quantum mechanics - If a photon truly goes through both slits (at the same time), then why can't we detect it at both slits (at the same time)?

I am not asking about whether the photon goes through both slits, or why. I am not asking whether the photon is delocalized as it travels in space, or why.

I have read this question:

Do we really know which slit the photon passed through in Afshar's experiment?

Which theory explains the path of a photon in Young's double-slit experiment?

Shooting a single photon through a double slit

Where John Rennie says:

The photons do not have a well defined trajectory. The diagram shows them as if they were little balls travelling along a well defined path, however the photons are delocalised and don't have a specific position or direction of motion. The photon is basically a fuzzy sphere expanding away from the source and overlapping both slits. That's why it goes through both slits. The photon position is only well defined when we interact with it and collapse its wave function. This interaction would normally be with the detector.

Lasers, Why doesn’t a photon go through the same slit every time?

Where ThePhoton says:

for example, if you put a detector after a two-slit aperture, the detector only tells you the photon got to the detector, it doesn't tell you which slit it went through to get there. And in fact there is no way to tell, nor does it even really make sense to say the photon went through one slit or the other.

In classical terms, this question might be obvious, because a classical billiard ball cannot be at two places in space at the same time. But this is not a billiard ball, this is a photon, a QM phenomenon. And this is not classical terms, but QM.

And if we truly accept that the photon travels through both slits, then it basically must exist in space at both places (both slits) at the same time.

But as soon as we interact with it (the wave function collapses), the photon becomes spatially localized, but only at a single location (at a certain time).

What is not obvious from QM, is how we can have these two things at the same time:

1. 
   

   the photon pass through both slits

   
   
   


2. 
   

   but we can only interact with it at one slit (not both)

   
   

What is that basic thing in QM, that will disallow for the photon to pass through both slits and be interacted with at both slits too? Somehow the QM world underneath will change to classical as soon as we measure, and interact with the photon. This change from QM to classical is where the possibility of the photon being at both places (both slits) at the same time gets disallowed somehow. This could be decoherence, as the QM entity gets information from the environment (because of the measurement), or just the fact that the wavefunction collapses and that has to have a single spatial location for the photon when measured.

So basically the photon goes through both slits, thus, it in some form exists at both slits at the same time. But when we try to interact with it, it will only be spatially localizable at one of the slits, not both at the same time.

Question:

1. If the photon truly goes through both slits (at the same time), then why can't we detect it at both slits (at the same time)?

Answer

If the photon truly goes through both slits (at the same time), then why can't we detect it at both slits (at the same time)?

Alright, let's play some word games:

This isn't a well-defined question. "Detect a particle" doesn't mean anything in quantum mechanics. Quantum mechanical measurements are always measurements of specific observables. There is no holistic act of "observing all properties of a system at once" like there is in classical mechanics - a measurement is always specific to the one observable it measures, and the measurement irrevocably alters the state of the system being measured.

People often use "detect a particle" as shorthand for "perform a position measurement of a particle". By definition, a measurement of position has as its outcome a single position, and interacts with the state of the particle being measured such that it now really is in the state in which it is at that single position and nowhere else. So if you could perform position measurements that yielded both slits as the position of the particle, this would mean you have performed an impossible feat - there are now two particles, each in the state of being at one slit and that slit only. Quantum mechanics may be weird, but it is hopefully clear it is not this weird - we cannot duplicate a particle out of thin air just by measuring it.

If you don't insist on "detect" meaning "performing a position measurement", then of course the standard double slit setup is a "detection" of the photon at both slits - the pattern on the screen is only explainable by the particle's wavefunction passing through both slits and interfering with itself. This is of course just indirect reasoning - there simply is no observable whose eigenstates would naively correspond to "we have detected the photon at both slits at once".

Lastly, you seem to confuse "interacting" with "measuring" or "detecting". Of course we can interact with the particle at both slits - we just cannot perform position measurements (or other "which-way" measurements) at both slits and expect them to yield the impossible result of the particle split in two. But if you look at more sophisticated setups like the quantum erasers, there certainly is interaction with the particle at both slits - just carefully set up to not destroy the interference pattern, and hence no obtaining useable which-way information.

thermodynamics - How might a resonant antenna and black body radiation interact?

How does an antenna behave when it is cooled so that its black-body radiation is emitting energy at its resonant frequency?

Edit: To clarify, its not how they're related in general, but how might thermal radiation and resonance interact with each other when their spectra are aligned well?

Edit: Also, I'm sure that the thermal radiation spectra that have a significant peaks are associated with incredibly high temperatures, and peak at incredibly small wavelengths, rendering such an antenna completely impractical to build. Still, I'm still interested in the theoretical concept.

Answer

OK, the simple answer: When there is a resonance in the antenna you have a coherent phenomenon. All the bands of electrons of the antenna are marching in tune.

The black body radiation is an incoherent phenomenon coming from the individual atoms of the antenna. Even if the peak of the black body radiation were sitting on the resonance of the antenna it is still an incoherent phenomenon that cannot couple to the coherent behavior of the electrons in the current that resonate.

Think of a single violin tune and a crowd of people talking. The noise of the people does not cover the clarity of the violin even in high volumes.

spacetime - What is the physical interpretation of the Wick rotation?

What is the physical interpretation of the Wick rotation?

How is it that we can just propose there's a new time coordinate tau? Are physicists saying time is modeled by an imaginary number? Isn't that at odds with the clocks we use everyday?

quantum mechanics - Dyson series derivation

$$ i\hbar \frac{dU_I(t, t_i)}{dt} = \hat{V}_I(t)\hat{U}_I(t,t_i) \tag{10.32} $$ The solutions of this equation, with the initial condition $\hat{U}_I(t_i,t_i)$, are given by the integral equation $$ \hat{U}_I(t,t_i) = 1 - \frac{i}{\hbar}\int_{t_i}^t\hat{V}_I(t')\hat{U}_I(t',t_i)dt' \tag{10.33} $$

In the derivation of Dyson Series please explain why in equation (10.33) $t$ is changed to $t'$ without integrating.

How is energy extracted from fusion?

I understand that combining deuterium and tritium will form helium and a neutron. There are three methods to do this (1) tokamak (2) lasers and (3) cold fusion. I would like to know after helium is formed. How is that energy extracted from tokamak and stored?

Answer

If you search the ITER site, ITER being the international prototype fusion reactor which will demonstrated the possibility of getting megawat useful energy from fusion, one sees that their main aim is to demonstrate this feasibility:

The main carrier of energy out of the plasma is the neutron, and methods to efficiently use this energy have not been developed yet, but wait for the commercial prototype.

The helium nucleus carries an electric charge which will respond to the magnetic fields of the tokamak and remain confined within the plasma. However, some 80 percent of the energy produced is carried away from the plasma by the neutron which has no electrical charge and is therefore unaffected by magnetic fields. The neutrons will be absorbed by the surrounding walls of the tokamak, transferring their energy to the walls as heat.

In ITER, this heat will be dispersed through cooling towers. In the subsequent fusion plant prototype DEMO and in future industrial fusion installations, the heat will be used to produce steam and—by way of turbines and alternators—electricity.

They have developed methods for cooling the system and dissipating the energy to the environment.

special relativity - How to motivate the importance of the spacetime interval

The spacetime interval is a rather important thing in Special Relativity. It allows us to define the separation between any two events as spacelike, timelike or lightlike and more importantly, the Lorentz transformations can be defined as the transformations which keep the spacetime interval fixed.

In that sense, as we know: lenghts and time intervals themselves are observer dependent. They are not absolute notions. On the other hand, the spacetime interval is one absolute notion.

Now, given events $(t_1,x_1,y_1,z_1)$ and $(t_2,x_2,y_2,z_2)$, its definitios is:

$$I = -c^2 \Delta t^2+\Delta x^2+\Delta y^2+\Delta z^2,$$

so that it is the difference between the distance that light traveled between the two events and the spatial separation of the events.

My problem here is that if we try to construct special relativity following the historical procedure there is a certain gap when introducing the spacetime interval.

We can usually go on like this: we start reviewing the problems in electrodynamics which motivated the theory as Einstein himself stated in his paper. After that, we can follow Einstein's procedure and starting with the postulates derive the relativity of simultaneity, the lengths contraction and the time dilation. From that we are able to get the Lorentz transformations.

The next natural step is to give more mathematical substance to this construction, and start labeling events with elements of $\mathbb{R}^4$ so that we finally get to the idea of spacetime. The problem is that in this point one usually just defines this formula for $I$, shows that it is preserved by the Lorentz transformations and shows that it allows us to classify the separation between events.

What I want is to be able to motivate why do we introduce the spacetime interval. It is just a certain object with certain properties, but how do we motivate its importance in the context of relativity and how do we motivate its definition?

After all, as far as I know, originally relativity is trying to solve the inconsistency between Newtonian Mechanics and Maxwell's Electrodynamics. The Lorentz transformations would seem to already to the job. How could one in this context motivate the definition of $I$?

homework and exercises - Magnetic force doesn't do work and therefore can't change the KE of a particle?

I am having a real hard time understanding the principles behind a question I've come across during test prep. Here is the question:

Q: A proton and an electron are traveling in the uniform magnetic field with identical velocities. If the movement of both particles is perpendicular to the magnetic field lines, is the following true?

The proton will experience a greater kinetic energy change than the electron.

On one hand I think if there is a $\mathrm{KE}$ change for the particles, they would have to be both changed equally, since an input of force to each particle would be converted into $\mathrm{KE}$. On the other hand, since $\mathrm{KE} = \frac{1}{2}mv^2$, and both particles are traveling at the same velocity, it seems the proton's much large mass would mean it has a greater $\mathrm{KE}$ change.

However, neither of my thoughts were on target, as the answer reads as follows:

False, because the magnetic force does not do work, so it cannot change the kinetic energy of either particle.

How does this make sense? Won't the particles change velocity when exposed to the magnetic field, and therefore change $\mathrm{KE}$?

Answer

You basically just need to be careful about the distinction between velocity and speed. In particular, you say that

Won't the particles change velocity when exposed to the magnetic field, and therefore change KE?

A change in velocity is not necessarily accompanied by a change in speed, and it's the speed that determines the kinetic energy. The magnetic field can change the directions of the motions of the charged particles, but it will not change their speeds.

The mathematical details of this are as follows. The force on a charged particle of charge $q$ moving in a magnetic field $\mathbf B$ is $$ \mathbf F = q\mathbf v\times\mathbf B $$ where $\mathbf v$ is the velocity of the particle. Now, note that Newton's second law says that the force on the particle equals the rate of change of its momentum; $\mathbf F = \dot{\mathbf p}$. So we get $$ \dot{\mathbf p} = q\mathbf v\times\mathbf B $$ Now take the dot product of both sides with $\mathbf p$. Note that the (non-relativistic) momentum of the particle is $\mathbf p = m\mathbf v$, so when we take the dot product of the right hand side with $\mathbf p$, we get zero. Putting this all together gives $\mathbf p \cdot\dot{\mathbf p} = 0$ which implies that $$ \frac{d}{dt}\frac{\mathbf p^2}{2m} = 0 $$ In other words, the kinetic energy is constant in time.

Addendum June 24, 2013.

The details of the argument above assumed the non-relativistic expression for the momentum of a mass particle. However, the argument still carries through in a relativistic context. To see this, note that the relativistic expression for the momentum of a particle is $$ \mathbf p = \gamma m\mathbf v, \qquad \gamma = (1-\mathbf v^2/c^2)^{-1/2} $$ and using this definition, Newton's second law can in this context still be written as $\mathbf F = \dot{ \mathbf p}$, so the equation of motion for a massive particle moving in a magnetic field is still $\dot {\mathbf p} = q\mathbf v\times \mathbf B$. Moreover, since the velocity and momentum are still proportional to one another, dotting both sides of this equation of motion still gives $\mathbf p\cdot\dot{\mathbf p} = 0$ and therefore that $$ \frac{d}{dt}\mathbf p^2 = 0 $$ Now recall that the kinetic energy of a relativistic particle is given by its total energy $E=\gamma mc^2$ minus its rest energy $mc^2$; $$ K = E - mc^2 $$ In particular, the time derivative of its total energy equals the time derivative of its rest energy; $\dot K = \dot E$. On the other hand, recall the relation $$ m^2c^4 = E^2 - c^2\mathbf p^2 $$ Combining this relation, and the fact that the time derivatives of the kinetic and total energies are equal, we find that the desired result $$ \dot K = 0 $$ holds in a relativistic context as well.