- #1

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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!

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- #1

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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

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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 When this happens, the theorem that the eigenstates are the same is no longer applicable, and thus you get no more guarantee that the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

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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

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

- #5

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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

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Simon Bridge

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

- #7

Vanadium 50

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

- #8

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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

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- #9

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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

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The +/- there doesn't bother you? It should.I think that my question is wrong, the momentum is quantized actually.

- #11

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I suggest reading posts #3 and #9 above.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

- #12

jtbell

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

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

- #13

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

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

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

"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/

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- #15

Vanadium 50

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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

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

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

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OK, it's an eigenfunction of pIt'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."

ψ

(d

p

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