Find the eigenstates and eigenvalues of the Hamiltonian.

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I don't think you need to do that as long as you have the <0 in the final line.)I guess I'm not sure what you are looking for. You already have \langle H(a)\rangle = \langle \frac{p^2}{2m} \rangle + \langle V \rangle. The only other thing you need to know is that \langle p^2 \rangle is a constant, independent of a. This is because \langle p^2 \rangle is an expectation value of an observable, and observables are time-independent quantities. Therefore, taking the derivative of \langle H \rangle with respect to a is equivalent to taking the derivative of \langle
  • #1
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Homework Statement


1.Consider a one dimensional attraction potential V(x) s.t V(x)<0 for each x.
Using the variational principle, show that such a potential has at least one bound state.
Hint: use a gaussian in x as a trial functio.

2. A particle with charge e and mass m is confined to move on the circumference of a circle of radius r. Find the eigenstates and eigenvalues of the Hamiltonian.



The Attempt at a Solution


1. Now sure where to start, I have calculated [tex]<H(a)>=<\psi|H|\psi>/<\psi|\psi>[/tex] where [tex]\psi(x)=exp(-ax^2)[/tex], and I know that it's bigger than E0, usually in order for us to find a strict upper bound for E0, we need to differentiate <H(a)> wrt to a and find its minimum, but here V(x) is not given explicitly so I guess I need to show somehow that <H(a)> is bounded, I've showed that it's bounded by [tex]\frac{a\hbar^2}{m}[/tex], is that enough to show that for this potential it has at least one bounded eigenstate.

2. I have no idea, I mean the potential is: [tex]V(r)=-mw^2r^2/2[/tex], how do I proceed from here?

Thanks in advance.
QM rulessssssssssssssss! :-)
 
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  • #2
For #2 you should start with Schrodinger's equation:

[tex]
-\frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2}+V(x)\Psi(x)=E\Psi(x)
[/tex]

And work from there.
 
  • #3
I know that, is the solution different than the harmonic oscillator equation?
 
  • #4
For #2, why are you saying that the potential is [itex]V(r) = -m\omega^2 r^2/2[/itex]? That doesn't appear to be part of the problem. In fact, there isn't any potential at all mentioned in the problem, which means that [itex]V = 0[/itex] everywhere. Unless you've left something out of the problem statement? (actually, I guess it wouldn't even matter for a purely radial potential)
 
  • #5
Well, if a particle is confined to move on a circular path then obviously a centriptal force acts upon it, which is m w^2 r, integrate this wrt r and multiply this by a minus sign, and you get the potential I wrote.
 
  • #6
Um... no. You're still thinking classically. In quantum mechanics, when a particle is confined to move along a circular path, it literally has to be confined to that path - you have to assume that the entire space in which the particle is able to exist is nothing more than that circular path. You're dealing with a one-dimensional system with a single position coordinate, let's say [itex]s[/itex], using periodic boundary conditions with a period of [itex]2\pi r[/itex] (so that [itex]s[/itex] and [itex]s + 2\pi r[/itex] represent the same point for any [itex]s[/itex]). You can, if you like, analyze the system using polar coordinates, i.e. use [itex]\theta \equiv s/r[/itex] instead of [itex]s[/itex] as the coordinate.

If you would like to solve the Schrödinger equation using the potential you came up with, [itex]-m\omega^2 r^2/2[/itex], by all means go ahead and do it. In fact, if I may venture a guess as to what point you're at in your physics education, it's a problem you will probably wind up doing at some point in the future - the two-dimensional harmonic oscillator. But you will find that the particle is not confined to a circle; there will be some nonzero probability of finding it at any point in space.
 
  • #7
So what is the potential in this specific question? zero?
 
  • #8
BTW, any help on the first question is welcomed as well.
 
  • #9
MathematicalPhysicist said:
So what is the potential in this specific question? zero?
Yeah, like I said, if the problem doesn't mention a potential, that generally means [itex]V = 0[/itex] everywhere. Keep in mind that "everywhere" for this problem is just a circle, a one-dimensional entity.

By the way, if a problem mentions some standard system with a known potential like "harmonic oscillator" or "infinite square well", that's a backhanded way of telling you what the potential is, so be on the lookout for such non-mathematical ways to give you information. But I don't think that's the case here. (Unless you are really insistent on treating it as a two-dimensional system, then you could try [itex]V = 0[/itex] at radius [itex]r[/itex] and [itex]V = \infty[/itex] everywhere else... but that really complicates things. It is, however, how you'd construct this kind of system experimentally.)
 
  • #10
For 1, can't you break it up as this:

[tex]
\langle H(a)\rangle=\langle \frac{p^2}{2m}\rangle +\langle V\rangle
[/tex]

(you can neglect the denominator terms because the wave function should be normalized.) Now what condition makes a bound state exist?
 
  • #11
If the Energy is negative.

i.e I need to show that <H(a)><=0.
Now I guess because [tex]<H(0)>=\int V(x)<=0[/tex] all I need to show is that
d<H(a)>/da=0 for a=0, and d^2<H(0)>/da^2<0.

Thanks.
 
  • #12
Ok, I solved it, because <H(a)>>=E0 for any a>0, then for a=0 0>=<H(0)>>=E0, so there is at least one bound state mainly the ground state.
 
  • #13
MathematicalPhysicist said:
If the Energy is negative.

i.e I need to show that <H(a)><=0.
Now I guess because [tex]<H(0)>=\int V(x)<=0[/tex] all I need to show is that
d<H(a)>/da=0 for a=0, and d^2<H(0)>/da^2<0.

Thanks.

Not quite. [itex]\langle p^2\rangle[/itex] has a value, so you need to show that

[tex]
\int \Psi^2V(x)dx>-\frac{1}{2m}\langle p^2\rangle
[/tex]

also, I know I did copy that part of it, but why are you looking at H(a)?
 
  • #14
I am not sure how to explictly show the inequality you wrote, but I think I solved it, look at post #12 in this thread.
 
  • #15
MathematicalPhysicist said:
Ok, I solved it, because <H(a)>>=E0 for any a>0, then for a=0 0>=<H(0)>>=E0, so there is at least one bound state mainly the ground state.

While this is technically true, I'm not sure that this will satisfy the answer for your teacher. You are making the claim that

[tex]
\langle H\rangle=\frac{\langle p^2\rangle}{2m}+\langle V\rangle \sim E_0
[/tex]

which generally is the case, but you are not at all showing that it is bounded. The expectation value of [itex]p^2[/itex] should give you a positive number (which means, given just this there is no bound state). What you have to do next is show what I wrote in post #13:

[tex]
\langle H\rangle=\frac{\langle p^2\rangle}{2m}+\langle V\rangle<0
[/tex]

So that

[tex]
\langle V\rangle=\int\Psi^2V(x)\,dx>\frac{1}{2m}\langle p^2\rangle
[/tex]

(where in this line I've taken [itex]V[/itex] to be negative already so that the negatives cancel).
 
  • #16
I don't see how to show this...
<p^2> depends on a, and I don't know what this value is?
 
  • #17
I'm assuming by [itex]a[/itex] you mean the [itex]a[/itex] in [itex]\Psi=\exp[-ax^2][/itex]. If this is the case, then no, the momentum does not depend on [itex]a[/itex]--even if this were the case, with your setting [itex]a=0[/itex], you should see that exp[0]=1 and then you'd have to show that [itex]\langle V\rangle >1[/itex].

If you use the Heisenberg uncertainty principle, you should find

[tex]
\langle p^2\rangle=\langle (\Delta p)^2\rangle=\frac{\hbar^2}{2\langle(\Delta x)^2\rangle}\simeq \frac{\hbar^2}{2d_0^2}
[/tex]

where [itex]d_0[/itex] is some characteristic length (possibly the Bohr radius, but it doesn't really matter in this problem). What you need to show now is that the expectation value of the potential is greater than this number using the equation I gave in the previous post (so that when going back to the expectation value of the Hamiltonian, you get a negative number so that there are bound states).
 
  • #18
jdwood983 said:
I'm assuming by [itex]a[/itex] you mean the [itex]a[/itex] in [itex]\Psi=\exp[-ax^2][/itex]. If this is the case, then no, the momentum does not depend on [itex]a[/itex]--even if this were the case, with your setting [itex]a=0[/itex], you should see that exp[0]=1 and then you'd have to show that [itex]\langle V\rangle >1[/itex].


Actually, what I should say is I think you're looking at this problem the wrong way. You should not be choosing [itex]a[/itex] to satisfy the problem by minimization via differentiation (particularly since we're told the potential depends on [itex]x[/itex], not [itex]a[/itex] and the momentum also does not depend on [itex]a[/itex]). You need to be solving the expectation values (both [itex]\langle p^2\rangle[/itex] and [itex]\langle V\rangle[/itex] should be non-zero) so that the total energy is negative. Since you know what the momentum expectation value is, you can solve the potential expectation value by using the inequality I wrote a couple posts ago.
 
  • #19
if I write a trial function f(x)=exp(-x^2), I get that:
<H>=<V>+hbar^2/2m sqrt(pi/2)
I don't see why this should be negative?
 
  • #20
The condition for there to be a bound state is that the energy must be less than 0. Recall that since [itex]\langle H\rangle\simeq E_0[/itex], then by showing that

(1) [itex]\langle V\rangle<0[/itex]

and

(2) [itex]\left|\langle V\rangle\right|>\left|\langle p^2\rangle\right|[/itex]

you are showing that there is a bound state.
 
  • #21
I know all that, but V(x) is not given explicitly, so how can I show that [tex]|<V>|=|\int V(x)exp(-2x^2)dx|[/tex] is smaller than hbar^2/2m=<p^2/2m>?
All I know is that V(x) is negative for every x.
 
  • #22
More or less the only way you can solve this is by approximation. If we assume that there is a range in the potential, extending out to [itex]R[/itex] (we're centered at the origin in this argument) and if the characteristic length [itex]d_0[/itex] is much much greater than [itex]R[/itex], we can assume that the wavefunction is constant in this limit. Then,

[tex]
\langle V\rangle\simeq\left|\psi\right|^2\int\,dx V(x)
[/tex]

But since the normalized wavefunction is essentially constant and proportional to [itex]1/\sqrt{d_0}[/itex], then we can say

[tex]
\langle V\rangle\simeq\frac{1}{d_0}\int\,dx V(x)
[/tex]

In which case, any negative potential [itex]V(x)[/itex] with a [itex]1/d_0[/itex] coefficient will always be larger than a positive number with a [itex]1/d_0^2[/itex] coefficient. Thus, there will be a bound state for any potential that is negative for any [itex]x[/itex].

There is a bit of hand-waving involved in this result, but it is usually a 'good enough' approximation to satisfy the problem. You should also note that in considering this approximation for a 3D potential, we find [itex]\langle V\rangle\propto d_0^{-3}[/itex], so this potential cannot have a bound state as it will never be more negative than the positive [itex]d_0^{-2}[/itex] term.
 
  • #23
jdwood983 said:
I'm assuming by [itex]a[/itex] you mean the [itex]a[/itex] in [itex]\Psi=\exp[-ax^2][/itex]. If this is the case, then no, the momentum does not depend on [itex]a[/itex]--even if this were the case, with your setting [itex]a=0[/itex], you should see that exp[0]=1 and then you'd have to show that [itex]\langle V\rangle >1[/itex].
If by momentum, you are referring to the expectation value of momentum, then yes, it will depend on [itex]a[/itex].

Also, if [itex]a=0[/itex], then [itex]\spi[/itex] must be zero everywhere for an attractive potential (do you see why), so a=0 is not a very good choice.

If you use the Heisenberg uncertainty principle, you should find

[tex]
\langle p^2\rangle=\langle (\Delta p)^2\rangle=\frac{\hbar^2}{2\langle(\Delta x)^2\rangle}\simeq \frac{\hbar^2}{2d_0^2}
[/tex]

where [itex]d_0[/itex] is some characteristic length (possibly the Bohr radius, but it doesn't really matter in this problem). What you need to show now is that the expectation value of the potential is greater than this number using the equation I gave in the previous post (so that when going back to the expectation value of the Hamiltonian, you get a negative number so that there are bound states).

Why are you assuming that the potential has a characteristic length? Doesn't this mean that outside of some region (say [itex]x\gg d_0[/itex]) the potential approaches infinity?...How can this be possible for an attractive, finite potential?
 
Last edited:
  • #24
MathematicalPhysicist said:
if I write a trial function f(x)=exp(-x^2), I get that:
<H>=<V>+hbar^2/2m sqrt(pi/2)
I don't see why this should be negative?

There's no reason it has to be negative for [itex]a=1[/itex]...just assume [itex]a[/itex] can be any positive real number, then start by normalizing your wavefunction and then calculating [itex]\langle H\rangle [/itex] using the fact that is [itex]V(x)<0[/itex] for all [itex]x[/itex], then [itex]V(x)=-|V(x)|[/itex]...you might find the mean value theorem for integration useful here...
 
  • #25
gabbagabbahey said:
Why are you assuming that the potential has a characteristic length? Doesn't this mean that outside of some region (say [itex]x\gg d_0[/itex]) the potential approaches infinity?...How can this be possible for an attractive, finite potential?

I sort of figured that it'd be obvious that for a finite attractive potential, that outside of the characteristic length, the potential would be zero and not infinity.
 
  • #26
jdwood983 said:
More or less the only way you can solve this is by approximation. If we assume that there is a range in the potential, extending out to [itex]R[/itex] (we're centered at the origin in this argument) and if the characteristic length [itex]d_0[/itex] is much much greater than [itex]R[/itex], we can assume that the wavefunction is constant in this limit. Then,

[tex]
\langle V\rangle\simeq\left|\psi\right|^2\int\,dx V(x)
[/tex]

I really don't buy this argument at all- you appear to be making unnecessary assumptions about the potential...try to envision some extremely complicated attractive potential with peaks and valleys of various heights all over the place (obviously, the peaks will still be negative)...how exactly would one define a characteristic length for such a potential?
 
  • #27
gabbagabbahey said:
I really don't buy this argument at all- you appear to be making unnecessary assumptions about the potential...try to envision some extremely complicated attractive potential with peaks and valleys of various heights all over the place (obviously, the peaks will still be negative)...how exactly would one define a characteristic length for such a potential?

I, in fact, made it quite clear in that post that this was only an approximation, that there is a bit of hand-waving, and that it should be a 'good enough' solution to satisfy this problem. You do not have to buy it at all because of your want for making overly-complicated assumptions, but this is pretty much how I solved the problem in my QM class and got full credit for it.
 
  • #28
Ok, after normalization I get:
[tex]<H(a)>=\frac{a\hbar ^2}{2m}+V(x_0)[/tex] where I have used here mean integral value theorem for <V>.
which means that [tex]d<H(a)>/da=\frac{\hbar ^2}{2m}[/tex] which is positive which means that <H(a)> > <H(0)>.

Now as I said, <H(0)> is obviously negative, and because E0 (the ground state energy) is at most <H(a)> for every a>=0, it should be lesser than the minimum <H(0)> which is negative.

I must say, you have complicated matters too much.
 
  • #29
MathematicalPhysicist said:
Ok, after normalization I get:
[tex]<H(a)>=\frac{a\hbar ^2}{2m}+V(x_0)[/tex] where I have used here mean integral value theorem for <V>.

EDIT: Looks good to me:

[tex]\langle H(a)\rangle=\frac{a\hbar ^2}{2m}-\int_{-\infty}^{\infty}|V(x)|\psi(x)\psi^{*}(x)dx=\frac{a\hbar ^2}{2m}-|V(x_0)|\int_{-\infty}^{\infty}\psi(x)\psi^{*}(x)dx=\frac{a\hbar ^2}{2m}-|V(x_0)|[/tex]

for some [itex]x_0[/itex] in the interval [itex](-\infty,\infty)[/itex]

Now as I said, <H(0)> is obviously negative,

You need to be a little careful here...your wavefunction is actually zero everywhere for [itex]a=0[/itex], so [itex]\langle H(a=0)\rangle[/itex] is pretty meaningless...however, you can choose [itex]a[/itex] to be arbitrarily small and look at the sign of [itex]\langle H(a)\rangle[/itex] in the limit that [itex]a\to 0[/itex]

and because E0 (the ground state energy) is at most <H(a)> for every a>=0, it should be lesser than the minimum <H(0)> which is negative.

If you mean, "since [itex]\langle H(a)\rangle\geq E_0[/itex] and there exists an [itex]a[/itex] such that [itex]\langle H(a) \rangle<0[/itex], then there must be at least one state for which [itex]E_0<0[/itex], and therfor there exists at least one bound state", then yes, you are correct.
I must say, you have complicated matters too much.

Are you referring to me or jdwood?
 
Last edited:
  • #30
The wave function is defined as exp(-ax^2), so for a=0 it's 1.
Now, [tex]<H(0)>=\int V(x)dx[/tex] Now if I assume that this integral converges then it mustn't be bigger than zero, if it were zero then the integrand would equal almost everywhere to zero, but V(x)<0 for every x, so this integral is negative.
 
  • #31
MathematicalPhysicist said:
The wave function is defined as exp(-ax^2), so for a=0 it's 1.

No it isn't, you are forgetting about the normalization constant...which depends on [itex]a[/itex]:wink:

Now, [tex]<H(0)>=\int V(x)dx[/tex] Now if I assume that this integral converges then it mustn't be bigger than zero, if it were zero then the integrand would equal almost everywhere to zero, but V(x)<0 for every x, so this integral is negative.

Again, [itex]\langle H(0) \rangle[/itex] is undefined, but [itex]\lim_{a\to 0^{+}}\langle H(a)\rangle[/itex] is well defined.
 
  • #32
Ok, thanks.
 
  • #33
wait a minute, how do I calculate this limit?
[tex]<H(a)>=a\hbar^2/2m + \sqrt(2a/\pi)\int exp-2ax^2)dx[/tex]
it looks to me it converges to zero.
 
  • #34
MathematicalPhysicist said:
wait a minute, how do I calculate this limit?
[tex]<H(a)>=a\hbar^2/2m + \sqrt(2a/\pi)\int exp-2ax^2)dx[/tex]
it looks to me it converges to zero.

Shouldn't there be a V in the integrand?
 
  • #35
Yes, there should, and square root is over (2a/pi).

I believe this question is the easiest from the questions from my HW.
:-)
 

Related to Find the eigenstates and eigenvalues of the Hamiltonian.

1. What is the Hamiltonian operator?

The Hamiltonian operator is a mathematical representation of the total energy of a physical system. It is commonly used in quantum mechanics to describe the energy of a particle or system of particles.

2. What are eigenstates and eigenvalues?

Eigenstates are the states of a physical system that correspond to definite values of a physical quantity, such as energy. Eigenvalues are the values that correspond to these eigenstates.

3. How do you find the eigenstates and eigenvalues of the Hamiltonian?

To find the eigenstates and eigenvalues of the Hamiltonian, you need to solve the Schrödinger equation for the system. This involves using mathematical techniques such as diagonalization or perturbation theory.

4. Why is it important to find the eigenstates and eigenvalues of the Hamiltonian?

Knowing the eigenstates and eigenvalues of the Hamiltonian allows us to understand the energy levels and behavior of a physical system. This information is crucial in many applications, such as predicting the behavior of atoms and molecules.

5. Are the eigenstates and eigenvalues of the Hamiltonian unique?

Yes, the eigenstates and eigenvalues of the Hamiltonian are unique for a given physical system. However, different systems can have the same eigenstates and eigenvalues, which can lead to interesting connections and insights in quantum mechanics.

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