Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

B Double slit & trajectories

  1. Nov 7, 2016 #1
    I have 2 questions

    1. I'm telling a friend that in the double slit experiment.. the electron has no trajectories between the emission and detection. But she commented the initial emission has trajectory. Is this correct? What is the term for this situation of the initial emission of the electron?

    2. In the detection, how do you change the observable from position to let's say energy. Or what kind of detector do you have to put so it won't measure the position of the particle but some other observable (and what are they?)

    Thank you.
     
  2. jcsd
  3. Nov 7, 2016 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Not in the sense that the source emits an electron with a predetermined path to the slits. The whole setup is arranged so there is no way to tell what path the electron took to get from the source to the detector. Now: epistomologically this may be considered to be the same thing as saying there is no such path ... but that is not what is actually being said here.

    The initial situation is described by the statistical geometry ... ie. it could be a (hemi-)spherical source: so the electron is equally likely to be emitted into any solid angle. We can, in principle, set up any profile we like. We could set up a profile so the source lets out a very narrow beam of electrons so tightly controlled that all of them will pass through a specific slit ... in which case, no diffraction pattern. There is also the issue that we cannot know exactly when the electron was emitted either... or the pattern vanishes.

    You use an "energy detector". note: energy and position can be measured simultaneously. You can work out the basics just by thinking of the electrons as bullets and the detectors are sand-bags.
    Adjust for the understanding that electrons are charged.
    How things get measured is a very important question so I want to answer in more detail if needed: what level do you need the answer at?

    My goto guy for this stuff is Richard Feynman:
    http://www.vega.org.uk/video/subseries/8
    ... it's 8 lectures aimed at a lay audience, see all 8 though.
     
  4. Nov 7, 2016 #3
    Usually what is the actual emission device.. is the electron in the atom pumped out and ejected outside?

    What is the actual device that can detect only the energy? Does it mean no position of the electron could be detected? Actual experiments I need.. intermediate explanations.. Thank you!

    My goto for this stuff is Richard Feynman:
    http://www.vega.org.uk/video/subseries/8
    ... it's 8 lectures aimed at a lay audience, see all 8 though.[/QUOTE]
     
  5. Nov 7, 2016 #4

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Anything that fires electrons can be used as a source - ie a slug of Strontium 90. Traditionally a large van der Graaf generator is the source, and the beam is shaped using electromagnetic lenses and fired down an evacuated steel pipe.
     
  6. Nov 7, 2016 #5

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    ... most any detector will do it. We use semiconductors (most often now) and crystals most of the time, though Geiger tubes are still in use.

    Technically, the semiconductor (or whatever) is manufactured so that a charged particle passing through it's layers will trigger a flow of current (think - photodiode, but for charges rather than photons). What this means is that none of these things detect the position of the particle directly ... all you know is that the particle hit someplace on the material surface.

    The detector itself is an earthed metal box with a hole in it - the semiconductor is mounted just behind the hole. It follows that only particles that pass through that hole will trigger the detector - which is how we have position information. If we just made a semiconductor sphere and put it around the source, we would not be able to tell the angular position of particles detected (we would know how far away the particle was when it was detected though). Since that is very expensive, we cannot help measuring position and energy at the same time. But you will notice that we do not measure exact positions. All this setup tells us is that some charged particle passed through the detector aperture somewhere.
     
  7. Nov 7, 2016 #6
    But this information that the particle hit someplace on the material surface means there is still position.. but is it not when you chose the energy observable. There is literally no longer any position at all so the particle shouldn't even hit someplace on the material surface?

    "know how far away the particle" means there is still position. But there should 100% be no position at all.. is it not.
    do you know of a site that has illustration of this pure energy detector.. why can't I find it in youtube.
     
  8. Nov 7, 2016 #7

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    You are not making sense.
    You do not know where I am, correct? But you agree that I do have a position, right?

    Not knowing what the position is, is not the same as not having a position.
    Yes - there is some information about position available to the experimenter - however vague.
    Nobody is saying that you can have zero information about position of a particle that has been detected. Just the opposite - you cannot help but find some position information. For instance: once the particle has been detected, then we know it must be inside the observable Universe ... that is as vague as position information can get.

    We can do that in theory - the wavefunction of a free particle is flat. We'd verify this by noticing it is equally likely to be detected anywhere and finding the probability of detceting one in as many places as we can think of. However there is no such thing as an absolutely free particle.

    ... what you are looking for does not exist. Why would you suspect such a device could exist?

    What has this got to do with quantum interference at slits?
    Where are you going with this?
     
  9. Nov 7, 2016 #8
    Where I am going with is this. A particle when not detected has no intrinsic position or energy or spin, they are state vector or the wave function is unitary (or a ray in Hilbert space). I just want to be clear I understood this so I expected you to say the particle has no intrinsic position when you choose the energy observable. So you are saying it has vague position but is the vague position really intrinsic in the particle meaning it has really vague position and not just due to our apparatus focusing on energy observable and not position?

    About quantum interference at slits. The detector either detects left or right hits. So if position is not an intrinsic property of particles. I'm imaging when the particle is emitted in the double slit experiment.. it can end up invisible. Do you know of other experimental setup where a particle exists first with position then it just vanishes into thin air?
     
  10. Nov 7, 2016 #9

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    No. Before measurement, we cannot say anything about any of it's state variables.
    That does not mean that it does not have a specific value for that variable. Just like you do not have not measured my position, body temperature, shoe size, whatever; but I do have a position, body temp, and shoe size.
    Not knowing anything about something is just lack of knowledge.

    It may be that I don't have feet - then the concept of "shoe size" does not apply to me... but you still do not know.
    You seem to be saying that QM being silent about something means that the something does not exist. That is sloppy thinking.

    You were asking about what happens in real physical experiments, not what can be described in the mathematics. I replied accordingly. If you want to ask about idealized situations, sure - start a new thread.
    What I am telling you is that knowing only that position has not been measured, we cannot say much about the position.
    Not knowing the position of something is not the same as something not having a position, and it does not mean we can say nothing at all about the position.

    That does not sound like a standard setup. I don't know what you mean by "The detector either detects left or right hits.". A detector usually detects [a particle] or it does not. In the source+slits setup, you can put the detector anywhere you like. If you put it completely blocking the left slit, then it detects particles at the left slit or it does not. It tells you nothing about particles it does not detect.

    I don't know what you mean by an intrinsic position.
    Maybe we are just using the words to mean different things. (See: "intrinsic spin" for example.)

    What you just described is nonsense. Quantum mechanics makes no such claims.
    You are trying to make the theory say something it does not.
    The QM description just says that we do not know how the particle got from emitter to detector. In fact, we can show the particle can be detected anywhere within the experiment by changing the detector position. But what path it took - QM is silent on that question.

    Logically, and by experimental confirmation, any particle that makes it to the other side of the slits must have passed through one or the other of the slits.
    What we find is that if we do any experiment where we can determine which slit an individual particle passes through, we must also disturb the setup in such a way that the diffraction pattern does not happen.
     
  11. Nov 7, 2016 #10

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    It also doesn't mean that it does. It depends on the state and the variable. If the state is an eigenstate of the operator associated with a particular variable ("observable" would be a better term), then it does have a specific value for that variable--more precisely, a measurement of that variable will yield a specific, definite value with certainty. If the state is not an eigenstate of the operator, however, then it doesn't; you can only make predictions about the probability of various values being measured.

    If the particle is not in a position eigenstate (which it won't be given the apparatus you are describing), then not having a specific, definite value for position is intrinsic to the particle. But "vague position" is not a good way of describing this. See my description above.
     
  12. Nov 8, 2016 #11

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    I agree.

    I think we can demonstrate that a particle has a position by measuring it - position exists, the particular value is a result of the measurement and what is being measured. Before the measurement, we don't know the position ... but we can know about the probability of the outcome of possible measurements. Do we have an example of a particle that exists and does not have a position?

    I am being a bit pedantic about this because of the protest earlier that finding the particle on the dome detector (post #2/3?) still tells me something about it's position, and OP wanted to know a way to measure energy alone - with no knowledge of position. IRL this cannot be done - though I guess we could measure the energy of a particle confined to a well if we make the "detector aperture" from my earlier description bigger than the well. But even then, the resulting energy eigenstate would have a different position distribution... which would result in knowing more about position. I don't think the conditions for a measurement of energy not leading to some information about position is possible or indicated in the theory...
     
  13. Nov 8, 2016 #12

    Zafa Pi

    User Avatar
    Gold Member

    Is there not an interpretations that says there is a path, but we don't know what it is, and an other that says there is no path? And, indeed they may be considered "epistomologically" the same.
    Is this true? I thought that the time and position operators commute.
     
  14. Nov 8, 2016 #13
    But without the Born Rule. Everything is just wave function. Lets' say born rule is overridden and suspended in an object. Then it won't collapse to any position or energy or spin.. so it would just literally vanish. A quantum object has no intrinsic position or energy or spin.. you make them appear by using an operator and Born rule. But if Born rule won't be deployed.. the properties would not even exist in our world or measurable. Come to think of it. Why do you deny this since this is what is taught in QM.

     
  15. Nov 8, 2016 #14

    Zafa Pi

    User Avatar
    Gold Member

    Are you saying that if we don't measure the position it still must exist (has some value x), we just don't know it? It seems that is the very logic, counterfactual definiteness, that leads to the Bell inequality, which we know is false.
     
  16. Nov 8, 2016 #15

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Sure - there are interpretations that say all kinds of things. There's some that say the electron travels all possible paths at the same time. I think there is an insights article about it someplace...
    What seems to have been suggested (bottom post #8) is that the particle ceases to exist when we are not looking at it.

    The commutation is irrelevant - I am talking about the conditions for a clear diffraction experiment.

    The source has a low uncertainty on the energy of the electrons - so the uncertainty in the time they are emitted is large.
    If we want to say precisely when they were emitted, then we have to accept a large uncertainty in the energy - and thus a wide range of overlapping diffraction patterns possibly washing the effect out. It's like giving the slits fuzzy edges.

    ie. For slit separation ##d\approx hc/E##, the interference maxima occur at ##d\sin\theta_n = n(hc/E)## right?
    For ##E\to E\pm\Delta E##, and propagate the uncertainties.

    No. I am saying that position exists without being measured, and our lack of knowledge of position does not mean the particle does not have a position.

    ... wait, are you saying that the Bell inequality is known to be false?
     
  17. Nov 8, 2016 #16

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Nope. Everything is itself. You are confusing the map for the territory.
    No - the mathematics we use to describe what can be known about it would no longer function in a useful way.
    The thing itself would still behave the way it always does.
    What has this got to do with interference at slits?

    What do you mean by "intrinsic" in this sense? Do you think that energy or spin do not exist unless they are measured? Have you not heard of the "intrinsic spin" of an electron?
    No - the model will just cease to function as a useful theory. It will stop being predictive.

    What you are saying does not make sense.

    ... but I did not say that so you appear to have gone into a tailspin. Is it your position that a article has no "intrinsic position" - please define that concept and present evidence to support it.
    You were the one asking questions - you have received answers that you do not agree with... sorry about that.
    Have you read @PeterDonis reply post #10?

    Please answer the questions from my previous posts. If you will not answer questions or define your terms I cannot help you.
     
  18. Nov 8, 2016 #17
    By intrinsic. I mean existing even without measurement. But you really believe a particle has position even when not measured. Just wondering. What is your specialty in physics? Does it mean different physicists have different ideas? Is it not part of the mathematical formalism that when a state is not an eigenstate of the operator associated with a particular observable (as Peterdonis put it), it doesn't have the properties? Or is the properties like position being there or not there when unmeasured a part of interpretations that different physicists have different way of understanding it?

    But then Simon point is that even when not measured, a particle has spin or energy for instance. Without this occurring, molecules would just collapse and objects cease to exist. This seems acceptable logic on his part.

    However Peterdonis wrote that "It depends on the state and the variable. If the state is an eigenstate of the operator associated with a particular variable ("observable" would be a better term), then it does have a specific value for that variable--more precisely, a measurement of that variable will yield a specific, definite value with certainty. If the state is not an eigenstate of the operator, however, then it doesn't; you can only make predictions about the probability of various values being measured."

    Peterdonis, is this true for all observables? For example.. position may not exist without the state being in an eigenstate of position.. but spin should definitely exist even without the state being in an eigenstate of spin or else atoms would cease to function. How do you resolve this?
     
  19. Nov 8, 2016 #18

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    I think we are getting bogged down in ontology... this can get very involved, see, for eg:
    By intrinsic. I mean existing even without measurement.

    Show me a particle whose position has not been measured and I can answer the question.
    What I have said repeatedly is that the lack of knowledge of the position of a particle is not the same as saying that the particle has no position.
    The particle exists even when we are not looking at it.

    In QM our possible knowledge of the position of a particle is given by the state vector.

    Quantum Mechanics :) Meaning nothing of course.
    Of course different physicists have different ideas ... otherwise science would not work.
    No.
    False dichotomy. Why can't it be neither?
    How would you determine if a particle has energy?
    Does the act of measuring the energy give the particle that energy or did the particle have energy all along?
    I am maintaining that the particle had energy all along and the act of measuring determined that energy even if it was prepared in a superposition of energy eigenstates to begin with.

    The quantum mechanics tells us the statistics of what is real - as for anything else, the theory is silent.
    Cause and effect is tricky.

    ... and I agreed with that statement.

    If the particle state vector is described by a superposition of eigenstates for an observable, then the particle does not have a determined value for that observable. That is not the same as saying the particle does not exist for the purposes of that observable... which appears to be your assertion. If that is your idea then of course you think a particle in a superposition of position eigenstates must "vanish" - but all particles are in position superpositions all the time, even after position has been measured, are you saying all particles vanish? (probability of finding a particle in an eigenstate of position is zero.)

    Remember how we get here - you asked about detectors for energy sans position ... when I gave an example where angular position was completely uncertain, you objected that radial position was measured to some uncertainty. Now you seem to be wanting to restrict position to an absolute value.

    Where were you trying to go with that? What does this have to do with diffraction at slits?
    What does this have to do with your original question about talking to someone ...
     
  20. Nov 8, 2016 #19
    This is puzzling how a person that specializes in quantum mechanics could still diverged from the mainstream views. Unless you are talking about Bohmian trajectories? Because in conventional QM, the particle has no position when unmeasured although it has spin and energy. Only the "position" is unreal when unmeasured. I hope other experts here can share this to you and resolve all this. I read that many Ph.Ds in physics are not familiar with Bell's Inequity. It shows that a particle has no properties displayed when unmeasured.. because if it has.. you can simultaneously measured it's position and momentum for two entangled particles that is sent away. So it has really no position when unmeasured.

    Double slit is the most convenient to imagine the state vector and its evolution. I'm sharing QM to someone about this article https://www.wired.com/2014/06/the-new-quantum-reality/ which starts:

    "For nearly a century, "reality" has been a murky concept. The laws of quantum physics seem to suggest that particles spend much of their time in a ghostly state, lacking even basic properties such as a definite location and instead existing everywhere and nowhere at once. Only when a particle is measured does it suddenly materialize, appearing to pick its position as if by a roll of the dice.

    This idea that nature is inherently probabilistic — that particles have no hard properties, only likelihoods, until they are observed — is directly implied by the standard equations of quantum mechanics."

    The article was written by Natalie Wolchover whose serious articles are permitted in physicsforums.
     
  21. Nov 8, 2016 #20

    Nugatory

    User Avatar

    Staff: Mentor

    This thread seems to have drifted away from the oroiginal questions for a while now.....
    At the moment of emission, the electron is very localized. There is a high probability that we we will find it very near a particular point in space, namely the point where the emitter is located. However, as time passes the wave function evolves as described by Schrodinger's equation; it spreads out and the probability of finding at any particular location becomes smaller. Eventually, however, the electron interacts with something (a detector, a droplet of vapor in a cloud chamber, a random dust particle, .....) and then the wave function collapses. Immediately after that interaction, the electron is highly localized again.

    So there's an initial position (the interaction with the emitter leaves the electron is a localized state) and there's a final position (the interaction with the detector leaves the electron in a localized state) but it doesn't follow that there is a trajectory in between. We often refer to the first interaction as "preparation" and the second as "measurement".

    You measure the energy of a charged particle by passing it through a magnetic field or between the plates of a parallel plate capacitor before you detect it.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Double slit & trajectories
  1. Double slit (Replies: 5)

  2. Double slit? (Replies: 3)

  3. Double Slit (Replies: 3)

  4. Double Slit (Replies: 40)

Loading...