- #1

- 17

- 5

I have a problem understanding the particle in a box (V=0 inside, V=∞ outside), how is it possible that momentum can vary continuously while the energy spectrum is discrete? Aren't they related by E=p

^{2}/2m? What I am missing? Thanks!

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter marcom
- Start date

- #1

- 17

- 5

I have a problem understanding the particle in a box (V=0 inside, V=∞ outside), how is it possible that momentum can vary continuously while the energy spectrum is discrete? Aren't they related by E=p

- #2

- 11

- 1

However, the Hamiltonian for a particle in a box [tex]\mathcal H = \begin{cases}\frac{p^2}{2m}&\text{if }0\leq x\leq L\\\infty&\text{otherwise}\end{cases}[/tex] is definitely NOT an analytic function of momentum. When this happens, the theorem that the eigenstates are the same is no longer applicable, and thus you get no more guarantee that the

Another example: the Hamiltonian for a harmonic oscillator is [tex]\mathcal H = \frac{p^2+(m\omega x)^2}{2m}.[/tex] This is a function of both the momentum and position operators, and thus is not purely an analytic function of either. This implies the same (lack of) results about eigenstates as above.

- #3

- 17,333

- 7,205

If you instead take a finite potential well, you will soon realise that the Hamiltonian and the momentum operators do not commute and as a result there is no common set of eigenstates.

- #4

Simon Bridge

Science Advisor

Homework Helper

- 17,874

- 1,655

The relation you are missing is: <E> = <p>

- #5

- 17,333

- 7,205

This is simply wrong, <p> = 0 for all eigenstates of the Hamiltonian in the case of the infinite potential well. Since H = p^2/(2m) + V, it follows that <E> = <p^2>/(2m) + <V>. For all states of the infinite potential well, <V> = 0, so <E> = <p^2>/(2m).<E> = <p>^2/2m

- #6

Simon Bridge

Science Advisor

Homework Helper

- 17,874

- 1,655

Oh typo ... <p^2>/2m, thanks.

- #7

Vanadium 50

Staff Emeritus

Science Advisor

Education Advisor

- 27,581

- 11,780

This is simply wrong, <p> = 0 for all eigenstates of the Hamiltonian in the case of the infinite potential well.

As is required by the symmetry of the system. In a stationary state, the system must spend as much time with p to the left as p to the right, so <p>=0.

- #8

- 17

- 5

I think that my question is wrong, the momentum is quantized actually.

E_{n}=n^{2}π^{2}ħ^{2}/2mL^{2}

P_{n}^{2}=2mE_{n}=2mn^{2}π^{2}ħ^{2}/2mL^{2}

P_{n}=±nπħ/L

E

P

P

Last edited:

- #9

- 18,839

- 9,710

To state it even more drastically: There is no such thing as momentum for a particle in a finite potential pot with infinitely large wells. It simply doesn't make sense. There is of course a complete set of energy eigenvectors, and the energy spectrum is entirely discrete.

If you instead take a finite potential well, you will soon realise that the Hamiltonian and the momentum operators do not commute and as a result there is no common set of eigenstates.

In reality, there's of course no such thing as infinitely high potential wells. It's just a simplified model system for a particle in a very deep potential well.

- #10

Vanadium 50

Staff Emeritus

Science Advisor

Education Advisor

- 27,581

- 11,780

I think that my question is wrong, the momentum is quantized actually.

The +/- there doesn't bother you? It should.

- #11

- 17,333

- 7,205

I think that my question is wrong, the momentum is quantized actually.

E_{n}=n^{2}π^{2}ħ^{2}/2mL^{2}

P_{n}^{2}=2mE_{n}=2mn^{2}π^{2}ħ^{2}/2mL^{2}

P_{n}=±nπħ/L

I suggest reading posts #3 and #9 above.

- #12

jtbell

Mentor

- 15,816

- 4,101

the momentum is quantized actually.

No. Take a look at the momentum-space wave functions ##\phi_n(p)## for the infinite square well (particle in a box):

http://physicspages.com/2012/10/04/infinite-square-well-momentum-space-wave-functions/

- #13

- 18,839

- 9,710

$$-\frac{\hbar^2}{2m} u_E''(x)=E u_E(x).$$

This has to be solved with the boundary conditions ##u_E(\pm L/2)=0##. The solutions are of the form

$$u_E(x)=A \sin(k x), \quad u_E(x)=A \cos(k x).$$

The boundary conditions lead to

$$k L/2=n \pi, \quad k L/2=(2n+1) \pi/2,$$

or

$$k_j=\frac{j \pi}{L}, \quad j \in \mathbb{N}=\{1,2,3,\ldots \}.$$

The energy eigenvalues are

$$E_j=\frac{\hbar^2 k_j^2}{2m}.$$

The eigenfunctions are

$$u_j(x)=A \begin{cases}

\cos(k_j x) \quad \text{for} \quad j=2n+1, \quad n \in \{0,1,\ldots\},\\

\sin(k_j x) \quad \text{for} \quad j=2n, \quad n \in \{1,2,\ldots \}.

\end{cases}$$

Note that these are NOT eigenfunctions of the would-be momentum operator, ##-\mathrm{i} \hbar \mathrm{d}_x##.

- #14

- 17

- 5

No. Take a look at the momentum-space wave functions ##\phi_n(p)## for the infinite square well (particle in a box):

http://physicspages.com/2012/10/04/infinite-square-well-momentum-space-wave-functions/

I found a page in the same site that proves my point:

"the momentum is equally likely to be in either direction. The magnitude of the momentum is a constant, since this is a state with fixed energy, and

http://physicspages.com/2012/09/13/infinite-square-well-momentum/

Last edited:

- #15

Vanadium 50

Staff Emeritus

Science Advisor

Education Advisor

- 27,581

- 11,780

The +/- should have been a clue. If I am in an Eigenstate of energy, and E and p commute, I should have one and only one value of momentum. This means I have at least two. So E and p don't commute, and I am not in a momentum eigenstate.

- #16

jtbell

Mentor

- 15,816

- 4,101

The magnitude of the momentum is a constant, since this is a state with fixed energy, and"

This is wrong. If this were true, then the momentum space wave function for the n'th energy eigenstate would be basically a pair of Dirac delta functions, with the "spikes" at ##p = \pm n \pi \hbar / a##. This is clearly not true for the n = 1 and n = 2 states, whose graphs are on the page that I linked to. Those graphs use the variable ##\rho = pa/\hbar##, so the "spikes" on those graphs would be at ##\rho = \pm n\pi##.

Note that the graph for n = 2 does have maxima at or near ##\rho = \pm 2\pi##. I once did graphs for some higher values of n, which illustrated that the peaks become sharper as n increases.

A sinusoidal wave function corresponds to an exact value of p only if it extends to infinity in both directions. A sinusoidal wave function that is "chopped off" to zero at the sides of the well, corresponds to a sum (actually an integral: a

- #17

- 17

- 5

It's talking about the mean momentum, not about what you would get when you measure p. If you go to that page and up a bit, it will say "Thusis not an eigenfunction of momentum, since the momentum operator doesn’t yield the original wave function multiplied by a constant."

OK, it's an eigenfunction of p

ψ

(d

p

Last edited:

Share: