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I'm reading a book about single particle detection and particle detectors in general. All detectors in the book rely on some energy transfer from the particle to the detector in form of ionisation, activation, photo effect, etc. I'm wondering if it is possible to detect incident particles of any form with a detector, without there being an energy transfer between particle and detector.

I want to explicitly rule out processes in which the presence of a particle is theoretically deducted from the measurement of another particle where energy transfer is allowed.

On a side note, it would be interesting to hear of concepts, where the energy transfer is negative (i.e. from detector to particle).

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  • $\begingroup$ Some remarks. (1) You probably want to talk about energy-momentum transfer, not energy transfer specifically. (2) If we can detect a particle without a transfer of energy-momentum, then what's stopping us from detecting a particle whose energy-momentum is zero? Such a particle would have zero energy-momentum in all frames. (3) If you can detect a particle without an energy transfer, then it seems that you can detect it without changing its phase. This creates paradoxes in a double-slit experiment where you put a detector at one slit. $\endgroup$ – Ben Crowell Aug 3 '17 at 16:17
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    $\begingroup$ This is somewhat related: Does a measurement always require the exchange of energy? and now has a tiny bounty too. $\endgroup$ – uhoh Aug 3 '17 at 19:20
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    $\begingroup$ Do elastic collisions count as energy transfer? $\endgroup$ – immibis Aug 4 '17 at 5:55
  • $\begingroup$ @BenCrowell (1): I wanted to keep the question as open as possible to (2) include such fancy cases as particles with zero momentum and energy (are there any? not that I know of) (3) how does a change of phase imply an energy transfer (and vice versa?)? $\endgroup$ – Dschoni Aug 7 '17 at 10:09
  • $\begingroup$ @immibis: Not that I know of. My books say, that the net energy transfer in this cases is zero and e.g. Rayleigh Scattering cannot be used to detect particles. $\endgroup$ – Dschoni Aug 7 '17 at 10:10
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We can use a qubit as a particle detector that doesn't change the particle's energy. This can be implemented as follows. We start out with a qubit initialized in the state $\left|0\right>$ and we apply the Hadamard gate $U$ that acts as follows:

$$ \begin{split} U\left|0\right> &= \frac{1}{\sqrt{2}}\left[\left|0\right\rangle + \left|1\right\rangle\right]\\ U\left|1\right> &= \frac{1}{\sqrt{2}}\left[\left|0\right\rangle - \left|1\right\rangle\right] \end{split} $$

Note that $U$ is its own inverse, so applying $U$ again will bring the qubit back to the state $\left|0\right\rangle$ we started out with. But now consider what happens if during the time the qubit spends being a superposition of $\left|0\right\rangle$ and $\left|1\right\rangle$ a particle collides with it, but such that no energy is exchanged. Then the qubit will get entangled with the particle, so the qubit-particle system will be in a state of the form:

$$\left|\psi\right\rangle = \frac{1}{\sqrt{2}}\left[\left|0\right\rangle \left|D_0\right\rangle + \left|1\right\rangle\left|D_1\right\rangle\right]$$

where the states $\left|D_{i}\right\rangle$ are the particle states after scattering off the qubit in state $\left|i\right\rangle$. You might think that because the qubit was not affected by the interaction at all, we cannot perform a measurement on the qubit's state to find out that it has interacted with a particle. But watch what happens if we apply the Hadamard gate again to the qubit:

$$U\left|\psi\right\rangle =\left|0\right\rangle\left|D^{+}\right\rangle+\left|1\right\rangle \left|D^{-}\right\rangle$$

where $D^{\pm} = \frac{1}{2}\left[\left|D_0\right\rangle\pm\left|D_1\right\rangle\right]$

So, had there been no interaction, the qubit would have returned to the initial state $\left|0\right\rangle$ but now we end up with an entangled state of the qubit and the particle such that there is now a finite probability of finding the qubit in the state $\left|1\right\rangle$, despite the fact that the collision with the particle happened in a purely elastic way such that it did not affect the physical state of the qubit in any way at the time of the collision. The probability of finding the qubit in the state $\left|1\right\rangle$ is $\frac{1}{2}\left[1-\operatorname{Re}\left\langle D_0\right|D_1\left.\right\rangle\right]$, so it depends on the overlap between the two particle states corresponding to scattering off the qubit in the two states of the superposition.

If the states $\left|D_i\right\rangle$ are orthogonal, then you have 50% probability of finding the qubit in the states $\left|0\right\rangle$ and $\left|1\right\rangle$; the density matrix after tracing out the particle state is $\frac{1}{2}\left[\left|0\right\rangle\langle 0| + \left|1\right\rangle\langle 1|\right]$.

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    $\begingroup$ There is this sort of ideal notion of entanglement. Your statement "the states $\left|D_{i}\right\rangle$ are the particle states after scattering off the qubit in state $\left|i\right\rangle$." employs the term scatter, which does involve an interaction or transfer of momentum and energy. The question might be whether one can form an entanglement completely "for free" without some transfer of momentum and energy in some interaction. $\endgroup$ – Lawrence B. Crowell Aug 3 '17 at 17:11
  • $\begingroup$ I'm not a professional. But for the simple terms I understand so far it sounds more like a detector that can confirm the state of an present particle more over really detecting a particle. $\endgroup$ – Zaibis Aug 4 '17 at 8:49
  • $\begingroup$ +1 to @LawrenceB.Crowell. Is there some energy/momentum transfer hidden in the quantum entanglement? Besides that, this is my favorite answer. However I feel the remaining question of an "energy transfer free entanglement" a little over the top in the scope of this question. I might open a new one on this topic. $\endgroup$ – Dschoni Aug 7 '17 at 10:19
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All detection requires a signal, which in turn requires energy. You can't detect without taking energy from the signal!

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    $\begingroup$ @Moonstroke but is it correct? $\endgroup$ – uhoh Aug 3 '17 at 19:12
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    $\begingroup$ That is correct but incomplete: you do not really need to take energy from the system you want to check its presence. See @Count Iblis answer for a detailed example or mine, for a short one. $\endgroup$ – ccorbella Aug 3 '17 at 23:13
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As a rule in order to detect a set of excited quantum states you have to input energy necessary to excite the harmonic oscillator modes of those states. If you want to detect the Higgs particle, which occurs at $125$GeV energy you need to impart that energy into the vacuum by exciting virtual modes through an interaction with at least that much energy. There is also the further consequences of measurement fidelity, which then requires far more energy to make the signal robust. The Fermilab Tevatron did appear to detect the Higgs particles at the TeV energy level, but the statistics were not good enough to claim detection. Now leap to the Planck scale where to detect quantum gravity you have to excite modes that are $10^{19}$GeV in energy gaps, say from vacuum to the generation of a quanta of black hole. To do quantum gravity experiments it means one must have interaction energy on this scale. This makes quantum gravity very tough or impossible to do in the lab.

There are some other more subtle ways of detecting things. The Lamb shift is a low energy detection of QED processes that split energy levels in the atom. The Lamb shift is in the microwave domain that is far in the IR domain from the energy gap of the $^2S_{1/2}$ and $^2P_{1/2}$ energy levels. This still involves the detection of an excited state, but with far lower energy than the main energy-gap of the theory. In the case of quantum gravity these gaps are the formation of Hawking radiation that is far more IR than the Planck energy. For standard model physics this might occur in subtle corrections to the electronic states of atoms as the wave function has small “exponential tails” in the nucleus so there could be weak neutral currents that perturb electronic states. There were proposals along these lines back in the 1970s, but I am not sure what came out of that.

There are other subtle measurements, such as weak quantum measurements. In the end though one must has a detector that registers a voltage. Some measurements involve no voltage, such as using the polarization of the electric vector in photons to do a “non-measurement” of a hidden bomb. However, if a bomb is not present a homodyne detector registers a voltage. It is probably not physically possible to measure something without having some coupling that can transfer energy or momentum to a detector or needle state.

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You can do that type of indirect measurements. Imagine you wanted to detect, for instance, a electron level decay: it would be as easy as to check if a photon has been emitted or not. You would not be taking energy from the electron, but you would from the photon.

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  • $\begingroup$ I explicitly wanted to rule out that kind of indirect measurements. Nevertheless: Good point. $\endgroup$ – Dschoni Aug 7 '17 at 10:14
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Re: negative energy transfer - for example an electron that is accelerated in electrical field takes measurable amount of energy from the field, which is detectable.

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    $\begingroup$ I don't understand how this answers the question. That does involve energy transfer. $\endgroup$ – JMac Aug 3 '17 at 13:30
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    $\begingroup$ OP was asking about negative energy transfer,too. $\endgroup$ – Juraj Aug 3 '17 at 14:34
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In order to make a detection you need to change the state of something. Energy is just an abstract quantity used to follow processes and state changes. So it is natural that you will always need energy to detect something.

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  • $\begingroup$ Energy is just an abstract quantity used to follow processes and state changes. Mm, not really. $\endgroup$ – Ben Crowell Aug 3 '17 at 21:31
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    $\begingroup$ @BenCrowell Would you elaborate, please? Maybe you convince me of the opposite. $\endgroup$ – Diracology Aug 3 '17 at 22:08
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    $\begingroup$ Energy is not at all abstract, is it? It is the cause of processes and state changes, but it is as real as matter is (because it's the same thing) $\endgroup$ – joH1 Aug 4 '17 at 13:43
  • $\begingroup$ @Moonstroke When we say that a particle decay and release energy there is actually a process occurring: one particle turning into other particle(s). Then we associate amounts of energy to each particle and make a balance. It is not obvious to me how energy cannot be abstract. $\endgroup$ – Diracology Aug 4 '17 at 13:54
  • $\begingroup$ well that kind of process involves quantum mechanics -- which is, I agree with you, totally abstract (at least to the human mind). $\endgroup$ – joH1 Aug 4 '17 at 13:58

protected by Qmechanic Aug 3 '17 at 14:23

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