Understanding Momentum Eigenfunctions in Quantum Mechanics

[snoopies622]

I'm enjoying this introductory essay about quantum mechanics found here

http://www4.ncsu.edu/unity/lockers/users/f/felder/public/kenny/papers/psi.html

and I have a question. About five-eighths of the way into it a wave function is given "at time t=0",

$$<br /> \psi = \sqrt { \frac {2} {L} } [ \frac {1}{2} sin (\pi x / L) + \frac { \sqrt {3} }{2} i sin (5 \pi x / L)]<br />$$

and some questions and answers follow. If I am understanding the authors, the answers imply that this wavefunction is a (normalized) linear combination of two momentum eigenfunctions, where the momenta are $h/2L$ and $5h/2L$.

My question is, shouldn't

$$<br /> \psi = \sqrt { \frac {2} {L} } [ \frac {1}{2} (cos (\pi x / L) + i sin (\pi x / L)) + \frac { \sqrt {3} }{2} (cos (5 \pi x / L)<br /> + i sin (5 \pi x / L))]<br />$$

or - more succinctly -

$$<br /> <br /> \psi = \sqrt {\frac {2} {L} } [ \frac {1}{2} e ^ {i \pi x / L } + \frac { \sqrt {3} } {2} e ^ {i 5 \pi x / L } ]<br />$$

?

 

[snoopies622]

Oops - I just looked backward and noticed that this is one of those infinite potential well situations. That explains why both the real and imaginary components of $\psi$ have to be sine waves here.

And yet, how do I square this fact with the postulate that

$$e ^ { ipx/ \hbar }$$

- not $sin (px / \hbar )$ - are the momentum basis states?

 

[sonoluminated]

Write the sin as exp(ipx) - exp(-ipx), so isn't the sine just a different basis?

 

[sweet springs]

Hi.

If I am understanding the authors, the answers imply that this wavefunction is a (normalized) linear combination of two momentum eigenfunctions, where the momenta are $h/2L$ and $5h/2L$.

No, the answers imply that this wavefunction is a (normalized) linear combination of two ENERGY eigenfunctions, not momentum.
Regards.

 

[snoopies622]

Write the sin as exp(ipx) - exp(-ipx), so isn't the sine just a different basis?

Ah, yes - that did it. Thanks sonoluminated.

No, the answers imply that this wavefunction is a (normalized) linear combination of two ENERGY eigenfunctions, not momentum.

When the potential energy is zero, don't those amount to the same thing?

 

[sweet springs]

When the potential energy is zero, don't those amount to the same thing?

Hi.
When the potential energy is not zero but infinite square well or particle in a box, that is I thought your case is ( Am I wrong? ) consideration of momentum is shown in the text of Schubert.
http://www.ecse.rpi.edu/~schubert/C...um mechanics/Ch03 Position&momentum space.pdf
Regards.

 

[snoopies622]

That's a nice source, thanks sweet springs.

What I meant was, since inside the well the potential energy is zero, there is a one-to-one relation between the discrete momentum states and the discrete energy states.

 

[sweet springs]

Hi, snoopies622.

What I meant was, since inside the well the potential energy is zero, there is a one-to-one relation between the discrete momentum states and the discrete energy states.

Shubert (3.22),(3.23) and Figure 3.3 show us that the discrete energy eigenstate is superposition of continuous momentum eigenstates.
Energy at somewhere e.g. inside the well means simultaneous consideration of energy E or Hamiltonian H and position X. Commutation relation [H,X] ≠0 does not allow such consideration.
Regards.

 

[snoopies622]

...the discrete energy eigenstate is superposition of continuous momentum eigenstates.

I haven't studied the math yet but conceptually I am confused. In a place where there is no force acting on a particle and the potential energy is zero, the total energy must be the kinetic energy, which is a function of the momentum. How then can discrete energy states not imply discrete momentum states, and vice versa?

 

[sweet springs]

Hi, snoopies622.
I say here not using mathematics so do not take it so strict. See ψ0 in Shubert Figure 3.2. that is the ground energy state. You see near the well walls particle rarely exists and around center of the well particle is most probably found. This behavior is different from classic particle whose probability density is uniform in the well. The particle seems to be knowing the distances from the walls though local information e.g. potential energy does not change as you stated;

In a place where there is no force acting on a particle and the potential energy is zero, the total energy must be the kinetic energy, which is a function of the momentum.

In quantum physics such a global behavior replaces classical localism. In quantum physics amount of energy cannot be distinguished to kinetic and potential parts according to where the particle is.
Regards.

 

[snoopies622]

By "in a place" I didn't mean at a particular location inside the well, just inside the well. That is, at any location inside the well, the potential energy is zero and therefore all the energy is kinetic energy. I think we agree on this.

 

[sweet springs]

Hi, snoopies622

By "in a place" I didn't mean at a particular location inside the well, just inside the well. That is, at any location inside the well, the potential energy is zero and therefore all the energy is kinetic energy. I think we agree on this.

In quantum mechanics, even if potential energy is zero anywhere inside the well, energy eigenstate is under the influence of the potential beyond the walls. I called it "global" in my previous post. Tunnel effect is on the same footing.
Regards.

 

[snoopies622]

...the discrete energy eigenstate is superposition of continuous momentum eigenstates.

The problem is that $\psi$ as described above

$$<br /> \psi = \sqrt { \frac {2} {L} } [ \frac {1}{2} sin (\pi x / L) + \frac { \sqrt {3} }{2} i sin (5 \pi x / L)]<br />$$

can be expressed equivalently as

$$<br /> <br /> \psi = \sqrt { \frac {2} {L} } [ \frac {i}{4} (e ^ {-i \pi x / L} - e ^ {i \pi x / L}) + \frac { \sqrt {3} }{4} (e ^ {5i \pi x / L} - e ^ {-5i \pi x / L})]<br /> <br />$$

which is a linear combination of four momentum eigenfunctions,

$$<br /> e ^ {-i \pi x / L} \quad<br /> e ^ {i \pi x / L } \quad<br /> e ^ {5i \pi x / L} \quad<br /> e ^ {-5i \pi x / L}$$

not a continuum of them.

 

[sweet springs]

Hi. Snoopoies622
You are right in case the formula stands from -infinity x to +infinity x. I am afraid you are not right in case the formula stands only within the well and turns to zero for the outside. Which one is your case?
Regards.

 

[Dickfore]

Oops - I just looked backward and noticed that this is one of those infinite potential well situations. That explains why both the real and imaginary components of $\psi$ have to be sine waves here.

And yet, how do I square this fact with the postulate that

$$e ^ { ipx/ \hbar }$$

- not $sin (px / \hbar )$ - are the momentum basis states?

Because when the particle is in an external potential, the momentum does not commute with the total Hamiltonian of the particle and, hence, it does not have a definite value in a stationary state.

 

[snoopies622]

You are right in case the formula stands from -infinity x to -infinity x. I am afraid you are not right in case the formula stands only within the well and turns to zero for the outside

I was wondering about that...

Because when the particle is in an external potential, the momentum does not commute with the total Hamiltonian of the particle and, hence, it does not have a definite value in a stationary state.

So when the energy of the particle (inside the well) is measured to be $E _n$ we cannot assume that its momentum is therefore $$<br /> \pm \sqrt {2m E_n }<br />$$ ?

 

[sweet springs]

Hi. snoopies622

So when the energy of the particle (inside the well) is measured to be $E _n$ we cannot assume that its momentum is therefore $$<br /> \pm \sqrt {2m E_n }<br />$$ ?

No, in case the system has potential energy so there is a well.
Yes, in case the system has no potential energy so there is no well.
Regards.

 

[snoopies622]

This surprises me. I always looked at the particle-in-a-one-dimensional-box situation like this:

The wave must fit neatly inside the box, so the wavelength must be 2L/n where n is integer. Therefore, the particle's momentum must be

$$<br /> p = \frac {h}{\lambda} = \frac {hn}{2L}<br />$$

and its energy

$$<br /> E_n = \frac {p^2}{2m} = \frac {h^2 n^2}{4 L^2} \frac {1}{2m} = \frac {\hbar ^2 \pi ^2 }{2 m L^2} n^2<br />$$

 

[sweet springs]

Hi, snoopies622.
Fig.1 is wave function of momentum eigenstates p=h'/λ and p=-h'/λ superposed in equal weight.
Fig.1
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Fig.2 is wave function of an energy eigenstate in infinite square well potential.
Fig.2
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Though Fig.1 and Fig.2 coincide within the well region, They are different states. Fourier transform of Fig.2 gives continuous momentum spectrum. See Schubert.
You can rewrite En as pn^2/2m where pn is a quantity whose dimension is momentum, but pn is not momentum of the system. However as classical limit taking n large, the state momentum tends to +-pn. See Schubert Fig3.2.
Regards.

 

[snoopies622]

I think I understand now. I have one more question: what does the Schrödinger equation look like in the momentum basis (instead of the position basis)?

 

[sweet springs]

Hi, snoopies622.
Shrodinger equations of stationary state for Hamiltonian H(p,x) are H(h'/i d/dx, x) ψ(x)=Eψ(x) in coordinate space and H(p, ih'd/dp) φ(p)=Eφ(p) in momentum space where φ(p) is Fourier transform of ψ(x).
Regards.

 

[snoopies622]

Thanks for that, sweet springs.

I'm still having a problem with the mathematical equivalence I mentioned in entry 13. I see how neither representation takes into account the boundaires of x=0 and x=L, and that this is why the momentum representation (the second equation) doesn't work. But why then does the first one? I assume that a function that is sinusoidal between two x values and then suddenly zero outside of that domain is much more complicated than the sum of either two or four sinusoidal terms.

 

[Dickfore]

Schroedinger Equation in momentum basis is:

$$\frac{p^{2}}{2 m} \, a(\mathbf{p}) + \int{\frac{d^{3}p}{(2 \pi \hbar)^{3}} \, \tilde{U}(\mathbf{p} - \mathbf{p}') \, a(\mathbf{p}') = E \, a(\mathbf{p})$$

where

$$a(\mathbf{p}) \equiv \int{d^3{r} \, \psi(\mathbf{r}) e^{-\frac{i}{\hbar} \mathbf{p} \cdot \mathbf{r}}}$$

is the wave function in momentum space. It's physical significance is that:

$$|a(\mathbf{p})|^{2} \frac{d^{3}p}{(2 \pi \hbar)^{3}}$$

is the probability that the particle will have a value of momentum inside an elementary cube of volume $d^{3}p$ around the value $\mathbf{p}$, and

$$\tilde{U}(\mathbf{p}) = \int{d^{3}r \, U(\mathbf{r}) \, e^{-\frac{i}{\hbar} \mathbf{p} \cdot \mathbf{r}}}$$

is the spatial Fourier transform of the potential energy.

If the potential depends only on one Cartesian coordinate (denoted by x), then:
$$\tilde{U}(\mathbf{p}) = \tilde{U}(p_{x}) \, (2 \pi \hbar)^{2} \, \delta(p_{y}) \, \delta(p_{z})$$

and we can choose a wave-function with a definite value of the components of the momenta in the y and z direction:

$$a(\mathbf{p}) = a(p_{x}) \, (2 \pi \hbar)^{2} \, \delta(p_{y} - p_{y0}) \, \delta(p_{z} - p_{z0})$$

Then,

$$|a(p_{x})|^{2} \frac{dp_{x}}{2 \pi \hbar}$$

is the probability that $p_{x}$ has a value in the interval $(p_{x}, p_{x} + dp_{x})$. The Schroedinger equation becomes one dimensional:

$$\frac{p_{x}^{2}}{2 m} \, a(p_{x}) + \int{\frac{dp_{x}}{2 \pi \hbar} \, \tilde{U}(p_{x} - p'_{x}) \, a(p'_{x})} = \epsilon \, a(p_{x}), \ \epsilon = E - \frac{p^{2}_{y} + p^{2}_{z}}{2 m}$$

 

[sweet springs]

Hi. snoopies622

I see how neither representation takes into account the boundaires of x=0 and x=L, and that this is why the momentum representation (the second equation) doesn't work. But why then does the first one?

Both the representation take into account the potential energy generating boundaries at x=0 and x=L. The coordinate representation is easier to solve because equation includes up to (d/dx)^2. In the momentum representation equation (d/dp)^n of any large n appear corresponding to power series expression of V(x). Dickfore discussed it in another way using Fourier transform of the potential energy.
Regards.

 

[snoopies622]

Wow, thanks Dickfore. I'll have to give all that some time.

The coordinate representation is easier to solve...

Actually, the two representations I had in mind in my previous question were not position vs. momentum. They were both in position representation. (See entry #13.) The first was a linear combination of energy eigenfunctions, and the second was a linear combination of momentum eigenfunctions. I was wondering why one could use the first one to find the particle's allowed energy values, but not the second one to find its allowed momentum values.

 

[sweet springs]

Hi. snoopies622 From your post #13

The problem is that $\psi$ as described above

$$<br /> \psi = \sqrt { \frac {2} {L} } [ \frac {1}{2} sin (\pi x / L) + \frac { \sqrt {3} }{2} i sin (5 \pi x / L)]<br />$$

inside the well and zero outside, I remind you again. This can be expressed equivalently as 1/h＾1/2∫a(p)e^ipx/h' dp where a(p)=1/h^1/2∫ψ(x)e^-ipx/h' dx where h=2πh', Fourier transformation.

not a continuum of them.

So continuum of momentum eigenfunctions, i.e. 1/h＾1/2∫a(p)e^ipx/h' dp.
You don't have to solve two different equations of different representation of an energy eigenstate independently. Once you get a solution in one representation, Fourier transform of it becomes the solution in other representation.

I was wondering why one could use the first one to find the particle's allowed energy values, but not the second one to find its allowed momentum values.

On the state ψ(x), when you want to know about energy, energy representation is convenient. When you want to know about momentum, momentum representation is convenient. When you want to know about position, coordinate representation is convenient, so on.
Regards.

 

[snoopies622]

...when you want to know about energy, energy representation is convenient. When you want to know about momentum, momentum representation is convenient. When you want to know about position, coordinate representation is convenient, so on.

I'm not sure I'm completely following you here. Both equations I mentioned in entry 13 are in the position basis, and both are linear combinations of eigenfunctions. The first one uses energy eigenfunctions, the second one momentum eigenfunctions. One can look at the first equation and see what the allowable energy values will be, but not the second equation and see what the allowable momentum values will be. Why the difference? I assume that they both take the boundary conditions into account.

 

[sweet springs]

Hi, snoopies622

Both equations I mentioned in entry 13 are in the position basis, The first one uses energy eigenfunctions,

Yes. You see energy eigenfunctions has non zero value only within the well region. This feature is kept even if we rewrite it in another form, so,

the second one momentum eigenfunctions.

No, the second one does not use momentum eigenfunctions. Eigenfunction of momentum is e^ikx for all the regions of x.　
Function that is e^ikx for some region and zero outside of it is not eigenfunction of momentum that is the case of your second one. If you want to rewrite the formula with momentum eigenfunctions, you should apply Fourier transform.
Regards.

 

[snoopies622]

Eigenfunction of momentum is e^ikx for all the regions of x. Function that is e^ikx for some region and zero outside of it is not eigenfunction of momentum...

Why does this restriction not apply to energy eigenfunctions as well? They are both sine waves after all, which by definition extend forever in both directions.

 

[sweet springs]

Hi, snoopies622

Why does this restriction not apply to energy eigenfunctions as well? They are both sine waves after all, which by definition extend forever in both directions.

Why not? Read remind to you in my post #26, please. Energy eigenfunctions are sinusoidal waves in the well and are zero outside.
Do not be bored to see the same Figures again. They are NOT like Fig.1
Fig.1 (superposition of two) momentum eigenstate(s)
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but like Fig.2.
Fig.2 an energy eigenstate
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Regards.