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

Electron diffraction

  1. Mar 28, 2006 #1
    An electron is diffracted like a wave when it passes through a double slit, which is narrower than the de Broglie wave length of electron.
    However, doesn't electron interacts with its surrounding ? (air molecules, screen, etc.) And since interaction equals to observation, isn't the electron observed? :confused: How can an electron diffracts eventhough it is involved in various interaction?:confused:

    These questions probably arise due to my lack of understanding of wave-particle duality. Any advise is welcomed.
  2. jcsd
  3. Mar 28, 2006 #2
    Interaction does not equal observation, but the presence of air molecules would make it possible in principle to find out which slit the electron took. For instance, the electron might "bump into" a molecule near the left slit, and this could make it possible to learn by an observation of the molecule that the electron took the left slit. As long as this is possible, no interference fringes are seen. That's why you need a vacuum in which to do this experiment. The interaction with the backdrop, on the other hand, is essential, since this contains the necessary detectors. The slit plate is treated as a classical boundary condition. This is one of the many approximations and simplifications we have to live with if we are to obtain any results.
  4. Mar 29, 2006 #3
    Thanks for your advice. If interaction doesn't necessarily mean observation, what is the accurate definition for observation?
  5. Mar 29, 2006 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor

    Please note that you don't have to observe the interference pattern only from "free" electrons moving in a vacuum. You can also observe an identical effect using supercurrent in a superconductor. This is what is observed in a Superconducting Quantum Interference Device (SQUID) experiments. The supercurrent itself is already "isolated" from its surrounding to maintain its coherence.

  6. Mar 29, 2006 #5
    Observation is just measurement. Interaction is harder to define since it is essentially a classical concept like force.
  7. Mar 29, 2006 #6


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Hehe, I like Harmony's question. So, what happens again in a measurement device ? Interaction, or measurement ? :biggrin:
  8. Mar 29, 2006 #7


    User Avatar
    Science Advisor

    Absolutely not. If you examine electrons in a two slit experiment done in air, you are dealing, formally, with a scattering experiment. For practical purposes, as is done in diffraction theory of light, each slit acts as a source, which means two spherical wavefronts propagate out from the slites -- cylindrical might be more appropriate. Collisions with air molecules can be reasonably approximated by the Born approximation. When you do this, you will see that there's no way the "slit of origin can be determined. The wave function is the sum of many terms :

    W == | 1;k,x> + |2,k,x> +

    (Sum over n) V(n,k, )*(| 1;k(n),x>*|n,P,x>+ |2,k'(n),x>*|n,P,x>)

    where the first two terms represent the wave function eminating from the slits, with momentum of magnitude k. The second set of states are the scattering states. After scattering, the electron from 1 has momentum k(n) and the associated air molecule has momentum P, and so on. V represents the interaction strength -- the appropriate matrix element. Sorry about my sloppiness-- I neglected to include the unperturbed molecular wave functions. But they are their in spirit.)

    If the air is of low density, then the scattering will simply deform or modulate the interference pattern. That is, the sum will be smaller than the first two terms. But, clearly, as the density grows, the multiple scattering will totally dominate, and the observed interference will vanish.

    If you can tell me how to get rid of the 1 or 2 wave function from molecular collisions, I'd be most grateful . I don't see how it could be done.

    Reilly Atkinson
  9. Mar 30, 2006 #8
    I took the formal approach according to which the electron emerging from the slits in vacuum is associated with a state proportional to |L> +|R> , while the electron + molecules system would be associated with a state proportional to |L>|ML> +|R>|MR>, where |ML> and |MR> are states of the system of molecules.
    With a bow to your track record I accept that it takes a lot of molecules to make the states |ML> and |MR> orthogonal and the interference fringes disappear as a result.
    If |ML> and |MR> are orthogonal then the electron state is the mixture |L><L|+|R><R|. We have gotten rid of the interference terms, not of |L><L| or |R><R|.
  10. Mar 30, 2006 #9
    For simplicity's sake let's assume that we have two initially uncorrelated systems each associated with a pure state. This means that the probability algorithm associated with the composite system at some initial time is |A>|B>.
    If they don’t interact then at a later time we have UAUB|A>|B>.
    If they do interact then we get a correlated state. It is then always possible to find a bi-orthogonal (Schmidt) decomposition for the probability algorithm associated with the composite system at a later time:
    SUMk ck |ak>|bk>.
    (If the ck all differ, the Schmidt decomposition is unique.)
    So much for interactions.
    In a measurement situation, one of the systems (say, the first system) is the apparatus. If you think of this algorithm as an ontological state then the statements "the apparatus has property ak" (k=1,2,3…) are neither true nor false but meaningless. But you are a good boy and think of it instead as an algorithm providing the following conditional information: if a measurement is made, the apparatus comes into possession of one of the properties ak, which indicates the possession by the system of the corresponding property bk. Moreover, the probability with which this particular outcome is obtained is |ck|2.
    So much for measurements.
  11. Mar 30, 2006 #10


    User Avatar
    Science Advisor

    koantum --Before I can comment, I'd like to know more about the two "M" states -- what are they, how are they defined?

    Reilly Atkinson
  12. Mar 31, 2006 #11
    Dear Reilly,
    I treat this as a 2 component system, electron and air. Because of the interaction between the electron and the air, the composite system will be associated with a probability algorithm proportional to
    |L>|ML> +|R>|MR>,
    where |ML> and |MR> are relative states (as in Everett's relative state formulation of quantum mechanics) of the air system.
    Measurement outcomes are represented by projectors. In discussions of foundations it is often assumed that every projector can in principle be tested. That is, one can in principle test whether or not the air has the property |MR><MR|, as well as whether or not it has the property |ML><ML|.
    I don’t have the faintest idea of how such a measurement could be made! (Is this what you wanted to hear?) However, I have faith in experimental physicists. One should never underestimate their ingenuity!
    If the electron-air interaction is strong, |ML> and |MR> will be virtually orthogonal. Finding |ML><ML| then makes finding |MR><MR| extremely unlikely, and vice versa. So one can conclude from finding |ML><ML| that the electron went through L and from finding |MR><MR| that it went through R.
    On the other hand, if the interaction is weak, |<MR|ML>|2 will be close to 1, and virtually nothing concerning the electron can be inferred from a test of either |ML><ML| or |MR><MR|.
  13. Mar 31, 2006 #12


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I was in agreement, up to the "you are a good boy" :smile: I just stop short of it, because the rest is circular: you use "if a measurement is made" and you were telling me how a measurement apparatus was working!
    So now I have to make a measurement of my measurement apparatus...
    And to explain *that*, I then have to make a measurement of the measurement apparatus that was making said measurement...

    Where does this end ? This is exactly what von Neumann realized already in his "mathematical foundations of quantum mechanics": there's an "infinite ladder" of these considerations which starts at the level of the microsystem being "measured" and ends at the conscious perception.
    Last edited: Mar 31, 2006
  14. Apr 1, 2006 #13
    Like von Neumann, you are traveling in the wrong direction. Instead of starting from the wave function (a probability algorithm transmogrified into What Exists) you should start from measurements, which are presupposed by the quantum formalism. After all, the wave function is nothing but an encapsulation of correlations between measurement outcomes.
    And no, I wasn’t telling you how a measurement apparatus was working. I was asked to say what I mean by a measurement.
  15. Apr 1, 2006 #14


    User Avatar
    Science Advisor

    koantum -- I don't have a clue about your relative states. (I've never studied Everett, I find it very hard to take his ideas seriously.) I would greatly appreciate your demonstrating how you compute the interference patterns. There is nothing in the air nor in the slits that allows a binary air state -- there's just air. a gas. And, of course, the air pretty much stays in the same "state" (If the molecules are Fermions, the Fermi sea will do.) I don't understand two air states. (How are they defined, starting from scratch. What are the initial conditions? I don't get it. how would you treat the energy loss of a high energy particle going through air? (Ordinary QM does a great job with this phenomena, as is discussed in many nuclear physics and/or scattering books and articles.) There's much more to QM than two-slit problems. In fact, I would imagine that this problem of interference in air could easily be made into a homework problem for an introductory QM course. It is a very straighforward problem, as I indicated above with a Born Approximation approach.

    Reilly Atkinson
  16. Apr 2, 2006 #15
    I find it impossible. A relative state, however, is a well-defined and useful concept (if by "state" we mean a probability algorithm). Too bad Everett used it in the title of his original MWI paper. If you’ve got a 2 component system, you can pick any pure state |a> of the first system and compute the state of the second given that the state of the first is |a>. The general state of the composite system is

    SUMi SUMk cik |ai> |bk>​

    where the kets |ai> and |bk> form bases in the respective Hilbert spaces Ha, Hb of the component systems, and the basis in Ha is chosen so that |a>=|a1>. The (unnormalized) state of the second system relative to |a> is then

    SUMk c1k |bk>.​

    Upon normalization, this is a conditional probability algorithm giving the probabilities of the possible outcomes of measurements performed on the second system (given that the probability of obtaining |a> by the appropriate measurement performed on the first system) is 1, or given that a measurement performed on the first system yields the property represented by the projector |a><a|.
    I didn’t invoke any binary air state. |ML> was the state of the system of air molecules relative to the state |L> of the electron. This state belongs to a Hilbert space that is the tensor product of the Hilbert spaces of the individual air molecules. The air isn’t in either |ML> or |ML>. But if the electron goes through L, then it is associated with the probability algorithm |ML>, and if the electron goes through R, then it is associated with the probability algorithm |MR>.
    Unless |<ML|MR>|2 equals 1, the two states (qua probability algorithms) differ, even if the probabilities they assign to the same measurements performed on the system of air molecules differ ever so little.
    "Starting from scratch" means different things to different people. :biggrin:
    I fully agree. But there is also much more to it than the Born approximation.
    I guess so, and it's even likely that it has been done somewhere, but I can't afford the time to do the problem or to search for the reference. The following (easily found) references are however pertinent:
    • Wootters and Zurek, "Complementarity in the double-slit experiment…", reprinted in Wheeler and Zurek, Quantum Theory and Measurement (Princeton University Press, 1983) p. 443.
    • Bartell, same title (except for the dots), reprinted in the same volume, p. 455.
    These papers show that if one doesn’t insist on 100% accurate measurements of the slits taken by the electrons (or photons, or whatever), a surprisingly strong interference pattern remains in evidence. In other words, if one considers the system of air molecules as an apparatus for measuring the slit taken by an electron, it is only if |<ML|MR>|2=0 (the two relative states are orthogonal) that finding |MR> (say) makes it 100% certain that the electron went through R and the interference pattern disappears completely.

    Ulrich Mohrhoff (aka koantum)
  17. Apr 5, 2006 #16


    User Avatar
    Science Advisor

    My take on your double sum -> into a fixed system 1 eigenstate is : it makes no sense to me -- what are the complete set of operators that define the states, why do you neglect interactions? What's the Hamiltonian? Does it make any difference if the interaction is mostly strong or weak, short range or long range, positive or negative, which could possibly produce bound states or resonances ? Could a single molecule destroy the interference pattern?

    How dependent upon the interaction is the detected distribution of electrons? Surely you are not going to say that the interaction between electrons and molecules is of no consequence.

    And, the experiments of 1. two open slits. and 2. one open slit are totally different -- two distinct scattering problems. That they would give differing results is quite reasonable. One of the key things in doing scattering experiments is to capture the experimental conditions in the cross-section -- basically, you integrate the wave function's absolute square over the detectors. (note that this can get a bit tricky for pure elastic scattering. But, except for purely electrically neutral particles, as Bloch and Nordseick showed, pure elastic scattering cross-sections are zero. Does your approach lead to this well-known conclusion? )

    I still don't have a clue what the M states are. I'll admit that I've never encountered such reasoning nor such concepts. To me, they seem totally disjoint from standard QM, scattering theory, wave packets. I look through such classics as Mott and Massey's book on Theory of Atomic Collisions, Bohm's Quantum Theory's discussion of Scattering Theory (1950s). Goldberger and Watson's Collision Theory (1960s) and Weinberg's Quantum Theory of Fields (Vol I) (1990s), and find nothing remotely like your take on the scattering in the double slit experiment done in air. Also, there's a huge literature, mostly associated with particle physics on energy loss of particles passing through matter, particularly for cloud chambers and Geiger counters, all of which shares little or nothing with your approach.

    If your way is better or even equal, then you have a lot of work to do to convince anybody that your approach is correct -- do the Rutherford cross section without waves, with projectors and whatever. There's a solid tradition, of some 70-80 years of scattering theory, which has been an extraordinarily powerful way to examine many different aspects of physics -- Compton scattering, pair production by means of electrons scattering from heavy atoms, neutrons scattering from crystals, pion -nucleon scattering and the 3-3 resonance, Bjorken and Feynman's partons, neutron proton scattering, nucleii-nucleii scattering, and ... If you can do these problems with your approach you probably might convince someone that you are on the right track. It's all in the details. And there is a remarkable consistency over the years of scattering theory and its applications -- the biggest differences are in notation, and the formulation of the Lippman-Schwinger Eq.(1950) which became the center piece and unifying approach to scattering theory.

    The biggest mystery to me still is the "M" states -- how would I recognize one? What's their mathematical structure? Does your method merge easily into a treatment of the air as a reservoir ? (Mandel and Wolf, Cohen-Tannoudji, et al on QED and quantum optics.) If you can, how do you deal with coherent states -- can you use your approach to calculate the quantum E&M electromagnetic field generated by a classical current? (Glauber (1963), see Mandle and Wolf for details)

    Reilly Atkinson
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook