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If you could measure the KE in the classically forbidden region

  1. May 25, 2005 #1


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    A particle in a stationary state with energy E (all kinetic) is shown to have some probability of tunnelling through a potential barrier. (say an electron in a potential well, where U = 0 in the well, and U_1 out of the well)

    My question is, if the particle ends up in the classically forbidden region, what does its energy look like in that region?

    My book says that no energy is lossed in the tunneling procces, so does that mean that no KE is lost to PE due to tunneling? how could that be?

    Classically the particle would have a negative Kinetic engergy which would be meaningless since it would imply an imaginary velocity or negative mass.

    If you could measure the KE in the classically forbidden region, what would it read?

    Thanks in advance
    Last edited: May 25, 2005
  2. jcsd
  3. May 25, 2005 #2


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    The energy of the particle in the forbidden region is negative -- that's why it's a forbidden region. The subtle point that you might be missing is that the particle cannot ever be measured actually in the forbidden region -- it can only be measured on either side of it, though it has a non-zero chance of tunneling through it. You can't measure the energy of the particle in the forbidden region.

    - Warren
  4. May 25, 2005 #3
    Your question is a bit nonsensical. By a stationary state with energy E, I take that to mean an eigenstate of (say) the potential well the particle's in. When you say 'what does its energy looks like', do you mean the functional form of the wavefunction? If so, it's pretty easy to solve the Schrodinger equation for these types of system. Say we have a case where the potential (in 1D) is:

    [tex]V(x) = \left\{\begin{array}{cc}0,&\mbox{ if} |x| \leq a\\b, & \mbox{ if} |x| >a\end{array}\right[/tex]

    We can use a few rules to work out the functional form of the eigenstates:

    1) The function must be normalised (ie [itex]\int\Psi dx = 1[/itex])
    2) The function must be continuous (ie differentiatable at [itex]|x| = a[/itex])

    So, we break the system down into the regions x < a, a<= x <= a, x >a, work out the eigenstate in each section with regards to some normalisation parameters (it's an inverse exponential in the regions with potential, and some form of oscillation in the region without potential). We then equate the differentials at x = ± a to ensure (2) is met, do some jiggery pokery, and integrate over the entire functional range to ensure (1) is met.

    Not the most elegant way, but it works, and you get a nice way of visualising the wave function as in this time independant case you get a real function.

    P.S. The wavefunction in a potential barrier is also an inverse exponential.
    Last edited: May 25, 2005
  5. May 25, 2005 #4


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    Why is it impossible to measure the particle in the forbidden region?
    We can measure it outside can't we?

    James Jackson:
    I'm taking an intro course to modern physics so the terms might be simplified. The example you provide is the same as what I'm thinking of, and the formula I have for the wave function is the same form, sinusoidal before and after the barrier, and decreasing exponentially in the barrier. But I'm just wondering if the particle tunells into the barrier, and stays there, then how could we explain what its energy should be. It seems as though the laws of physics as far as Energy is concerned only applies some of the time.

    Also if it tunnels completely through the barrier, does it emerge on the other side with the same Energy it had before tunneling?

    Just so you guys know, I realize that the term tunneling doesn't litterally mean tunnelling, I understand that I have to think in terms of probabilities. But if the particle is initially to the left of the barrier, and at some time later is inside the barrier or has gon throught the barrier, then doesn't it have to have get there continuously?

    Thanks for your help and your patience!
  6. May 25, 2005 #5
    Hmm, I asked my professor a very similar question about this today and he said that you could detect the particle in the barrier area just as you could in the other areas. :confused: In particular I asked about a transistor type problem with U=0, then U > E of the particle, then back down to U=0, He said you could detect it in the 2nd area(the area with U > E), and then a little while later in region 3. My professor is quite a brain so it's odd that on something that is a pretty basic principle of QM he would lead me wrong. Any insight anyone?
  7. May 25, 2005 #6


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    If we know that an electron with energy E was initially in a potential well and later found anywhere out of the well, we know it has penetrated the barrier.

  8. May 26, 2005 #7
    I think its like this: you can detect the particle inside the barrier however when doing this you will give the particle energy so that it has a positive total energy and thus can exist in the barrier.
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