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How do you define the Eigenstates for the number operator?

  1. Oct 6, 2015 #1
    I was studying quantum states in quantum field theory and I came across the formula for defining eigenstates:

    |n> = [(a)n / sqrt(n!)] * |0>

    However, my book did not actually define ground state |0> (meaning the book did not give some function or numbers or anything like that to define what |0> is equal to).

    I know that the number operator acting on a state |n> should equal n. That should look like this:

    N|n> = n|n>

    Therefore, the number operator acting on |0> should yield 0|n> which is just 0.

    Knowing this, I set up a differential equation to solve for |0>. The number operator when expanded out is as follows:

    N = mωx2/2 - (1/2) + p2/2mω (where p is the momentum operator and not just momentum)

    Now in natural units where ħ= 1, p= -i (∂/∂x) so p2 = -∂2/∂x2

    Therefore the number operator becomes:

    N = mωx2/2 - (1/2) - (1/(2mω))(∂2/∂x2)

    Since N|0> = 0 and I wanted to solve for |0>, I set the above equation equal to 0 and got the differential equation:

    mωx2/2 - (1/2) - (1/(2mω))(∂2(|0>)/∂x2 = 0

    I solved the equation and derived a solution of:

    |0> =esqrt(m2ω2x2 - mω) * x + e-sqrt(m2ω2x2 - mω) * x

    (Of course the exponentials could be multiplied by some arbitrary constants, but I ignored those since I had no initial or boundary conditions).

    Now that I have derived |0>, I would like to verify if this is the correct way to define |0>. I want to verify this for two reasons:

    1. While my logic makes sense to me, I don't definitively know if the eigenstates of the number operator are even supposed to be continuous eigenstates such as the function I derived or if they are supposed to be discrete eigenstates.

    2. Whenever I try to plug in my solution into my differential equation, I always find that it is either too tedious to plug in to check my solution, or for whatever reason, as I am plugging back in it never looks like the final result will be 0. Also, differential equations in physics have this tendency to have one right form of a solution (for a physical context) even though mathematically speaking, there are multiple ways the solution could appear (depending on how you algebraically manipulate the differential equation).

    That is why I am asking two things here:

    1. Are the number states |n> supposed to have continuous eigenstates like the one I derived above, or are they supposed to have much simpler discrete eigenstates?

    2. If the eigenstates are continuous like the one I derived above, can someone check my work and verify my solution to my differential equation. I solved it multiple times in the same way that you solve any 2nd order homogeneous differential equation and got the same answer, but I'd really like to feel more confident about my solution. If my solution is right, then is it in the right form for the actual physics context, or should I only use one of those exponentials as my solution? Should I use a different form entirely (for example, should the solution appear trigonometric)?

    Please help. Thanks to anyone who helps.
     
  2. jcsd
  3. Oct 6, 2015 #2

    vanhees71

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    The groundstate is simpler to calculate if you define it as the eigenstate of the annihilation operator with eigenvalue 0:
    $$\hat{a} |0 \rangle=0.$$
    Then it's of course also an eigenvector for ##\hat{N}## with eigenvalue 0, because ##\hat{N}=\hat{a}^{\dagger} \hat{a}##. This is only a first-order linear differential equation. Try to solve it yourself. It's a nice and not too tedious exercise. For you reference, the resulting ground state in the position representation is [corrected definition of ##a##]
    $$\psi_0(x)=\frac{1}{(\pi a^2)^{1/4}} \exp \left (-\frac{x^2}{2a^2} \right), \quad a^2=\frac{\hbar}{m \omega},$$
    where I gave also the proper normalization factor (which is determined up to a phase factor only, of course), i.e., such that
    $$\int_{-\infty}^{\infty} \mathrm{d} x |\psi_0(x)|^2=1.$$
    The spectrum of the harmonic oscillator is entirely discrete. The potential is so strong that there are only bound states and no scattering states!
     
    Last edited: Oct 14, 2015
  4. Oct 7, 2015 #3
    Why are you saying the number operator is continuous? Of course, the function itself is continuous under the position [itex]x[/itex], but for each state, such as [itex]|0>, |1>, \cdots |n> [/itex], its discrete.
     
  5. Oct 13, 2015 #4
    Why did you square the term a when you derived that wave function? Applying the annihilation operator (in natural units where ħ= 1) to |0> should yield the differential equation:

    [(1/mω) * sqrt(mω/2) * (∂|0> / ∂x)] + [x * sqrt(mω/2) * |0> ] = 0

    The solution to this differential equation is:

    |0> = C e-mωx2/2

    The way you have it, squaring a would yield the solution |0> = C e-m2ω2x2/2

    I checked my solution to the differential equation by solving the equation in two different ways and also by plugging in my solution to the original differential equation, so I am confident that m and ω should not be squared.

    This of course would lead to the normalization integral being: $$\int_{-\infty}^{\infty} \mathrm{d} x |\psi_0(x)|^2=1.$$

    where

    $$|\psi_0(x)|^2$$ = C2e-mωx2

    The solution to the integral is C2sqrt(π/mω) and when setting this equal to 1, you should get:

    C= (mω/π)¼

    Therefore, shouldn't the solution be:

    ψ = (mω/π)¼ * e-mωx2 which if I substitute the term a into the solution would yield 1/(πa)¼ exp(-x2 / 2a)

    instead of the way you have it (which is the same as the way I have it, but the term a is squared in yours unlike in mine)?
     
  6. Oct 14, 2015 #5

    vanhees71

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    Sorry, that was a typo in the definition of ##a##. It's corrected now!
     
  7. Oct 14, 2015 #6
    Number state is the discrete energy level of a harmonic oscillator, to which the electromagnetic field in a single mode cavity is supposed to equavalent. You can find its wavefunction in the chapter of harmonic oscillator I suppose.
     
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