ as $\Delta t$ goes to $0$. Similarly, the one-dimensional spatial translation operator is defined by $\hat{T}(x_1,x_2)|x_1\rangle = |x_2\rangle$ and the momentum operator $\hat{p}$ is defined as the limit of $i\hbar\frac{\hat{T}(x,x+\Delta x)-1}{\Delta x}$ as $\Delta x$ goes to $0$. My question is, why is it that the Hamiltonian operator can be a function of the time parameter $t$, but the momentum operator cannot be a function of the position parameter $x$?

The only good answer I've gotten to this question is that time is not an operator in non-relativistic quantum mechanics, whereas position is an operator, so momentum being a function of position would spoil the position-momentum commutation relation. But this explanation doesn't make sense to me, because consider the case of spin angular momentum. If $\hat{R}_z(\theta_1,\theta_2)$ denotes the rotation operator for intrinsic rotations about the z-axis (as opposed to orbital rotations), then the spin angular momentum operator $\hat{J}_z$ (as opposed to Beyoncé) is defined as the limit of $i\hbar\frac{\hat{R}_z(\theta,\theta+\Delta\theta)-1}{\Delta\theta}$ as $\Delta\theta$ goes to $0$. And yet $\hat{J}_z$ is not a function of the angle $\theta$, even though there is no operator in quantum mechanics corresponding to $\theta$. (There is another operator called $\hat{\theta}$, which is one of the position operators in spherical coordinates, but that has nothing to do with spin and the $\theta$ that I'm talking about; it's related to orbital angular momentum.) So "the parameter has a corresponding operator" doesn't seem like the right explanation, since it doesn't explain why spin angular momentum can't be a function of angle.

Note that I'm not looking for an ad hoc explanation like "that wouldn't make physical sense in terms of how energy and momentum work classically". I want a first principles explanation in quantum mechanics.

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    $\begingroup$ This question contains many misconceptions. The translation operator is defined as $T(a)|x\rangle=|x+a\rangle$, where $a\in\mathbb R$. The momentum operator is defined as $T'(0)$, so it is manifestly independent of $x$: not even before taking $a\to 0$ did $T$ depend on $x$. Position is not a parameter, it is the eigenvalue of an operator (and therefore it makes no sense to claim that a second operator may or may not depend on it). Momentum does not depend on position in classical mechanics either, so it is unclear why it should in QM. $\endgroup$ – AccidentalFourierTransform Sep 8 '18 at 14:41
  • $\begingroup$ Similarly, the rotation operator is defined as $\hat R(\vec\alpha)|\vec x\rangle=|\mathcal R(\vec \alpha)\vec x\rangle$, where $\mathcal R(\vec\alpha)$ denotes a rotation matrix around $\hat\alpha$ with angle $|\vec \alpha|$. The angular momentum operator is defined as $\nabla\hat R(0)$, so it is manifestly independent of $\theta$: not even before taking $\vec\alpha\to0$ did it depend on $\theta$. Also, there is an operator that can be identified with $\hat\theta$, and the pair $\hat J_i,\hat \theta_j$ behaves very similarly to $\hat p_i,\hat x_j$ (although the picture is more subtle). $\endgroup$ – AccidentalFourierTransform Sep 8 '18 at 14:44
  • $\begingroup$ @AccidentalFourierTransform But the question is, why does the spatial translator operator not depend on $x$, whereas the time translation operator can depend on $t$? And the answer cannot be that it’s because position is an operator in nonrelativistic quantum mechanics whereas time is not, because it doesn’t explain why the intrinsic rotation operator doesn’t depend on $\theta$, given that $\theta$ isn’t an operator either. $\endgroup$ – Keshav Srinivasan Sep 8 '18 at 21:49

I think the answer comes down to causality. The typical problem that we address in nonrelativistic quantum mechanics is "Given an initial condition $\psi(\vec{x}, t = 0)$, what is the wave function $\psi(\vec{x}, T)$ at a later time $T$?" The fact that the Hamiltonian $H$ generating the time translation/evolution is allowed to depend explicitly on time corresponds to the fact that we, the experimenters, are free to externally drive the system in any way we want, and the external drive cannot be "predicted" endogenously within the system. Phrased differently, the causal influences from the external drive propagate forward in time to affect the future wavefunction.

But if the momentum operator $P$ which generates translations were allowed to depend nontrivially on space, then by analogy with the Schrodinger equation (and simplify to one spatial dimension) we could set up a problem with an "initial condition" $\psi(x = 0, t)$ and consider the problem of "space-evolving" the wavefunction in $x$ according to the differential equation $-i\, \partial \psi / \partial x = P(x)\, \psi(x)$. If the experimentalist were free to externally change $P(x)$, then the influences of that change would need to propagate in a spacelike direction in order to affect the wave function at the same $t$ but larger $x$, violating causality.

Therefore, the fact that the $P$ operator must be "space-independent" while the $H$ operator can depend explicitly on time is a non-relativistic reflection of the fact that the full quantum theory is relativistic and causal influences can only propagate in timelike directions. In a completely nonrelativistic universe, the momentum operator $P$ probably could logically depend explicitly on position - just as bosons and fermions could logically have any spins, but in the real world they "inherit" the spin-statistics relation from the underlying relativistic theory.

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    $\begingroup$ What a beautiful answer!!!! I literally paused in awe when i was reading "But if the momentum operator P which generates translations were allowed to depend nontrivially on space........" $\endgroup$ – Prasad Mani Oct 30 '16 at 13:26
  • $\begingroup$ But then why can't the spin angular momentum operator $\hat{J}_z$ depend on the angle $\theta$, the parameter of the intrinsic rotation operator $\hat{R}_{z}(\theta)$? $\endgroup$ – Keshav Srinivasan Oct 30 '16 at 22:03
  • $\begingroup$ @KeshavSrinivasan Again, it's logically and mathematically possible, but such a dependence would explicitly break symmetries that are assumed to hold for the full relativistic theory. Just as an explicit $t$ dependence in the Hamiltonian breaks time-translational symmetry, an explicit $\theta$ dependence for the $J$ operator would break the rotational invariance of space - there would be a "special direction" and physics might, say, behave differently for systems facing that direction vs. facing perpendicular to that direction. It's completely logically possible, though inelegant, ... $\endgroup$ – tparker Oct 31 '16 at 7:37
  • $\begingroup$ in a completely nonrelativistic universe and cannot be ruled out from first principles. But since special relativity mixes rotations and boosts, such a dependence would completely ruin the entire Poincare symmetry of the relativistic theory. If the Hamiltonian described a subsystem under the control of an experimentalist, you could even arrange to simulate a nontrivial $\theta$ dependence by, say, manually adding energy into the system every time it rotated to point north or something. But it's technically difficult and unnatural to build a "fundamental" relativistic theory that isn't ... $\endgroup$ – tparker Oct 31 '16 at 7:41
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    $\begingroup$ Very clear and eloquent; this should be the accepted answer. @KeshavSrinivasan, I think that the answer to your objection is the following. If the only Hamiltonian that anyone every studied was the Hamiltonian where the entire universe is taken as the system under study, then (1) this Hamiltonian would be time-independent, and (2) time translation would work just like spatial translation, so your question would vanish. But since we like to study systems that are smaller than the whole universe, where an experimentalist can control the environment, then we treat time differently. $\endgroup$ – sasquires Oct 10 '18 at 20:50

Your idea of what is defined in terms of what is a bit misleading. Usually, the physicist takes the infinitesimal generators $H,p,x$ as given self-adjoint operators and defines the finite transformations to be $U(t) = \exp(-\mathrm{i}tH),T(\xi) = \exp(\mathrm{i}\xi p),S(\pi) = \exp(\mathrm{i}\pi x)$ following Stone's theorem. If $H$ is time-dependent then $U(t)$ turns into the Dyson series for $U(t,t_0)$ in the interaction picture.

As to the definition of the momentum operator itself: It's simply defined to be the operator with $[x,p] = \mathrm{i}\hbar$. By the Stone-von Neumann theorem, all possible ways to realize operators with that commutation relation are essentially the same as the one on $L^2(\mathbb{R})$, where $x$ is multplication by the variable and $p$ is differentiation. The commutation relations also encode that the transformation $T(\xi)$ acts as a translation on the position and that $S(\pi)$ acts as a translation on the momentum, see also this answer of mine. But crucially, $p$ is by definition a single fixed operator. It's just not allowed to depend on anything.

Finally, your confusion seems to basically arise from writing all those transformations with two parameters, i.e. $U(t_0,t_1),T(x_1,x_2)$. Only the time evolution is allowed to depend on two parameters in that way, and only in the case of a time-dependent Hamiltonian. All other transformations are one-parameter groups as in Stone's theorem, generated by a single self-adjoint operator. This is not shown, but assumed. We assume that the rotation operator $R(\theta_1,\theta_2)$ really only cares about the difference between the two angles, that is, it's really just a function $R(\theta_1 - \theta_2)$ and we assume that the translation $T(x_1,x_2)$ is really just $T(x_1-x_2)$. You could assume differently, but that's not what we do in standard quantum mechanics.

We assume that for all those transformation because we want the $T(x_1,x_2)$ to actually be a (unitary) representation of the translation group $\mathbb{R}$, and the $R(\theta_1,\theta_2)$ to be a representation of the rotation group $\mathrm{SO}(3)$. And those groups don't contain the transformations "rotate from angle $\theta_1$ to $\theta_2$", but "rotate by angle $\theta$", so the operator will also only depend on the difference, and not the start/endpoints of the transformation.

The case for the time evolution is different - although one might say there's a "time translation group", what we actually want is an operator that encodes the evolution of a dynamical system. And in a dynamical system we can easily imagine that at some point in time $t_0$ something is "switched on/off" that alters the dynamics of the system after that point, so that $U(t_1,t_2)$ is different depending on whether both $t_1,t_2$ are before or after $t_0$.

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    $\begingroup$ I don't agree. It is perfectly standard to take the unitary operators as primitive and to define the infinitesimal generators in terms of those. That's what countless textbooks do, including Townsend and (if I recall) Sakurai. And then the commutation relation is derived, not assumed. And similarly the Dyson series is also derived, not assumed. It's in the context of that approach that I'm asking my question. $\endgroup$ – Keshav Srinivasan Oct 22 '16 at 15:43
  • $\begingroup$ @KeshavSrinivasan Then my last paragraph still applies: That the translation operator $T(x_1,x_2)$ depends actually only on $x_1-x_2$ and is hence a one-parameter subgroup to which we can apply Stone's theorem to get a single independent momentum operator is an assumption. $\endgroup$ – ACuriousMind Oct 22 '16 at 15:45
  • $\begingroup$ Well, that's precisely what my question is: why is it that we assume that $T(x_1,x_2)$ depends only on $x_1-x_2$, and $R_z(\theta_1,\theta_2)$ depends only on $\theta_1-\theta_2$, but we allow $U(t_1,t_2)$ to depend directly on $t_1$ and $t_2$? $\endgroup$ – Keshav Srinivasan Oct 22 '16 at 15:51
  • $\begingroup$ @KeshavSrinivasan I added a reasoning for that. $\endgroup$ – ACuriousMind Oct 22 '16 at 15:56
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    $\begingroup$ @KeshavSrinivasan Because this way of describing physical systems works? I'm not sure what kind of reason you're really looking for - in the end, the assumptions made in our physical theories are there because the resulting theory correctly predicts the results of experiments. Also, naively, wouldn't you agree that putting something at position $x_1$ to $x_1+ b$ and something at $x_2$ to $x_2+b$ should be the same operation no matter what $x_1$ and $x_2$ are? $\endgroup$ – ACuriousMind Oct 22 '16 at 16:57

I think the notion that either the position or momentum operator is a function of the other is a bit ill-defined. I realize that you do not want explanations from classical physics, so please excuse for the moment the analogy with Hamiltonian mechanics.

In Hamiltonian mechanics we deal with a $2n$-dimensional space, on which there should be a 2-form, i.e., an anti-symmetric 2-tensor $\omega_{ij}$, that should be non-degenerate, $\omega_{ij} v^j \neq 0$ unless $v^j = 0$, and closed, $\partial_{[i} \omega_{jk]} =0$ where the brackets mean anti-symmetrization. These conditions are all coordinate independent. Since $\omega_{ij}$ is non-degenerate, it has an inverse $\omega^{ij}$. If $f,g$ are functions on this $2n$-dimensional space, we can define an operation called the Poisson bracket of $f$ and $g$ by $$\{f, g\} = \omega_{ij} (\omega^{ik} \partial_k f)(\omega^{jl} \partial_l g).$$ Note that the Poisson bracket is defined in a coordinate-independent way. The other thing we need to do Hamiltonian mechanics is a Hamiltonian function $H$, that defines the dynamics through $$\dot f = \{f, H \}.$$

The components of the 2-form $\omega_{ij}$ are of course coordinate-dependent. Now by a theorem of Darboux, it is always possible to (locally) find what is called canonical coordinates $x^i, i = 1, \ldots, 2n$ such that $\omega_{ij}$ takes the following form $$\omega_{\mathfrak{ij}} = \begin{bmatrix}0 & I_n \\-I_n & 0\end{bmatrix}$$ where $I_n$ is the $n\times n$ identity matrix. Let $q^i = x^i, i = 1, \ldots, n; p^i = x^{i+n}, i = 1, \ldots, n.$ Then you can work out that $$\{f,g\} = \frac{\partial f}{\partial q^i} \frac{\partial g}{\partial p^i} - \frac{\partial f}{\partial p^i} \frac{\partial g}{\partial q^i}.$$ In particular, $$\{q_\mathfrak{i}, p_\mathfrak{j}\} = \delta_{\mathfrak ij} \quad \{q_\mathfrak{i}, q_\mathfrak{j}\} = \{p_\mathfrak{i}, p_\mathfrak{j}\} = 0 $$ and I can easily recover Hamilton's equations in the canonical form.

In a particular coordinate system like the one described, a canonical coordinate system, the $q^i$ are called the positions or coordinates, and the $p^i$ are called the momenta. Neither is a function of the other. But that is true for any coordinate system: all the coordinates are mutually independent. One could equally well use $v^i = p^i - eA^i(q^j)$ as the second half of the coordinates. The components of the 2-form will be more complicated and the formula for the Poisson bracket won't be as nice, but $(x^i, v^i)$ is a perfectly fine coordinate system.

Now let's go over to quantum mechanics. In quantum mechanics we have operators instead of we use commutators instead of Poisson brackets. The dynamics are given by $$\hat{ \dot O} = \frac{i}{\hbar} [\hat O, \hat H]$$ and by analogy with Hamiltonian mechanics we introduce as the observable operators $\hat x_i, \hat p_i$ that satisfy $$[\hat x_i, \hat p_j] = i\hbar \delta_{ij} \quad [\hat x_i, \hat x_j] = [\hat p_i, \hat p_j] = 0.$$ We say that these operators form a basis of a Lie algebra. But just as I am free to change coordinates in the classical case, I am free to change basis in the quantum case. It doesn't make sense to say that the position operator is a function of the momentum operator or vice versa because they are all mutually independent basis elements in the Lie algebra.

The quantum case can also directly be formulated in terms of a change of coordinates. Then instead of the Poisson bracket on $2n$-dimensional phase space, one has the Moyal bracket. Like the Poisson bracket, the Moyal bracket acts like a differential operator, and the expression for it is particularly simple in special coordinate systems, but it is not necessary to use such coordinates.

  • $\begingroup$ But this doesn't explain why the spin angular momentum operator isn't a function of the angle $\theta$ that the intrinsic rotation operator $\hat{R}_z$ is a function of. $\theta$ does not have any operator or observable corresponding to it, so it's like $t$ in that respect. And yet the Hamiltonian operator $H$ is allowed to be a function of $t$. $\endgroup$ – Keshav Srinivasan Oct 22 '16 at 16:27
  • $\begingroup$ @KeshavSrinivasan Why do you say that there is no operator for $\theta$ and that it's not an observable? Is $\arccos(\hat{z}/\hat{r})$ not good enough for you? Do you think quantum mechanics cannot describe rigid bodies? (How then, can classical mechanics?) $\endgroup$ – Robin Ekman Oct 22 '16 at 16:33
  • $\begingroup$ That's a different $\theta$, as I made clear in my question. That $\theta$, which is one of the position operators in spherical coordinates, is a parameter of the orbital rotation operator. Whereas the $\theta$ I'm talking about is a parameter of the intrinsic rotation operator, i.e. the rotation operator connected to spin. The orbital angular momentum operator has a commutation relation with the $\theta$ that you're talking about. But there is no similar angle operator that the spin angular momentum operator has a commutation relation with. $\endgroup$ – Keshav Srinivasan Oct 22 '16 at 16:45
  • $\begingroup$ Then why would such an operator exist? To the extent that a classical theory of spin, i.e., intrinsic angular momentum, exists, it isn't a theory of a rigid rotor. See for example here physics.stackexchange.com/questions/108387/… $\endgroup$ – Robin Ekman Oct 22 '16 at 16:54
  • $\begingroup$ Yes, that's precisely my point, such a $\theta$ operator does not exist. So why can't $\hat{J}_z$ be a function of $\theta$, just as $\hat{H}$ is a function of $t$? $\endgroup$ – Keshav Srinivasan Oct 22 '16 at 17:17

Actually, there are scenarios where the momentum operator could depend on position. Consider for instance the propagation of light through a random medium. If we incorporate the effect of this random medium into the unitary evolution of the field through the medium, then the momentum operator that one would derive from that would depend of position due to the randomness of the medium.

When the momentum operator does not depend on position, it reflects the fact that the system under investigation obeys spatial translation invariance, and therefore supports the conservation of momentum. In a random medium, momentum is not conserved, because the medium could cause the scattering of light, which implies a change in the momentum.

However, the same applies for the Hamiltonian. If the Hamiltonian has an explicit time-dependence, then the system is not invariant with respect to time translations and then energy conservation breaks down.

At the fundamental level we know that both momentum and energy are conserved. This is reflected in the fact that neither the Hamiltonian, nor the momentum operator depends explicitly on time or position. For instance, in quantum field theories (such as QED) the dependence on the space-time coordinates is restricted to that of the fields and does not appear explicitly in the Lagrangian. The implied translation invariance in space-time coordinates leads to a Noether current, the energy-momentum tensor, from which one obtains the expressions for the Hamiltonian and the momentum operators, purely expressed in terms of the fields and their derivatives.

  • $\begingroup$ As a simplest example, can we consider the particle in a box system ? Here the momentum operator can only be defined in a small interval (hence position dependent), although there are other problems concerning the self-adjointness of momentum but that is another topic of interest. $\endgroup$ – user35952 Oct 26 '16 at 8:26

First think of the Hamiltonian.The Hamiltonian is a generator of time translations which is why it can be a function of $t$. It can be time dependent or time independent.This encodes how the system evolves as a function of time. Now mathematically speaking the momentum is the generator of space translations so in principle it can be a function of a position parameter. So it could be position dependent or position independent. This would tell us how the system evolves in space. Most experimentalists do not do experiments moving quantum systems around in space but they do, do experiments moving systems in time. It is conceivable in the future when techniques in quantum control are highly advanced and quantum systems can be protected from decoherence that one day an experimentalist would need to know a momentum operator as a function of a position parameter. Also remember that this is non-relativistic physics, position and time are not on the same footing. There is no reason to think they should behave the same way. So to answer your question directly the momentum operator can in principle be a function of a position parameter.

  • $\begingroup$ I'm pretty sure that if momentum was a function of position, it would be easily observable experimentally, since the position-momentum commutation relation would not hold, and thus you would observe violations of Heisenberg's uncertainty principle, which we simply do not observe. $\endgroup$ – Keshav Srinivasan Oct 22 '16 at 1:01
  • $\begingroup$ The Heisenberg commutation relation is true if the momentum operator is independent of a position parameter but that says nothing about if it is possible for momentum to be dependent on a position parameter. There could well be a differential equation to solve for momentum in some complicated system and as long as the momentum operator remained hermitian the operator acting on the quantum state got by exponentiating the momentum operator would remain unitary and therefore a legitimate quantum operator. $\endgroup$ – Amara Oct 22 '16 at 14:01
  • $\begingroup$ This would need to be checked but I am pretty sure that any Hamiltonian that coupled position with momentum in a sufficiently complicated way would give you momentum as a function of position. $\endgroup$ – Amara Oct 22 '16 at 14:54

why is it that the Hamiltonian operator can be a function of the time parameter t, but the momentum operator cannot be a function of the position parameter x?

A momentum operator can be a function of $x$. I'll quote extensively from pages 57-58 of Aitchison & Hey's "Gauge Theories in Particle Physics, 2nd Ed." so this is too long to be a comment:

The essential point is that (in one dimension, say) $\hat p$ is defined ultimately by the commutator $(\hbar = 1)$

$$[\hat x, \hat p] = i$$

Certainly, the familiar choice

$$\hat p = -i \frac{\partial}{\partial x}$$

satisfies the commutation relation. But we can also add any function of $x$ to $\hat p$, and this modified $\hat p$ will still be satisfactory since $x$ commutes with any function of $x$. More detailed considerations by Dirac showed that this arbitrary function must actually have the form $\frac{\partial F}{\partial x}$, where $F$ is arbitrary. Thus

$$\hat p ' = -i \frac{\partial}{\partial x} + \frac{\partial F}{\partial x}$$

is an acceptable momentum operator.

  • $\begingroup$ There might be ten thousand operators that satisfy the commutation relation, but there's presumably only one operator that's equal to the limit of $i\hbar\frac{\hat{T}(x,x+\Delta x)-1}{\Delta x}$ as $\Delta x$ goes to 0. I'm interested in why that operator cannot be a function of position. $\endgroup$ – Keshav Srinivasan Oct 23 '16 at 0:47
  • $\begingroup$ @KeshavSrinivasan, fair enough but allow me to point out the following in your question: "so momentum being a function of position would spoil the position-momentum commutation relation". $\endgroup$ – Alfred Centauri Oct 23 '16 at 1:18

A lot of mathematical answers here. I'm not sure how satisfactory these will "feel" as the ones I've read do not appear to give any intuition for why momentum is not the derivative of position.

I think the intuitive answer is simply that momentum for a wave is not the same as momentum for a particle. This momentum operator ($\frac{\hbar}{i} \frac{d}{dx}$) is more like a "wave-momentum-operator" and measures the momentum of the wavepacket associated with the quantum state. But nothing stops you from keeping track of the "particle-momentum" and explicitly calculating $m\frac{dx}{dt}$.

Here's a decent answer for building intuition for what the momentum of a wave is:

To clarify the problem, let us consider a simplified model of the string: The string extends along the x-direction and is made up of masses connected by springs. For conceptual clarity, suppose these masses can only move up and down (along y; this could be enforced in a mechanical model by having the masses sliding up and down on little wires). In that case, the mechanical momentum is clearly only in the y-direction (all motion is along y), and there is no mechanical momentum in the direction of propagation of the wave (x). Depending on the shape of the wave packet, the total momentum in y-direction might also be zero, if some masses are moving up while others are moving down.

This "wave-momentum" operator turns out to be particularly useful particularly when describing the energy of the system (and can also be used to "generate" motion in position by applying the operator multiple times, but this motion is the motion of the entire wavepacket).

Now why is it the derivative of x specifically? We can gain some intuition by looking at the plane wave solution to schroingers equation:

$$ \psi (x,t)=e^{{\frac {i}{\hbar }}(px-Et)} $$

Let's try to solve for p and figure out what p is specifically. We can get the p to "come down" from the exponential by applying a partial derivative.
where p is

$$ \frac{\partial \psi}{\partial x} = \frac{\partial}{\partial x} e^{{\frac {i}{\hbar }}(px-Et)} = \frac {i}{\hbar} p e^{\frac {i}{\hbar }(px-Et)} $$

$$ p \psi = \frac {\hbar}{i} \frac{\partial}{\partial x} \psi $$

Therefore we can infer what this p operator is by moving all of the terms around to explicitly show what p must be doing as an operator:

$$ \hat {p}=-i\hbar \frac{\partial }{\partial x} $$


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