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GR9277 #53
Problem
 GREPhysics.NET Official Solution Alternate Solutions
\prob{53}
A particle of mass m is confined to an infinitely deep square-well potential:
$\begin{eqnarray} V(x)&=&\infty,\;\;x\leq0,x\geq a \\ V(x)&=&0,\;\;0\lt x \lt a\end{eqnarray}$

The normalized eigenfunctions, labeled by the quantum number n, are $\psi_n=\sqrt{\frac{2}{a}}\sin \frac{n\pi x}{a}$

A measurement of energy E will always satisfy which of the following relationships?

1. $E\leq\frac{\pi^2\hbar^2}{8ma^2}$
2. $E\leq\frac{\pi^2\hbar^2}{2ma^2}$
3. $E=\frac{\pi^2\hbar^2}{8ma^2}$
4. $E=\frac{n^2\pi^2\hbar^2}{8ma^2}$
5. $E=\frac{\pi^2\hbar^2}{2ma^2}$

Quantum Mechanics$\Rightarrow$}Energy

If one forgets the energy of an infinite well, one can quickly derive it from the time-independent Schrodinger's Equation $-\frac{\hbar}{2m}\psi^{''}+V\psi = E\psi$. However, since $V=0$ inside, one has $-\frac{\hbar}{2m}\psi^{''} = E\psi$.

Plug in the ground-state wave function $\psi = A \sin(kx)$, where $k=n\pi/a$. Chunk out the second derivative to get $E=\frac{k^2 \hbar^2}{2m}$. Plug in k to get $E=\frac{\hbar^2 n^2 \pi^2}{2ma^2}$.

Note that $k=n\pi/a$ can be deduced from boundary conditions, i.e., the wave function vanishes at both ends ($\psi(0)=0$ and $\psi(a)=0$). The second boundary condition forces the n's to be integers. Since one can't have a trivial wave function, $|n| \geq 1$, and thus $n^2 \geq 1$. One finds that $E \geq \frac{\pi^2 \hbar^2}{2m a^2}$, since $n=1,2,3...$, as in choice (E).

Alternate Solutions
 panos852007-10-30 11:31:21 The energy depends on n so it cannot be a constant. This eliminates C, D, E. Furthermore it cannot be bounded from above as it increases with n. (You can easily remember that.) So A is out. Choice B remains.Reply to this comment
francesco
2017-09-14 20:26:58
if you look at the eigenfunctions (in particular, at the argument of the sine) you can deduce that $k = \\frac{n\\pi}{a}$, plug it in $E = \\frac{\\hbar^2 k^2}{2m}$ and find choice B
syreen
2013-09-18 13:47:54
The Schrodinger EQ should have hbar^2, not just hbar.

Also, as others have noted, the typed question has a typo. B. should have E>=, NOT E<=
Kabuto Yakushi
2010-09-04 08:58:08
As long as you remember (which it is probably a good idea to memorize for this test) the energy in a potential well to be

E = $\frac{\pi^2\hbar^2}{2ma^2}$

with a note that using $\hbar^2$ results in a two not an eight on the bottom, choice (B) is obvious. (C),(D), and (E) are to precise to satisfy the "always" criteria.
Fizzics
2009-10-30 11:44:19
Well I tried to post that last thing as a typo.
Fizzics
2009-10-30 11:43:33
The Answer should be "B" not "E". And the answer shown in the problem as "B" is wrong. Should be E is GREATER THAN or EQUAL TO.

Typos.
jw111
2008-11-04 11:45:30
Since $E=\frac{p^2}{2m}$
and $p=H*k$ where H = h bar

so $E=\frac{H^2k^2}{2m}$

with the solution given by question,
$k=\frac{n\pi}{a}$

thus
$E=\frac{n^2\pi^2H^2}{2ma^2}$
Monk
2008-10-13 21:46:07
(hbar)^2 is missing in the original presentation of the TISE, it just shows (hbar)...though most of you should know that anyhow :P
2007-11-02 12:49:53
B should read E >= ....
 evanb2008-06-23 19:07:46 yes.
 evanb2008-06-23 19:11:02 Dammit, the preview de-TYPO-ified my entry. Yes, (B) should be >= and not <= and it should be the answer, not (E)
panos85
2007-10-30 11:31:21
The energy depends on n so it cannot be a constant. This eliminates C, D, E. Furthermore it cannot be bounded from above as it increases with n. (You can easily remember that.) So A is out. Choice B remains.
 hoyas082008-06-16 19:52:47 Unfortunately, choice D has an $n^2$ term in it, so you can't eliminate it by reasoning that the energy must depend on n. However, if you can remember that the energy of a particle in free space (V=0) is E=$\frac{\vec{p}^2}{2m}$, then the 8 in the denominator of D does not make sense.
 hoyas082008-06-16 19:55:05 Unfortunately, choice D has an $n^2$ term in it, so you can't eliminate it by reasoning that the energy must depend on $n$. However, if you can remember that the energy of a particle in free space (V=0) is E=$\frac{\vec{p}^2}{2m}$, then the 8 in the denominator of D doesn't make sense, eliminating that answer.
 Imperate2008-10-16 10:18:38 This is not true for D, D has E=n^2... which is the correct form.rnrnOne can throw away A becayse there is no upper bound to energy.rnOne can throw away C and E because they [b] are [/b] constants, which is unacceptable for the energy.rnrnThis leaves B and D, and D can only be eliminated by working out the correct coeff, as others have posted.rnrnI think this question is a bit harsh though because from elementary QM infinite square wells always have n^2 dependence so you see D whilst doing a high speed test like the GRE then you are going to pick D, well I did anyway....rnrnI
 ramparts2009-08-06 19:24:55 How does that eliminate D? D is dependent on n.
 ramparts2009-08-06 19:24:55 How does that eliminate D? D is dependent on n.
antithesis
2007-10-05 18:17:17
Aside from typos mentioned, the question itself has a typo, B should have equal or greater than
Mexicana
2007-10-04 17:46:32
It is much easier to derive this by getting the zero point energy via Heisenberg Uncertainty Principle $\delta x \delta p = h/2\pi$ and then square this expression. Now just substitute for $\delta x= a$ where $a$ is the width of the square well and then the usual expression for momentum $(\delta p)^2 = 2m E$.
 Tommy Koulax2007-10-29 09:53:59 doesn't that lead to E=$\frac{(h bar)^2}{2*m*a^2}$ which is not an option provided?
georgi
2007-08-26 20:20:13
as everyone is posting the answer should be choice b, not e as written
welshmj
2007-07-12 15:51:47
Choice B is the correct answer not E, but choice B on this page has a typo. It should be greater than or equal to, not less than or equal to.
JB
2005-12-07 02:32:15
 jcain62005-12-08 19:20:24 It should be choice (B) because the question requires E to ALWAYS satisfy the relationship. Choice (E) only works when the potential is 0 between o and a.
 Andresito2006-03-27 15:43:35 I also think it is (B).
 sirius2008-06-25 13:05:18 the potential is 0 from 0 to a. it says so in the problem. this is a particle in an infinite square well, and answer (E) is the only answer that makes sense.
 sirius2008-06-25 13:09:59 nevermind, i misread which had the >=, i agree it is (B)
jcain6
2005-11-22 19:06:10
Your missing a (hbar)^2 in your numerator.
 yosun2005-11-23 02:00:51 jcain6: thanks for the typo-alert; it's been corrected.

LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces $\alpha^2_0$. type this... to get...$\int_0^\infty$$\int_0^\infty$$\partial$$\partial$$\Rightarrow$$\Rightarrow$$\ddot{x},\dot{x}$$\ddot{x},\dot{x}$$\sqrt{z}$$\sqrt{z}$$\langle my \rangle$$\langle my \rangle$$\left( abacadabra \right)_{me}$$\left( abacadabra \right)_{me}$$\vec{E}$$\vec{E}$$\frac{a}{b}\$ $\frac{a}{b}$