De Broglie Wavelength interpretationI've just started learning about the double slit experiment (just in the short appendix section in Schroeder's Thermal Physics), and I'm extremely confused by this one thing:In it, out of basically nowhere he pulls out the De Broglie equation, that λ = h/p.I've studied double slit diffraction before, and I've been trying to connect them in order to understand what this wavelength actually means.In double slit diffraction, when the wavelength is larger, the diffraction "stripes" that form on the wall appear further apart. They also appear larger.y = m λ L d (approx, considering the distance to the screen is really large and thus almost parallel rays (drawn out waves) can have a path difference and interfere)If we were to make the wavelength extremely small, that would mean that anything a little off-center would interfere, so the smaller the wavelength, the closer together the "stripes" on the wall would be.Now, when we connect the 2 equations, this means that the faster the electrons are moving (the smaller their wavelength) the more places they will interfere on the wall, and therefore there will be a lesser distance between adjacent places where the electrons hit (bright spots) and places where they don't (dark spots).The way I'm interpreting this is that the smaller the "wavelength" of the electron, the more the probability it has to have been in different places at the same time, that is, the less we can know its position. That's why more stripes will appear on the detecting screen because there are more positions which the electron could've been in, and since its technically in all of them at the same time while it travels, it can interfere with itself more.Is this interpretation correct? Does a faster momentum (a smaller wavelength) mean that the electron literally is at more places at the same time while it travels from the electron gun, through the slits, and to the wall? Thank you!

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De Broglie Wavelength interpretationI&#039;ve just started learning about the double slit experiment (just in the short appendix section in Schroeder&#039;s Thermal Physics), and I&#039;m extremely confused by this one thing:In it, out of basically nowhere he pulls out the De Broglie equation, that λ = h/p.I&#039;ve studied double slit diffraction before, and I&#039;ve been trying to connect them in order to understand what this wavelength actually means.In double slit diffraction, when the wavelength is larger, the diffraction &quot;stripes&quot; that form on the wall appear further apart. They also appear larger.y =                               m          λ          L                d             (approx, considering the distance to the screen is really large and thus almost parallel rays (drawn out waves) can have a path difference and interfere)If we were to make the wavelength extremely small, that would mean that anything a little off-center would interfere, so the smaller the wavelength, the closer together the &quot;stripes&quot; on the wall would be.Now, when we connect the 2 equations, this means that the faster the electrons are moving (the smaller their wavelength) the more places they will interfere on the wall, and therefore there will be a lesser distance between adjacent places where the electrons hit (bright spots) and places where they don&#039;t (dark spots).The way I&#039;m interpreting this is that the smaller the &quot;wavelength&quot; of the electron, the more the probability it has to have been in different places at the same time, that is, the less we can know its position. That&#039;s why more stripes will appear on the detecting screen because there are more positions which the electron could&#039;ve been in, and since its technically in all of them at the same time while it travels, it can interfere with itself more.Is this interpretation correct? Does a faster momentum (a smaller wavelength) mean that the electron literally is at more places at the same time while it travels from the electron gun, through the slits, and to the wall? Thank you!

TettetoxDetnhte5 · Accepted Answer

The uncertainty that matters is transverse. Imagine an monochromatic plane wave (of infinite extent, wavenumber             k      →        =  k            z      ^        =  (  2  π      /    λ  )            z      ^      ) impinging normally on the slit apparatus (one slit, width   w in the   x-direction).The uncertainty in the transverse momentum is:  Δ      p    x    =  0Now it goes through the slit. We have now localized an infinite plane wave within a region of extent   w:  Δ  x  =  wIt acquires an uncertainty in transverse momentum such that:  Δ      p    x    Δ  x  ≈  ℏ      /    2Or:  Δ      p    x    =      ℏ          2      w      Hence:That means there is an angular spread in the wave emanating from the slit:  Δ  θ  =            Δ              k        x                            |                    |            k              |                    |              =      λ          π      w      That is, of course, diffraction. Diffraction can be viewed as a consequence of the uncertainty in position at the slit.With 2 slits separated by   d, the fringe-spacing (or rate of change of phase difference) can be computed with trig without appealing to the uncertainty principle.

Daphne Fry · Accepted Answer

The de Broglie wavelength is the wavelength associated with a particle with a fully-specified spatial momentum (which is impossible in reality): namely, the position-space wave function of a moving particle in a momentum eigenstate       |        p    =            p              g      i      v      e      n        ⟩ is, at any given time,  ψ  (  P  )  =  A      e          i      [                        p                          g          i          v          e          n                    ⋅      (      P      −      O      )      ]              /            ℏ      where   O is the origin and   A describes the phase. The periodicity is in the direction of             p              g      i      v      e      n       and has spatial length  λ  =      ℏ                  |                              p                          g          i          v          e          n                            |            which is the de Broglie wavelength.For particles in more complicated states, the de Broglie wavelength is less evident, but it still pops up in many cases as a characteristic natural scale, particularly when we can treat a particle&#039;s momentum as relatively well-specified (having a relatively high information content).The reason it appears in the double-slit experiment is it is useful to model the incoming particles, on the gun-facing side of the slit, as plane waves of this form.

De Broglie Wavelength interpretation I've just started learning about the double slit experiment (

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