- #1

jimmy.neutron

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Thanks

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- Thread starter jimmy.neutron
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- #1

jimmy.neutron

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Thanks

- #2

fermi

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- #3

michael879

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Given a finite potential barrier, the particle has some probability of being "inside" the barrier. The probability decays exponentially the farther inside you go, but it is always finite. Therefore you could view tunneling as the particle making it to the very end of the barrier. In classical terms, if the particle tunnels then it did actually go

As for interacting with the barrier, it really doesn't work like that. The barrier isn't some physical thing. It is a potential barrier, which means that the potential energy inside the barrier is higher than the particles kinetic energy. There isn't really anything to interact with in the idealized "barrier" since it is nothing more than some unchanging potential.

In a real situation, like in a hydrogen atom, the potential barrier is caused by some other particle. However, in the case of the hydrogen atom (and every other example I can think of), the particle isn't actually located inside the "barrier". With hydrogen, the potential is a -1/r curve with the proton situated in the center. In this case, although the electron can't "tunnel", it could theoretically be detected outside of the classical limit (i.e. at some distance further than where its kinetic energy should allow it to be). After it is detected there, the proton/electron system does change and you could say that the electron has "interacted" with the proton. However, its really your measurement than interacts with the 2 particle system.

- #4

jimmy.neutron

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- #5

f95toli

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Here N means normal metal, I insulator and S superconductor (a SIS structure is just a Josephson junction, the supercurrent is carrier by Cooper pairs).

The insulating layer must be very thin; in some cases it is just a few atomic layers thick. Note that the length-scale here (i.e. what defines "thin") is just the coherence lenght in the material, this is quite natural since the coherence length is just essentially the "size" of the wavepacket.

Hence, in an a tunnel junction electrons do really tunnel through a "wall", although the latter can always be modelled as an energy barrier (which -after all- is just what matter is to an electron)

- #6

kasse

- 382

- 1

Any experimental setup that can detect which slit the photon goes through destroys the interference pattern.

The particle knows that it's being observed? How is that possible?

Another question: in a tunneling setting, is it possible that one particle (matter wave) can be tunneled and refracted simultaneously?

- #7

jimmy.neutron

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- #8

Lojzek

- 249

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we find the particle in the barrier.

- #9

kasse

- 382

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- #10

Lojzek

- 249

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The only case when we may say that the particle is on a particular point is the case when it's wavefunction is zero everywhere except at that point. Otherwise we can measure it's possition to be on any point where the wavefunction is nonzero. In case of tunneling, the wavefunction is nonzero even in the area of the barrier, so it is possible (but not certain) that the measurement will tell us that the particle is in the barrier. You can also imagine a large number of particles flowing through the barrier (then some of them will probably be inside the barrier).

- #11

ice109

- 1,714

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The only case when we may say that the particle is on a particular point is the case when it's wavefunction is zero everywhere except at that point. Otherwise we can measure it's possition to be on any point where the wavefunction is nonzero. In case of tunneling, the wavefunction is nonzero even in the area of the barrier, so it is possible (but not certain) that the measurement will tell us that the particle is in the barrier. You can also imagine a large number of particles flowing through the barrier (then some of them will probably be inside the barrier).

umm there are no such wave packets, dirac delta wave packets, right? they're not normalizeable...

- #12

Lojzek

- 249

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Dirac delta function is defined by integral(f(x)dx)=1, so it is not normalized. I was talking about a function for which integral(f*(x)f(x))=1 (and is zero everywhere except at one point). Of course eigenfunctions of operator x must exist, since they form the base of x representation.umm there are no such wave packets, dirac delta wave packets, right? they're not normalizeable...

- #13

Hurkyl

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Such functions don't exist. It's a rather trivial exercise to prove if f satisfies that latter condition, then that integral is zero.I was talking about a function for which integral(f*(x)f(x))=1 (and is zero everywhere except at one point).

That would be true... if the word 'base' was meant in exactly the same way as the word 'basis' is meant in elementary linear algebra. (It's not)Of course eigenfunctions of operator x must exist, since they form the base of x representation.

- #14

ice109

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Dirac delta function is defined by integral(f(x)dx)=1, so it is not normalized. I was talking about a function for which integral(f*(x)f(x))=1 (and is zero everywhere except at one point). Of course eigenfunctions of operator x must exist, since they form the base of x representation.

i never said the eigenfunctions/eigenstates don't exist, they do and they are the dirac delta functions, which are not in the [itex] L^2[/itex] space but physicists don't care about that. what i said is that there are no dirac delta wave packets, as in you will never measure a particle to be in one of those eigenstates. anyway the generalized statistical interpretation says that for an operator with a continuous spectrum a measurement returns a value within a range of values centered about the eigenvalue of the eigenstate, not the eigenvalue.

- #15

Lojzek

- 249

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I did not mean f is a strict mathematical function with a clearly defined value at all points. I think we could define eigenfunction of x similary as delta function, except that we prescribe a value of integral(f*(x)f(x)dx) instead of integral(f(x)dx). We could defineSuch functions don't exist. It's a rather trivial exercise to prove if f satisfies that latter condition, then that integral is zero.

f=limit of f

f

- #16

Hurkyl

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In the space of Schwartz distributions, that limit is zero.We could define

f=limit of f_{n}, where

f_{n}=n^{1/2}(on interval [0,1/n] and zero everywhere else)

In particular, you can check directly that

[tex]\int_{-\infty}^{+\infty} f(x) \varphi(x) \, dx = 0[/tex]

where [itex]\varphi[/itex] is any test function.

- #17

ice109

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you that's an interesting question though what is the dirac delta convolved with itself?

- #18

Lojzek

- 249

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In the space of Schwartz distributions, that limit is zero.

In particular, you can check directly that

[tex]\int_{-\infty}^{+\infty} f(x) \varphi(x) \, dx = 0[/tex]

where [itex]\varphi[/itex] is any test function.

But the wavefunction f does not need to be a distribution! f*f must be a distribution, since this is how probability density is defined. You can check that f*f is in fact Dirac delta function.

For example: if we wanted to calculat the average potential energy of a state defined by f, we would get:

<V>=[tex]\int_{-\infty}^{+\infty} f*(x)V(x)f(x), dx = V(0)[/tex]

which is the correct value for a particle in an eigenstate x=0.

- #19

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Zz.

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