Bound state problem

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Homework Statement


Consider [itex]V(x)=-aV_{0}δ(x)[/itex]. Show that it admits a bound state of energy [itex]E=-ma^2V_{0}/2\hbar^{2}[/itex].Are there any other bound states?Hint:solve Schrodinger's equation outside the potential for E<0, and keep only the solution that has the right behavior at infinity and is continuous at x=0.

Draw the wave function and see how there is a cusp, or a discontinuous change of slope at x=0. Calculate the change in the slope at equate it to [itex]\int_{-ε}^{ε}( \frac{d^2\psi}{dx^2}) dx[/itex]
(where ε is infinitesimal) determined from Shrodinger's equation.

Homework Equations


Shrodinger equation in X basis : [itex]\frac {d^2\psi}{dx^2}+(2m/\hbar)(E-V)\psi=0[/itex].

The Attempt at a Solution


First how can we prove that [itex]E=-ma^2V_{0}/2\hbar^{2}[/itex] admits a bound state, shall I substitute it in the shrodinger equation then prove that the obtained wave function goes to zero when x goes to infinity?
For are there any other bound states?
I tried to solve the Shrodinger's equation with E<0 and outside the potential, what I got is this :
Let E=-A where A is a positive integer.
[itex]\frac {d^2\psi}{dx^2}+(2m/\hbar)(E)\psi=0[/itex]
[itex]\frac {d^2\psi}{dx^2}-(2m/\hbar)(A)\psi=0[/itex]
[itex]\psi=Ae^{-\sqrt{2mA/\hbar^2}x}+Be^{\sqrt{2mA/\hbar^2}x}[/itex]
[itex]Be^{\sqrt{2mA/\hbar^2}x}[/itex] blows up to zero when x goes to infinity, so B=0.
what we finally get is [itex]\psi=Ae^{-\sqrt{2mE/\hbar^2}x}[/itex]. I reached here and I don't know how to complete.
If anybody can help.Thanks.
 
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Answers and Replies

  • #2
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x can go both to positive and negative infinity, you have to consider both cases separately.
Then consider x=0 separately again, with the given hint.

Schrödinger (or Schroedinger), by the way.
 
  • #3
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x can go both to positive and negative infinity, you have to consider both cases separately.
Then consider x=0 separately again, with the given hint.

Schrödinger (or Schroedinger), by the way.
Let E=-U where U is a positive number.
Take case where x<0
the solution is[itex] \psi=Ae^{-\sqrt{2mU/\hbar^2}x}+Be^{\sqrt{2mU/\hbar^2}x}[/itex]
as x goes to -infinity the first part will blow up so A must be zero.
when x>0
the solution is [itex] \psi=Ce^{-\sqrt {2mU/ \hbar^2}x}+De^{\sqrt{2mU/ \hbar^2} x } [/itex]
as x goes to +infinity the next part will blow up so D=0.

now for case x=0
the solution is [itex] \psi=E e^{-\sqrt {2m (U+V)\hbar^2}x}+Fe^{\sqrt{2m(U+V)/ \hbar^2} x} [/itex]
here when x=0 V=infinity but 0*infinity is not defined. So how we can determine the wave function here?
 
  • #4
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The wave function has to be continuous.
You don't need x-dependent functions to consider the case x=0 as the wave function there has a single value only.
 
  • #5
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The wave function has to be continuous.
You don't need x-dependent functions to consider the case x=0 as the wave function there has a single value only.
Let [itex]\psi[/itex]at x=0 be a number called it F.
for x<0 [itex]\psi=Be^{\sqrt{2mU/\hbar^2}x}[/itex]
for x>0[itex]\psi=Ce^{-\sqrt{2mU/\hbar^2}x}[/itex]
As the wave function must be continuous at x=0,then we have [itex]\lim_{x\rightarrow 0^{-} } {B e^ {\sqrt {2mU/ \hbar ^ 2} x }}=[/itex] [itex]\lim_{x\rightarrow 0^{+} }{Ce^{- \sqrt {2mU/ \hbar^2} x }}= [/itex] [itex]\psi(0)=F[/itex]
then B=C=F.
using the normalization condition at [itex]-\infty<x<0 and 0<x<+\infty[/itex]
we have as B=C [itex]C\int_{-\infty}^{0^-} e^{\sqrt{2mU/ \hbar^2}x}dx +C\int_{0^+}^{+\infty}e^{-\sqrt{2mU/ \hbar^2}x}dx=1[/itex]
this will lead that C=[itex]\sqrt{2mU}/2\hbar[/itex]
but as V=infinity at x=0 then [itex]\psi=0[/itex]at x=0,so F=0.
and in this case U must be zero!!!as C=B=0.
which means that there is no bound states,so what goes wrong?
 
  • #6
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but as V=infinity at x=0 then [itex]\psi=0[/itex]at x=0,so F=0.
Why should the wave function be zero there?

Everything before that line was fine.
 
  • #7
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Why should the wave function be zero there?

Everything before that line was fine.
OK.
then B=C=F=[itex]\sqrt{2mU}/2\hbar[/itex]
and so for x<0 [itex]\psi=(\sqrt{2mU}/2\hbar )* e^{\sqrt{2mU/ \hbar^2}x}[/itex]
and so for x>0[itex]\psi=(\sqrt{2mU}/2\hbar)* e^{-\sqrt{2mU/ \hbar^2}x}[/itex]
and for x=0[itex]\psi=\sqrt{2mU}/2\hbar[/itex]
But this means that all negative energies are allowed, since their is no restriction on negative energy.Is this right?
 
  • #8
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The restriction comes from the change of the first derivative that happens at x=0.
 
  • #9
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The restriction comes from the change of the first derivative that happens at x=0.

for x<0 [itex]\left.\frac{d\psi}{dx}\right|_{x=0}=mU/ \hbar^2[/itex]
for x=0 [itex]\left.\frac{d\psi}{dx}\right|_{x=0}=0[/itex]
for x>0 [itex]\left.\frac{d\psi}{dx}\right|_{x=0}=-mU/ \hbar^2[/itex]
The change of the first derivative at x=0 is [itex]-mU/ \hbar^2[/itex]
Now as I calculated the change of slope, I will equate to [itex]\int_{-ε}^{ε} (\frac {d^2\psi}{dx^2})dx[/itex]
[itex]\int_{-ε}^{ε} (\frac {d^2\psi}{dx^2})dx=-2mU/ \hbar^2 e^{-\sqrt{2mU/ \hbar^2}ε} +2mU/ \hbar^2[/itex]
but I didn't see that equating the change of slope to this integral leads to any restriction in energy?
what is the trick here?
 
  • #10
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for x=0 [itex]\left.\frac{d\psi}{dx}\right|_{x=0}=0[/itex]
There is no derivative if your function has a single point only.

The delta potential at x=0 allows to calculate the change of the first derivative. Set it equal to ##\frac{2mU}{\hbar^2}## and you get the constraints on the energy.
 
  • #11
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mmmm ok.
I will use now the Schroedinger equation [itex]-\hbar^2 /2m\int_{-ε}^{ε}\psi' '(x)dx+\int_{-ε}^{ε}V(x)\psi(x)dx=E\int_{-ε}^{ε}\psi(x)dx[/itex]

Then [itex]-\hbar^2 /2m[\psi'_{x>0}(0)-\psi'_{x<0}(0)]-aV_{0}\psi(0)=0 (asE\int_{-ε}^{ε}\psi(x)dx=0)[/itex]
so [itex]-\hbar^2/2m[(-2mU/2\hbar^2)-2mU/2\hbar^2]=aV_{0}\psi(0)[/itex]
then [itex]U=aV_{0}\sqrt{2mU}/2\hbar[/itex]
[itex]U^2=a^2V_{0}^2mU/4\hbar^2[/itex]
then [itex]U(U-a^2V_{0}^2m/2\hbar^2)=0[/itex]
which means that[itex] U=0 or U=a^2V_{0}^2m/2\hbar^2[/itex]
and so the two allowed bound states are [itex]E=0[/itex] and[itex]E= - a^2V_{0}^2m/2\hbar^2[/itex]
 
  • #12
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U=0 leads to a contradiction elsewhere.
Do the units match for the final result? The approach looks fine.
 
  • #13
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Hint:solve Schrodinger's equation outside the potential for E<0
As the hint says, I must suppose E<0 out side the potential, so there will be one bound state which is [itex]E=-a^2V_{0}^2m/2\hbar^2[/itex].I think this is right because it matches what the question wants,as
Show that it admits a bound state of energy [itex]E=-ma^2V_{0}/2\hbar^{2}[/itex].Are there any other bound states?
By the way, it is [itex]E=-a^2V_{0}^2m/2\hbar^2[/itex] and not [itex]E=-a^2V_{0}m/2\hbar^2[/itex] I copied it wrong in the first post.
shall I suppose E<0 for simplicity or things would go wrong if I have dealt with E>0?as the hint restricted me to deal with E<0 out side the potential.
 
  • #14
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E>0 cannot be a bound state. There are states with positive energy, but they are not bound.
 
  • #16
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The wave function is bound and the size of the integration range goes to zero in the limit ##\epsilon \to 0##.
 
  • #17
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but why then the approach of letting the integration of
∫ε−ε(d2ψdx2)dx
equals to 2mU/h^2 led to fine answer?
shouldn't it also be equal to zero?
 
  • #19
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But why the second derivative is not bound?
It goes to zero for both sides.
 
  • #20
vela
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When you integrate the second derivative, you get
$$\int_{-\varepsilon}^\varepsilon \psi''(x)\,dx = \psi'(\varepsilon)-\psi'(-\varepsilon).$$ If ##\psi'## is continuous, then
$$\lim_{\varepsilon\to 0^+} \psi'(\varepsilon) = \lim_{\varepsilon\to 0^+} \psi'(-\varepsilon)$$ so that
$$\lim_{\varepsilon\to 0^+}\int_{-\varepsilon}^\varepsilon \psi''(x)\,dx = 0.$$ But that's not what you have in this problem. The delta function potential results in a discontinuity in ##\psi'## so
$$\lim_{\varepsilon\to 0^+} \psi'(\varepsilon) \ne \lim_{\varepsilon\to 0^+} \psi'(-\varepsilon)$$ and
$$\lim_{\varepsilon\to 0^+}\int_{-\varepsilon}^\varepsilon \psi''(x)\,dx \ne 0.$$
 
  • #21
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When you integrate the second derivative, you get
$$\int_{-\varepsilon}^\varepsilon \psi''(x)\,dx = \psi'(\varepsilon)-\psi'(-\varepsilon).$$ If ##\psi'## is continuous, then
$$\lim_{\varepsilon\to 0^+} \psi'(\varepsilon) = \lim_{\varepsilon\to 0^+} \psi'(-\varepsilon)$$ so that
$$\lim_{\varepsilon\to 0^+}\int_{-\varepsilon}^\varepsilon \psi''(x)\,dx = 0.$$ But that's not what you have in this problem. The delta function potential results in a discontinuity in ##\psi'## so
$$\lim_{\varepsilon\to 0^+} \psi'(\varepsilon) \ne \lim_{\varepsilon\to 0^+} \psi'(-\varepsilon)$$ and
$$\lim_{\varepsilon\to 0^+}\int_{-\varepsilon}^\varepsilon \psi''(x)\,dx \ne 0.$$
Yes, this makes it zero. But it is still bound doesn't it?
Besides the antiderivative of the wave function is not continuous at zero but the integral is equal to zero.
 
  • #22
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The second derivative diverges as ##\epsilon \to 0##. It is not bound (as function of epsilon).
 
  • #23
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The second derivative diverges as ##\epsilon \to 0##. It is not bound (as function of epsilon).
but for x<0 ##\psi(x)=\frac {\sqrt {2mU}} {2\hbar}*e^{\frac {\sqrt{2mU}} {\hbar}x}##
for x>0##\psi(x)=\frac {\sqrt {2mU}} {2\hbar}*e^{-\frac {\sqrt{2mU}} {\hbar}x}##

The second derivative of the wave function will converge to##(1/2) \frac {\hbar} {\sqrt {2mU}}## when ##\epsilon \to 0## in both cases x>0 and x<0.
and when ##x \to \infty## or ##-\infty## the second derivative of the wave function will converge to zero.
 
  • #25
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So I conclude that a wave function to be bound it is a must to be continuous, and once it is bound the definite integral of it from ##-\epsilon \to \epsilon## will approach zero if epsilon approaches zero(even if its antiderivative is not continuous like the case of this problem) .
 

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