Conservation laws and the foundations of quantum mechanics

It is useful to start by recalling the basic experiment in more detail.

At the core of our examples is the function $f(x)$

with $N$ an integer that we will take as large as we want and with $\alpha$ a real constant with the crucial property that $\alpha >1$.

This function has the property of “superoscillations”; see refs. 1 and 2) as well as (3–5) and the extensive review (6). It is a widespread belief that a function cannot have features smaller than the smallest wavelength of its Fourier components. Yet this is not true, as shown in refs. 1 and 2. Superoscillatory functions can oscillate on arbitrary long intervals with much shorter wavelengths than any of their Fourier components. Or, in frequency terms, they can oscillate with much higher frequency than the highest Fourier component. In particular, the function $f(x)$ in Eq. 1 is a superposition of frequencies strictly limited to the interval $[-1,1]$. However, for large $N$, in the region $|x|$ of order of $N^{\frac{1}{2}-ϵ}$ where $ϵ$ is an arbitrary small fixed positive constant, the function is

that is, it oscillates with frequency $\alpha >1$, faster than the maximal Fourier component.

To prove the above claims, first, by expanding the binomial, we see that $f(x)$ is a superposition of Fourier components of frequencies $\frac{2n-N}{N}$, with $n$ integer, $0\le n\le N$, which are limited to the interval $[-1,1]$:

with $C_n$ constants

However, consider now this function in the region $|x|\le L$ where $L$ is some fixed number, and let $N$ increase. (This is a smaller region than $N^{\frac{1}{2}-ϵ}$ where the superoscillations hold, but here, the proof is immediate. For the larger region of order $N^{\frac{1}{2}-ϵ}$, see ref. 1 as well as SI Appendix, Supporting Information I.) For large $N$, up to corrections of order $\mathcal{O}(1/N)$, we can approximate the exponentials by their first-order Taylor expansion and obtain

which means that in this region, the function oscillates with frequency $\alpha$.

Finally, note that since the size of the superoscillatory region increases with $N$, the number of superoscillatory wavelengths $\lambda =\frac{2\pi }{\alpha }$ in the superoscillatory region can be made as large as we want by taking $N$ large enough.

In the present paper, we will focus on angular momentum conservation, since it is somewhat simpler than the energy nonconservation example of ref. 1, though it is constructed along very similar lines.

For simplicity, we will also consider the mass of the particle of interest to be large enough and the time scales short enough so that the free Hamiltonian of the particle can be neglected.

We now construct our basic experiment using a version of the superoscillatory function $f$. Consider a particle moving on a circle and let $\theta$, $-\pi \le \theta \le \pi$, denote its position. Let the wavefunction be $|\mathrm{\Psi }⟩$ which in the angle representation is

where $a$ is an integer larger than 1 and $\mathcal{N}$ is a normalization factor. (The restriction of $a$ to integers is made for mathematical simplicity; the conclusions will not change if we take $a$ to be any rational or real number larger than 1).

Consider now the region of small angles $|\theta |$ of order $1/N^{1/2+ϵ}$. As we increase $N$, in this region, the state $\mathrm{\Psi }(\theta )$ presents superoscillations. The proof is immediate:

Since we will only be interested in angular momentum, once the angular dependence of the wavefunction is given, the value of the radius of the circle is irrelevant. However, it is convenient (also with an eye to possible experiments) to think of a circle of radius $R_N=N{r}_0$ with $r_0$ a fixed length. The region of interest, $|\theta |=O(1/N^{1/2+ϵ})$, corresponds to an arc length $L$ along the circle of order $N^{1/2-ϵ}{r}_0$. In the variable $x=\theta R_N$ that denotes the position along the circle, the situation maps immediately to that investigated in Eqs. 1–2; we can re-express it afterward in term of angles.

Following the same calculation as in Eq. 1–2, we find that $\mathrm{\Psi }(\theta )$ is a superposition of angular momentum eigenstates $e^{im\theta }$ of the angular momentum operator $\widehat{L}$ of eigenvalues $m$, with $-N\le m\le N$ (where we take $ħ=1$):

where $C_n$ is given by Eq. 4 with $\alpha =a$ and where $c_m=C_{2n-N}$.

On the other hand, for $|\theta |$ of order $1/N^{1/2+ϵ}$, in the limit of large $N$,

a much higher angular frequency, mimicking in this region a particle with angular momentum $\mathit{aN}$, $a$ times higher than the highest angular momentum in the superposition. Note also that if we take $N$ large enough, we can have the superoscillatory region to extend for as many superoscillatory wavelengths as we want.

The experiment consists in determining if the particle is in the superoscillatory region. In technical terms, we want to measure the projector on that region. When we find the particle there, it has angular momentum of approximately $\mathit{aN}$, which is much higher than all the angular momentum components that it initially had. The question is where did it take supplementary angular momentum from?

It is tempting to think that the answer is trivial and straightforward: There are two systems in the problem—the particle and the measuring device used for measuring the position. They interact when the measurement takes place. As the particle increased its angular momentum, surely, it must have received a kick from the measuring device. In return, the measuring device must have decreased its own angular momentum. Hence, this seems nothing else than a trivial example of conservation of the total angular momentum. Only that it is not so: The measuring device could not have provided the necessary angular momentum. In fact, in the limit of large $N$, the measuring device did not change its state at all.

The argument is simple. Let us compare what happens in two cases: i) when measurements are made on the particle prepared in the superoscillatory state $|\mathrm{\Psi }⟩$ which mimics the high angular momentum $\mathit{aN}$ in the region of interest and ii) when measurements are performed on a particle prepared in an angular momentum eigenstate of eigenvalue $\mathit{aN}$, i.e., when the high angular momentum is genuine. The interaction between the measuring device and the particle takes place in the region of superoscillations. In this region, the particle’s wavefunction $|\mathrm{\Psi }⟩$ looks (up to perturbations that we can make as small as we want by choosing $N$ large enough) like the eigenstate of angular momentum with value $\mathit{aN}$. Suppose now that we perform a measurement of the projection operator on the superoscillatory region. Since the difference between the wavefunction $\mathrm{\Psi }$ and the true eigenstate of eigenvalue $\mathit{aN}$ is only encoded in the wavefunction outside the superoscillatory region and since we took the free Hamiltonian of the particle to be zero which means that this information cannot propagate, this information is unavailable to the measuring device and cannot affect the result. Hence, the measuring device will behave in the same way in both cases.

Let us compare these two cases. In the latter case, the wavefunction is a monochromatic wave of wavelength $\lambda =2\pi R_N/aN=2\pi r_0/a$ (corresponding to the angular momentum $\mathit{aN}$). All that happens when we find the particle in our region of interest, $|x|$ of order $N^{1/2-ϵ}{r}_0$, is that the measuring devices simply cuts out of this monochromatic wave a wave-train of length equal to the lengths of this region. Since for increasing $N$, this region contains more and more wavelengths, this wave-train approximates better and better the original monochromatic wave. In other words, if we find the particle in the region of interest, the final wavefunction has angular momentum closer and closer to the original high angular momentum, and hence, the measuring device does not give it any supplementary angular momentum at all. But then, the measuring device could not give angular momentum in the first case either, since in both cases, it must behave in the same way. This is the paradox.

To model the measurement of the projection operator onto the region of interest in such a way as to enable us to see the exchange of angular momentum between the system and the measuring device, we proceed as follows. We take the measuring device to be a second particle which also moves on the circle and which has an internal degree of freedom of this particle, the “pointer.” The location of the measuring device on the circle determines the location of the region where we would like to perform the projection. Let its coordinate be denoted by ${\theta }_M$, and let its initial wavefunction be $\varphi ({\theta }_M)$. Let the pointer consist of an internal degree of freedom described by a 2-dimensional Hilbert space. Initially, we take the pointer in the state $|\uparrow ⟩$. The initial state of the particle and measuring device is therefore

What we want to arrange is an interaction that a) conserves the total angular momentum and b) the pointer flips to state $|\downarrow ⟩$ if and only if the particle is in the superoscillatory region. We can do this by using the interaction Hamiltonian

where $g(\theta ;\phi )$ denotes a square pulse of width $\phi$, with $0<\phi <\pi /2$, i.e.,

and where the ${\sigma }_x$ operator flips $\uparrow$ to $\downarrow$ and vice-versa.

Note first that the angular dependence of the Hamiltonian is only via $(\theta -{\theta }_M)\mathrm{mod}2\pi$, the relative angle between the particle and the measuring device. This ensures conservation of the total angular momentum. Second, the Hamiltonian is zero for $(\theta -{\theta }_M)\mathrm{mod}2\pi \ge \phi$ and proportional to ${\sigma }_x$ for $(\theta -{\theta }_M)\mathrm{mod}2\pi \le \phi$. In other words, it flips the pointer only when the particle is within a distance $\phi$ from ${\theta }_M$, the location of the measuring device, and leaves the pointer unchanged when the particle is further away. All that is left is to take $\phi$ and ${\theta }_M$ such as to ensure that this region is within the superoscillatory region and captures as much as possible of this region. To do this, we take $\phi$ of the order of the superoscillatory region and prepare the measuring device in a wavepacket $\varphi ({\theta }_M)$ centered on ${\theta }_M=0$ and with an appropriately small spread $\mathrm{\Delta }$. For example, we take (up to normalization) $\varphi ({\theta }_M)=g({\theta }_M;\mathrm{\Delta })$ with $\mathrm{\Delta }<<\phi$, so that the interval $|\theta |\le |\phi +\mathrm{\Delta }|$ is still in the superoscillatory region.

The corresponding unitary time evolution operator over an infinitesimal time $\tau$ is

where in the last equality, we used in the trigonometric functions the definition of the function $g$.

For any arbitrary initial state $\psi (\theta )$ of the particle, the state of the particle and the measuring device is a superposition of two terms, one with the pointer flipped (for $(\theta -{\theta }_M)\mathrm{mod}2\pi \le \phi$) and one with the pointer not flipped:

As expected, the term with the flipped pointer is (up to normalization) simply the arbitrary original state $\psi$ truncated to a region of size $\phi$ centered on ${\theta }_M$, that is, the projection of the original state onto this region.

Let’s now take the arbitrary state $\psi$ to be the superoscillatory state $\mathrm{\Psi }$. In this case, when the pointer has flipped, the state of the particle and measuring device is (up to normalization)

using the approximation of $\mathrm{\Psi }$ for the small angles and the fact that for larger angles $g(\theta -{\theta }_M;\phi )=0$.

The crucial thing, as we explained above in the previous section, is that when the particle was found in the superoscillatory region, particle plus measuring device state is, up to approximations of $1/N$ which can be made as small as we want, the same as when the initial state of the particle was the genuine high angular momentum eigenstate $e^{iaN\theta }$. So whatever happens to the measuring device when the initial state is the bona fide high angular momentum state, happens also when the initial state is our “fake” high angular momentum state.

Let us now study the particle-measuring device angular momentum exchange. Consider first what would happen if the particle were to start in the high angular momentum state $\mathit{aN}$ and the measuring device pointer is flipped, meaning that the particle is found in the $|\theta |$ region where our function $\mathrm{\Psi }$ superoscillates. To see the angular momentum exchange in this situation, we decompose the interaction term $g(\theta -{\theta }_M;\phi )$ in its Fourier transform and obtain

where we used

Here $e^{i(aN+k)\theta }$ is an eigenstate of the angular momentum of the particle corresponding to the eigenvalue $aN+k$ and $e^{-ik{\theta }_M}\varphi ({\theta }_M)$ if a state that is identical to ${|\mathrm{\Phi }⟩}_M$, the initial state of the measuring device, but shifted down in angular momentum by $k$.

In the Dirac notation that will be more convenient in the next sections, for the cases when the particle is found in the superoscillatory region (i.e., the measuring device pointer is flipped), the measurement time evolution is given by

where by ${|\mathrm{\Phi }-k⟩}_M$, we denote a state that is identical to ${|\mathrm{\Phi }⟩}_M$, the initial state of the measuring device, but shifted down in angular momentum by $k$.

What Eqs. 15 and 17 show is that in the case when initially the particle was prepared in the high angular momentum $\mathit{aN}$ and it emerges from the measurement with exactly the same angular momentum $\mathit{aN}$, (that is, corresponding to $k=0$ Eq. 17) the measuring device does not change its angular momentum at all. This is very much expected: Since the particle did not change its angular momentum, there is no reason for the measuring device to provide it any supplementary angular momentum, so no change of the angular momentum of the measuring device should occur. When the particle emerges with other angular momentum value $aN+k$ with $k\ne 0$, which occurs due to the truncation of the initial monochromatic wavefunction, the measuring device changes its angular momentum by $-k$ to compensate for the change. In other words, nothing changes in the measuring device when the particle emerges with precisely the high angular momentum $\mathit{aN}$; there are only changes to account for the deviations of the particle’s angular momentum from this value. Incidentally, we also note that these truncation disturbances become relatively less and less important with increasing $N$, since while the central angular momentum value increase as $\mathit{aN}$, the only significant deviations from it are only of order $|k|=O(N^{1/2-ϵ})$.

Now, had the particle started in our initial state $\mathrm{\Psi }$, the measuring device would have reacted in the same way, up to corrections of $1/N$ as described in Eq. 14, so when found in the superoscillatory region, we have

This means, in particular, that up to $1/N$ corrections, when the particle emerges with the high angular momentum value $\mathit{aN}$ (with $a>1$), the measuring device does not provide any angular momentum to the particle, despite the fact that maximal initial angular momentum of the particle is only $N$. Where does the difference $aN-N$ come from in this case? This is our paradox.

The following sections give the solution.

Let us now discuss a seemingly unrelated issue: How is this special state of the particle prepared in the first place? The reason the preparation is not completely trivial is that we are dealing with a conserved quantity, and that imposes some constraints.

In our original paper (1), we started by discussing a box containing a photon prepared in a similar special superposition of energy eigenstates. Presumably, a laser should have been employed to deliver the photon into the box. But the total energy is a conserved quantity; hence, when the laser emits the photon with a certain energy, the laser must lose the same amount. (Note: Actually, in general, the laser is connected to other objects—optical table, external power supply. It is the energy of this whole system that changes to compensate the change in the energy of the photon. For simplicity, here, we call “laser” this entire system).

As the photon is in a superposition of energy eigenstates, the laser will get entangled with it, each energy eigenstate of the photon being correlated to another state of the laser, shifted down in energy by the energy lost to the photon. The state of the photon is thus not a pure state, as we desired. However, if the state of the laser has a wide and smooth enough distribution of energy, a shift in energy does not modify the state considerably, and the entanglement is minimal. This is how nonmonochromatic but almost pure states of light are routinely prepared. In our present case, we could similarly imagine a “preparing device,” moving on the same circle as the particle, and isolated from everything else, that releases the particle on the circle that we consider. Or, for greater simplicity, imagine that we start with the particle in the initial state of zero angular momentum ${|0⟩}_p$ and the preparer boosts its angular momentum appropriately. Angular momentum being a conserved quantity, the state of the preparer also changes. Again, the particle and the preparing device get entangled. Nevertheless, by choosing the initial state of the preparation device appropriately, we can make this entanglement arbitrarily small.

In other words, instead of preparing the pure state

what the actual preparation procedure does is to take the initial particle-preparer state ${|0⟩}_p{|\mathrm{\Phi }⟩}_P$ to the final, entangled state ${|\chi ⟩}_{p,P}$, via the angular momentum conserving transformation

where ${|\mathrm{\Phi }-m⟩}_P$ is the initial state of the preparer shifted down in angular momentum by $m$.

As discussed above, to make the reduced density matrix of the photon in the two-party entangled state [20] approximate that corresponding to the pure state $\mathrm{\Psi }$, we need to choose the initial state of the preparer, ${|\mathrm{\Phi }⟩}_P$, as a wide and smooth enough superposition of angular momentum so that for all $m$, with $-N\le m\le N$ the shifted states ${|\mathrm{\Phi }-m⟩}_P$ are almost identical to the initial state ${|\mathrm{\Phi }⟩}_P$ (i.e., ${}_P{⟨\mathrm{\Phi }-m|\mathrm{\Phi }⟩}_P\approx 1$). To achieve this goal, we take ${|\mathrm{\Phi }⟩}_P$ to be narrow in its angular distribution. Indeed, the angle ${\theta }_P$ that denotes the location of the preparer is conjugated to its angular momentum, so a narrow distribution of ${\theta }_P$ ensures a wide and smooth distribution of angular momentum.

Going beyond the above purely mathematical argument of why we want to take ${\mathrm{\Phi }}_P({\theta }_P)$ with a narrow distribution of ${\theta }_P$, here are two physical reasons.

First, since total angular momentum is conserved, the Hamiltonian cannot depend on the absolute value of the angle of the particle or of the preparing device but only on their relative angle $\theta -{\theta }_P$. Hence, to each value of ${\theta }_P$ corresponds a wavefunction of the particle appropriately shifted in space. This, of course, means entanglement between the particle and the preparing device, which we want to avoid. To limit this entanglement, we need to reduce the uncertainty in ${\theta }_P$, hence to take for the state of the preparer $\mathrm{\Phi }({\theta }_P)$ a wavepacket narrow in angular distribution.

Another way to look at the problem is to note that we want to place the state of the particle at a precise position on the circle—if there is uncertainty in its positioning, as if it would be the case if ${\theta }_P$ were to have a large uncertainty, it will wash out the finer details of its space dependence, so we will not be able to see the short wavelength oscillations we are interested in. To ensure this, we need, again, to take for $\mathrm{\Phi }({\theta }_P)$ a wavepacket narrow in angular distribution.

For an explicit example, let $\mathrm{\Phi }({\theta }_P)$ be a top-hat function with spread $2\eta$, i.e.,

Then, the scalar product

where we have used the angular representation $e^{-i{m}^{\prime }{\theta }_P}\mathrm{\Phi }({\theta }_P)$ of $|\mathrm{\Phi }-m^{\prime }{⟩}_P$.

Since in our case the angular momenta of interest are limited to the interval $-N\le m,m^{\prime }\le N$, the difference is bounded, $|m-m^{\prime }|\le 2N$. Hence, by choosing $\eta$ small enough, we can make all the scalar products [22] as close to 1 as we wish and hence, from [20], the state of the particle as close to $\mathrm{\Psi }$ as we wish.

We have now arrived at the crucial point of our paper. Although in our preparation stage we have not prepared the pure state $\mathrm{\Psi }$, the state that we prepared has all the ingredients essential for our purpose:

Due to the above, in the subsequent stage of the experiment—the measurement stage—all will be as close as we want to what happens when starting with $\mathrm{\Psi }$. That is, when the particle is found in the superoscillatory region, its angular momentum is close to $\mathit{aN}$, much larger than its maximal initial value of $N$. However, the measuring device does not provide the angular momentum difference to compensate for the increase of the particle’s angular momentum from its maximal initial value of $N$ to around $\mathit{aN}$. More precisely, when the particle is found with angular momentum $\mathit{aN}$, the state of the measuring device does not change at all, so it does not give the particle any angular momentum whatsoever. When the particle is found with angular momentum different from $\mathit{aN}$ (which occurs with nonnegligible probability only for deviations of order $N^{1/2}$), the measuring device provides angular momentum to compensate for this difference, but not for the increase from the maximal initial value of $N$ to $\mathit{aN}$. Hence, our paradoxical effect stands: The particle can emerge with an angular momentum much larger than any angular momentum in its initial state without the measuring device—the only system with which it interacted—to give it the difference.

Let’s now see what happens to the preparer.

At first sight, as far as angular momentum is concerned, the preparer’s role is rather trivial: It provides the angular momentum gained by the particle from its initial zero angular momentum state, ${|0⟩}_p$ to the desired superposition. Indeed, that was the whole point of the previous section. As Eq. 20 shows, if after the preparation, we measure the angular momentum of the particle and find it to be $m$, meaning that the particle gained angular momentum $m$, then, the preparer is in the state ${|\mathrm{\Phi }-m⟩}_P$, whose angular momentum probability distribution is identical to that in the initial state ${|\mathrm{\Phi }⟩}_P$, but shifted by $-m$. This is a trivial case of angular momentum conservation, as discussed above.

Importantly for us here, since the particle’s state only has components limited to the interval $[-N,N]$, the preparer never provides more than angular momentum $N$ (i.e., the maximal down-shift of the angular momentum distribution of the preparer is $-N$).

Our experiment, however, requires that after preparation, we first check whether the particle is in the superoscillatory region, and only then we measure its angular momentum. So suppose we find the particle in a large spatial interval in the superoscillatory region and that the subsequent measurement of the particle’s angular momentum yields the value $k$. The preparer’s state in this situation is (up to normalization)

where $\widehat{\mathrm{\Pi }}$ is the particle projection operator onto that interval.

To evaluate the preparer’s state, it is convenient to express ${|\chi ⟩}_{p,P}$ as

where $\widehat{L}$ is the particle angular momentum operator and ${\widehat{\theta }}_P$ the particle angle operator.

Using [24] and the crucial fact that ${|\mathrm{\Phi }⟩}_P$, the initial state of the preparer, has only a very narrow angular distribution around ${\theta }_P=0$, we obtain

where we used first-order approximations in the small angle ${\theta }_P$.

The last line of Eq. 25 shows that (up to normalization), the final state of the preparer is equal to its initial state but displaced down in angular momentum by ${}_p⟨k|\widehat{\mathrm{\Pi }}\widehat{L}{|\mathrm{\Psi }⟩}_p/{}_p⟨k|\widehat{\mathrm{\Pi }}{|\mathrm{\Psi }⟩}_p$. Let us evaluate this quantity.

Hence, the (normalized) state of the preparer, given that the particle was found in the superoscillatory region and that a subsequent measurement of its angular momentum found it to be equal to $k$ is, up to $O(1/N)$, equal to

This is the key result of our paper. We will discuss its significance in the next sections, but before that, let us note a few more details. First, the final state of the preparer is not affected by the precise localization of the projected space interval in the superoscillation area. (That’s why we denoted the integration domain simply as “super” and not by indicating an exact location). This is important for our further analysis where we consider the projection being made by the interaction with the position-measuring device. As discussed before, the projection realized by the position measurement is not on some precise location on the circle, but on an interval centered around ${\theta }_M$, the location of the measuring device. Even though we can make ${\theta }_M$ as precisely localized on the circle as we want, we can never achieve infinite precision, so the projection intervals will differ slightly from one another. Since it is not sensitive to the precise location of this space interval, our result about the final state of the preparer applies to all these particular projections.

Second, we note that the final state of the preparer is independent of the particular angular momentum value $k$ with which we find the particle after localizing it in the superoscillatory region. In all the cases when the particle is found in the superoscillatory region, the final state of the preparer is the same, i.e. state given in Eq. [27].

Let us now analyze our key result. What Eq. 27 shows is that in all of the cases where the particle was later found to be in the superoscillatory region, the final state of the preparer is equal to its initial state but shifted down in angular momentum by $\mathit{aN}$. That is, in all these cases, the preparer loses $\mathit{aN}$ angular momentum. Yet, the particle started in the state of zero angular momentum, ${|0⟩}_p$ and the preparation brings it in a state [20] whose angular momentum components are limited to the interval $[-N,N]$. Thus, the angular momentum lost by the preparer is not transferred to the particle. We have found therefore another instance of angular momentum nonconservation in our cases of interest.

This “preparation nonconservation” is, if anything, more subtle than the nonconservation that occurs during the subsequent position measurement. The crucial point is that we cannot see this nonconservation right away but we have to wait until after the position measurement. Indeed, while there are, of course, many things we can find out immediately after the preparation, without any need to wait, they are not enough for spotting the preparation nonconservation. For example, we can, if we wish, certify that the particle has only angular momentum components in the interval $[-N,N]$. In other words, we certify that at this stage, the particle did not receive any large momentum such as $\mathit{aN}$. Importantly, we can certify this without affecting the experiment. This can be done by simply measuring the projector on this low angular momentum subspace. Since the state of the particle is an eigenstate of this projector, this measurement will not affect our experiment at all. Furthermore, we also can, without waiting, measure the angular momentum that the preparer has after the preparation. Since the preparer finished its interaction with the particle, and never interacts with it again, this measurement does not affect our experiment either. What we cannot do right away after the preparation and have to wait until after the position measurement is to know whether the particle will be found in the superoscillatory region or not. Not knowing this, we do not know which of the individual runs of the experiment correspond to our case of interest; hence, we do not know which results of the measurement of the angular momentum of the preparer should be included.

Let us now put everything together. Our entire experiment consists of the following:

To these we may add, between the preparation and position measurement, a verification that the particle angular momentum is restricted to the interval $[-N,N]$, i.e., measuring the projector on this interval.

The corresponding time evolution, restricting ourselves only to the cases when the particle is found at the end of the experiment in the superoscillatory region, is

where the simple arrow “$\to$” describes unitary evolution while the double arrow “$\Rightarrow$” describes unitary evolution plus selection, and $\phi$ is the size of the angular interval within the superoscillatory region onto which the position projection measurement is performed.

What the above equation shows is that the particle starts with zero angular momentum (i.e., in state ${|0⟩}_p$) and ends up gaining angular momentum $aN+k$, while the preparer provides $\mathit{aN}$ and the position measuring device provides $k$. Yet, the preparation stage alone does not provide the particle with the high angular momentum $\mathit{aN}$: at the end of the preparation stage, the particle’s maximal angular momentum is only $N$.

This is the main result of our study:

Let us now discuss the above result. First, there are three simple observations. Given that the particle is found in the superoscillatory region:

More importantly, this is not a simple conservation—it is the composite of two nonconservations. It also has a “holistic” time character: As we noticed before, by doing or not doing the position projection measurement, we can change the overall angular momentum transfer from the preparer to the particle, despite that a) at this stage, the particle and the preparer no longer interact and b) that we can certify that after the interaction with the preparer, the particle does not have any high angular momentum at all. This leads us to ask the following: Despite the overall conservation, where from and how precisely did the particle get the high angular momentum, if it did not have it before interacting with the position measuring device and the position measuring device did not change its state? Where has the high angular momentum lost by the preparer been stored until it finally materialized in the particle?

Note that both the preparer and the position measuring device play crucial roles: The preparer is the source of the high angular momentum gained by the particle, while the position measuring device gives an “umbrella” under which this transfer can occur at a time later than the one when the particle and the preparer interacted.

It is also instructive to contrast the above story with what happens in a related, but fundamentally different, case. Suppose that instead of preparing the particle in our special superposition, i.e., instead of the transformation [20], we simply use the preparer to prepare a state of well-defined angular momentum $m$, with $-N\le m\le N$, and then subject it to the same experiment as before. That is, after preparation, we measure the projection operator on what previously was the superoscillatory region and, after that, if we found the particle in this region, we measure the angular momentum of the particle. In this region, the particle’s state oscillates with spatial frequency corresponding to the angular momentum $m$. Yet, after the truncation of the wavefunction, it is still possible to find the particle with angular momentum $\mathit{aN}$, much larger than its initial angular momentum $m$. This happens with a small probability, corresponding to the tail of the frequency probability due to truncation, but still, with nonzero probability. (Contrast this with the probability of finding the particle with momentum close to $\mathit{aN}$ in the superoscillatory case given that we found the particle in this region which is close to 1.) Where did the particle get its angular momentum from now?

with $e_k=\mathrm{sinc}((aN+k-m)\phi )$.

What we see here is a trivial case of angular momentum conservation. The particle gains angular momentum $aN+k$, by a two-stage process. In the first stage, it gains $m$, and the preparer provides it, by shifting its state to ${|\mathrm{\Phi }-m⟩}_P$, i.e., down in angular momentum by $m$. There is no nonconservation now at this stage. In the second stage, the position measuring device disturbs this intermediate particle state and brings its angular momentum from $m$ to $aN+k$, while losing itself $aN+k-m$, hence a trivial conservation again.

Coming back to our example, one can ask how is it possible that the position measurement and associated selection can affect what has happened to the preparer at an earlier time? Moreover, why does the preparer matter at all? After all, once a state, say $|\mathrm{\Psi }⟩$, is prepared, it does not matter how it was prepared. The answer is that in our modified experiment, we did not prepare a pure state. We have not prepared the special state $|\mathrm{\Psi }⟩$. What we have prepared is an entangled state ${|\chi ⟩}_{p,P}$, [20], between the particle and the preparer, with the property that the reduced density matrix of the particle can be made as close as we want to the pure state $\mathrm{\Psi }$. Nevertheless, the particle is entangled with the preparer. True, ${|\chi ⟩}_{p,P}$ is a very weakly entangled state: Each angular momentum component ${|m⟩}_p$ is correlated with a corresponding preparer state ${|\mathrm{\Phi }-m⟩}_P$ which is identical to the initial state of the measuring device shifted in angular momentum by $m$. As discussed in the preparation section, we can make these states of the preparer as close to each other as we want, ${}_P{⟨\mathrm{\Phi }-m|\mathrm{\Phi }-m^{\prime }⟩}_P\to 1$, by making them wider and wider spread in angular momentum, so that ${|\chi ⟩}_{p,P}\to {|\mathrm{\Psi }⟩}_p{|\mathrm{\Phi }⟩}_P$. This means that the state of the particle can be made as close as we want to the pure state $\mathrm{\Psi }$. Hence, for its subsequent behavior, the particle behaves as close as we want to how it would in the pure state $\mathrm{\Psi }$. Yet, for questions related to angular momentum conservation, there is an enormous difference between the entangled state ${|\chi ⟩}_{p,P}$, and the pure state ${|\mathrm{\Psi }⟩}_p{|\mathrm{\Phi }⟩}_P$. Indeed, although the states ${|\mathrm{\Phi }-m⟩}_P$ can be made as close to each other as we want, their average angular momenta remain finitely different from each other, $L_0-m$ versus $L_0-m^{\prime }$ with $L_0$ being the average angular momentum of ${|\mathrm{\Phi }⟩}_P$. This is the reason why this entanglement, while minimal, is however essential. It is essential even for the simple case when we just prepare the initial state ${|\chi ⟩}_{p,P}$ and then we immediately verify that the preparation conserves angular momentum. It is even more important in our special case. Postselecting for the particle being in the superoscillatory region and then for a particular final angular momentum produces a superposition of these shifted wavefunctions, which, by interference, lead to a wavefunction with a much larger shift in angular momentum, as discussed in the previous sections.

Of course, while the entanglement with the preparer explains the mathematical machinery that allows our results to exist, it is by no means, by itself, an explanation for why this effect happens. It does not tell why it is the case that this entanglement is exactly such that angular momentum conservation is valid for the individual cases when the particle is found in the superoscillatory region, while the only thing demanded by the standard conservation laws is statistical conservation over the entire ensemble. The main principle, as we see it, is that we should demand more of the conservation laws than the statistical conservation.

One could, of course, try and bring a counterargument to our general line of thinking. One could ask what would happen if the particle were initially exactly in the pure state $\mathrm{\Psi }(\theta )$? Then, there would be no preparer to provide the required supplementary angular momentum, and we would not have angular momentum conservation in our individual case. Would this not invalidate our arguments?

Yes, if the particle could start in the initial state $\mathrm{\Psi }(\theta )$, there would be genuine nonconservation in the individual cases presented in our original experiment, as described in ref. 1. It would then follow that conservation does not need to hold in individual cases but only statistically, no matter how unpalatable this may seem to us. But there is a way out. This brings us to the last key element of our paper. We claim that

Arguing that $\mathrm{\Psi }(\theta )$ is unphysical and does not exist in nature seems to contradict one of the basic postulates of quantum mechanics, namely that any normalized wavefunction is a legitimate state for a quantum particle. The answer is that the issue is one concerning frames of reference. The wavefunction $\mathrm{\Psi }(\theta )$ has no meaning because the angle $\theta$ has, strictly speaking, no meaning.

It is standard in quantum mechanics to consider wavefunctions of variables $x$, $y$, $z$ or of angle $\theta$ and so on. But it is well understood that what they do represent are not some absolute values, but they are coordinates measured relative to some frame of reference, say the walls of the laboratory. The reason why one does not always mention this explicitly is that in general this does not significantly affect the results and can be ignored. In our case, however, it matters significantly, as we have shown. The preparer acts as a reference frame, and this is why it should always be present.

Lest it appears that the statement that “we need to always use a frame of reference in quantum mechanics in order to correctly describe conservation laws” is trivial, we want to emphasize that this is by no means so. Far from it. It is in fact a very subtle issue, which only now is becoming apparent. Indeed, there is absolutely nothing in standard quantum mechanics that requires us to refer to a frame of reference to ensure that the usual, statistical conservation laws hold. Indeed, in the standard quantum mechanical formalism, nothing prevents us to write an arbitrary wavefunction of two particles, say $\mathrm{\Psi }(x_1)\mathrm{\Phi }(x_2)$, and as long as they interact via a potential that depends only on their relative distance, $V(x_1-x_2)$, the standard statistical momentum conservation law holds exactly. Nothing else is needed. There is no need to refer to any other frame of reference or any “preparation” device, etc. Similarly, no consideration of a reference frame is needed to ensure the statistical conservation of angular momentum or energy, etc., and they apply to all wavefunctions. We could, if we fancy doing so, refer all the position or time coordinates to some other physical system, but there is no need whatsoever to do it. What we found here is therefore a fundamental aspect of conservation laws:

Moreover, and crucially important, the mere thing that one needs a frame of reference, and hence a preparer, is only a necessary but by no means sufficient condition to imply conservation at the level of individual cases. This is by no means the “explanation” for the effect. What is required is the entire subtle interplay between the preparer and the particle and their nonlocal-in-time angular momentum transfer via the measuring device which acts as an umbrella that makes this transfer possible by covering otherwise inevitable causality violations.

Let us zoom out from the particular example discussed here. We have argued that while the standard formulation of conservation laws in quantum mechanics, which is of a statistical nature, is perfectly valid as far as it goes, it is incomplete and needs to be revisited, specifically that we need to go beyond the statistical formulation and consider individual cases as well. In particular, we have argued that if we find in one particular run of an experiment, that, say, when we open a window of a box containing a single particle of energy strictly smaller than 1 eV and the particle emerges with energy of order of millions of GeV, it is legitimate to ask where did this increase in energy come from in this individual case. Had we ignored this question on grounds that the standard, statistical, conservation law is all there is, we would have not identified the strange conservation effect for the individual case presented here, and hence missed a lot of interesting physics. We would have also missed the implications for the need to explicitly take into account the issue of the frames of reference involved in the preparation of the initial state, an issue which does not arise when only the standard statistical conservation law is concerned but which arises when the conservation in individual cases is concerned, neither the fact that certain quantum states are unphysical, while there would be considered perfectly valid in standard quantum mechanics. We take this as a good indicator of the validity of this line of enquiry and believe these results are only the tip of an iceberg, part of a more general structure concerning conservation laws. In particular, we conjecture that when such a full analysis of any conservation experiment is performed conservation is obeyed in any individual case, not only statistically.

S.P. acknowledges the support of the Advanced European Research Council (ERC) Grant FLQuant.