Separable Hamiltonian for central potential

In summary: I'm not understanding you, I'm trying to prove that the Hamiltonian is separable, before separate the eigenfunction. Like in the 3D harmonic oscillator where the total Hamiltonian is equal to the sum of three Hamiltonians that commute with each other.The idea is to find three compatible observables, whose corresponding self-adjoint operators define a complete orthonormal system. If you want energy eigenstates, one of these operators is the Hamiltonian. Then, because your potential is a central potential, the Hamiltonian is symmetric under rotations and thus the angular momentum commutes with it. So the three compaible observables in this case are ##\hat{H}##, ##\hat{\vec
  • #1
Salmone
101
13
In a central potential problem we have for the Hamiltonian the expression: ##H=\frac{p^2}{2m}+V(r)## and we use to solve problems like this noting that the Hamiltonian is separable, by separable I mean that we can express the Hamiltonian as the sum of multiple parts each one commuting with the other, so for example: ##H=H_1+H_2+H_3## and ##[H_1,H_2]=0## ##[H_2,H_3]=0## ##[H_1,H_3]=0##. In the case of the central potential, how can we separate the Hamiltonian? In which Hamiltonians does it separate?
 
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  • #2
Perhaps in a part with angles in it and a part with ##r## in it ? :smile:

##\ ##
 
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  • #3
I guess but I don't know how to do it.
 
  • #4
Try spherical coordinates
 
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  • #5
What I get is ##H=\frac{-\hbar^2}{2m}(\frac{\partial^2 }{\partial r^2}+\frac{2}{r}\frac{\partial }{\partial r})+\frac{L^2}{2mr^2}+V(r)## now the angular dependence is in ##L^2## but what I also have an ##r^2## there, this is my problem.
 
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  • #6
What about the angular part ?
##\ ##
 
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  • #7
##L## comes later. First try ##\Psi(r,\theta,\phi) = R(r)P(\theta)F(\phi) ##

##\ ##
 
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  • #8
I'm not understanding you, I'm trying to prove that the Hamiltonian is separable, before separate the eigenfunction. Like in the 3D harmonic oscillator where the total Hamiltonian is equal to the sum of three Hamiltonians that commute with each other.
 
  • #9
The idea is to find three compatible observables, whose corresponding self-adjoint operators define a complete orthonormal system. If you want energy eigenstates, one of these operators is the Hamiltonian. Then, because your potential is a central potential, the Hamiltonian is symmetric under rotations and thus the angular momentum commutes with it. So the three compaible observables in this case are ##\hat{H}##, ##\hat{\vec{L}}^2##, and ##\hat{L}_3##. It's further clear that the right variables for the eigenfunctions are spherical coordinates, and that the eigen functions can be written as a product as said in #7.
 
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  • #10
Why the eigenfunctions can be written as a product? I know that in general if I have two commuting operators their simultaneous eigenfunctions are not the product. ##H,L^2,L_z## commute so we can find simultaneous eigenfunctions, what's next? How do we obtain ##\psi(r,\theta,\phi)=R(r)\Theta(\theta)\Phi(\phi)##?
 
  • #11
Can you at least try ?
Write out the SE in spherical coordinates ...
(hint: Laplacian)
 
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  • #12
Ok. ##[\frac{\partial^2 }{\partial r^2}+\frac{2}{r}\frac{\partial }{\partial r}-\frac{L^2}{\hbar^2r^2}-\frac{2m}{\hbar^2}V(r)+\frac{2mE}{\hbar^2}]\psi(r.\theta,\phi)=0##
 
  • #13
What about ##L^2## in spherical coords?
 
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  • #14
##[\frac{-\hbar^2}{2m}(\frac{\partial^2 }{\partial r^2}+\frac{2}{r}\frac{\partial }{\partial r}+\frac{1}{r^2}(\frac{\partial^2}{\partial \theta^2}+\frac{cos(\theta)}{sin(\theta)}\frac{\partial }{\partial \theta}+\frac{1}{sin^2(\theta)}\frac{\partial^2 }{\partial \phi^2}))-\frac{2m}{\hbar^2}V(r)+\frac{2mE}{\hbar^2}]\psi(r,\theta,\phi)=0##

But I have an ##r^2## under the angular part.
 
  • #15
Or...spelled out as in #7 it looks llike this :smile:
 
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  • #16
I didn't mean this, what I mean was: if we are using the method of "separable Hamiltonians" don't we have to prove that the Hamiltonian, not the SE, splits in a sum of terms commuting with each other? Like I've said before, if we consider the 3d isotrope harmonic oscillator what we get is ##H=H_x+H_y+H_z## and ##[H_x,H_y]=0 [H_y,H_z]=0 [H_x,H_z]=0## and this lead to a total eigenfunction to be ##\psi(x,y,z)=\psi(x)\psi(y)\psi(z)## without calculating the TISE.
 
  • #17
Salmone said:
I guess but I don't know how to do it.

 
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  • #18
Salmone said:
I didn't mean this, what I mean was: if we are using the method of "separable Hamiltonians" don't we have to prove that the Hamiltonian, not the SE, splits in a sum of terms commuting with each other?
That is too restrictive. The thing to note is that if you have a linear differential operator on the form
$$
\hat L = \hat L_x + f(x) \hat \Lambda
$$
where ##\hat \Lambda## is a Sturm-Liouville operator in some variable ##y## and ##\hat L_x## is a linear differential operator, then you can expand any function ##g(x,y)## in the eigenfunctions ##u_n(y)## of ##\hat\Lambda##:
$$
g(x,y) = \sum_n g_n(x) u_n(y).
$$
For the term ##g_n(x) u_n(y)##, ##\hat L## becomes
$$
\hat L g_n(x)u_n(y) = u_n(y) [\hat L_x + \lambda_n f(x)] g_n(x).
$$
If ##\hat L_x + \lambda_n f(x)## is a Sturm-Liouville operator, then any function ##g_n(x)## can be expanded in its eigenfunctions ##v_{nm}(x)## as
$$
g_n(x) = \sum_m G_{nm} v_{nm}(x).
$$
It is clear that the products ##\psi_{nm}(x,y) = v_{nm}(x) u_n(y)## are eigenfunctions of the full differential operator ##\hat L## and that any function ##g(x,y)## can be expressed as a linear combination
$$
g(x,y) = \sum_{n,m} G_{nm} \psi_{nm}(x,y).
$$

Note that the operator ##\hat L_x + \lambda_n f(x)## is not generally independent of the eigenfunctions of the ##\hat\Lambda##, but different ##n## lead to different operators in the ##x## direction through different ##\lambda_n##.
 
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  • #19
It's pretty easy to see, why in spherical coordinates the common eigenfunctions of ##\hat{H}##, ##\hat{\vec{L}}^2##, and ##\hat{L}_z## separate (provided the potential is a central potential).

In spherical coordinates,
$$\hat{L}_z \psi(r,\vartheta,\varphi)=-\mathrm{i} \hbar \partial_{\varphi} \psi(r,\vartheta,\varphi).$$
Now let's call the common eigenfunctions ##u_{nlm}##. Then you have
$$\hat{L}_z u_{nlm} =-\mathrm{i} \hbar \partial_{\varphi} u_{nlm} =m \hbar u_{nlm},$$
which is solved uniquely by
$$u_{nlm}=v_{nl}(r,\vartheta) \exp(\mathrm{i} m \varphi).$$
The general theory for the orbital (!) angular momentum tells you that ##m \in \mathbb{Z}##.
Further you can write (with ##\mu## denoting the mass of the particle, distinguishing it from the "magnetic quantum number", ##m##)
$$\hat{H} \psi=-\frac{\hbar^2}{2 \mu} \partial_r^2 (r \psi) + \frac{1}{2\mu r^2} \hat{\vec{L}}^2 \psi + V(r) \psi.$$
Applying this to ##u_{nlm}## you get
$$\hat{H} u_{nlm} = -\frac{\hbar^2}{2\mu} \partial_r^2 (r u) + \frac{\hbar^2 l(l+1)}{2 \mu r^2} u + V(r) \psi.$$
Further ##\hat{\vec{L}}^2## is a differential operator acting only on the angles ##\vartheta## and ##\varphi##. Thus you have
$$u_{nlm}=R_{nl} (r) \Theta_{lm}(\vartheta) \exp(\mathrm{i} m \varphi).$$
Of course at the end it results
$$\Theta_{lm}(\vartheta) \exp(\mathrm{i} m \varphi) = \text{Y}_{lm}(\vartheta,\varphi),$$
with the ##\text{Y}_{lm}## being the "spherical harmonics".

The "radial wave function" thus fulfills
$$-\frac{\hbar^2}{2 \mu} \partial_r^2 (r R_{nl}) + \frac{\hbar^2 l(l+1)}{2 \mu r^2} R_{nl} + V(r) R_{nl} = E R_{nl},$$
where ##E## is the energy eigenvalue, depending (in general) on the "main quantum number" ##n## and ##l##.

Due to the dynamical SO(4) symmetry of the Kepler problem (additional conserved quantity being the Laplace-Runge-Lenz vector) for the hydrogen atom (in its simplest form neglecting relativistic and spin effects) ##E## only depends on ##n##.
 
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  • #20
@vanhees71 I have two more questions:
1. the solutions of the eigenvalues equation for ##\hat{L_z}## is ##u_{nlm}=v_{nl}(r,\vartheta) \exp(\mathrm{i} m \varphi)## because we get a costant times the exponential and since ##L_z## depends only on ##\phi## the constant is, in general, a function of ##r,\theta##?
2. again sorry but I can't understand it, we don't need that the Hamiltonian splits in a sum of terms commutating with each other in order to apply "separation of variables" method, if that happens we jump to the conclusion that the total eigenfunction is factorized but if it doesn't happen maybe we can still use separation of variables method?
 
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  • #21
ad 1: Yes.

ad 2: The important point is that the three self-adjoint operators you choose are mutually commuting, because then (and only) then you can find a complete set of common orthonormal eigenvectors. This alone, however doesn't make the eigenfunctions in the position representation factorizing in functions each depending on only one coordinate. For this you must choose the appropriate coordinates (and its is not a priori clear that such a set of coordinates exists).

Here you are, however, dealing with a problem of high symmetry, and the description of the observables by self-adjoint operators in quantum mechanics are all understood by using Noether's theorem as known from classical mechanics. If you haven't heard about this, I highly recommend to have a look at it, because it helps tremendously to understand the "rules" behind quantum theory. In this case you have symmetry under time-translations, which implies according to Noether that the energy is conserved. The condition for that is simply that the Hamiltonian does not depend explicitly on time. Further you have symmetry under arbitrary rotations around the origin, because the potential is a central potential, i.e., it depends only on ##r=|\vec{r}|##. This strongly suggests to use spherical coordinates to describe the position vector and thus the wave functions as functions of these coordinates, ##(r,\vartheta,\varphi)##.

Obviously we can choose the three operators ##\hat{H}##, ##\hat{\vec{L}}^2##, and ##\hat{L}_z## as three compatible (i.e., commuting) complete set of operators and seek for their common eigenvectors to find a complete orthonormal set of energy eigenstates.

Next we have to express the operators in terms of the spherical coordinates. After some algebra you get
$$\hat{H} \psi =-\frac{\hbar^2}{2 \mu r} \partial_r^2 (r \psi) + \frac{1}{2 \mu r^2} \hat{\vec{L}}^2 \psi + V(r) \psi,$$
$$\hat{\vec{L}}^2 \psi = -\hbar^2 \left (\frac{1}{\sin \vartheta} \partial_{\vartheta}(\sin \vartheta \partial_{\vartheta} \psi) + \frac{1}{\sin^2 \vartheta} \partial_{\varphi}^2 \psi \right).$$
$$\hat{L}_3 \psi=-\mathrm{i} \hbar \partial_{\varphi} \psi.$$
Now take the common eigenfunctions for our three operators:
$$\hat{H} u_{nlm} = E_{nl} u_{nlm}, \quad \hat{\vec{L}}^2 u_{nlm} = \hbar^2 l(l+1) u_{nlm}, \quad \hat{L}_3 u_{nlm}=\hbar m u_{nlm}.$$
The last equation is immediately solved by
$$u_{nlm}(r,\vartheta,\varphi)=v_{nl}(r,\vartheta) \exp(\mathrm{i} m \varphi).$$
Now plug this into the 2nd eigenvalue equation
$$\frac{1}{\hbar^2} \hat{\vec{L}}^2 u_{nlm} = -\frac{1}{\sin \vartheta} \partial_{\vartheta} (\sin \vartheta \partial_{\vartheta} u_{nlm}) + \frac{m^2}{\sin^2 \vartheta} u_{nlm}=l(l+1) u_{nlm}.$$
Since the common factor ##\exp(\mathrm{i} m \varphi)## cancels you have the same equation for ##v_{nl}(r,\vartheta)##,
$$-\frac{1}{\sin \vartheta} \partial_{\vartheta} (\sin \vartheta \partial_{\vartheta} v_{nl}) + \frac{m^2}{\sin^2 \vartheta} v_{nl}=l(l+1) v_{nl}.$$
Obviously you can fulfill this equation by another separation ansatz,
$$v_{nl}(r,\vartheta)=R_{nl}(r) \Theta_{nl}(\vartheta),$$
with the common factor, ##R_{nl}(r)##, cancelling on both sides. So for ##\hat{L}_z## and ##\hat{\vec{L}}^2## the separation ansatz was already successful, but now it's also easy to see that it is also working for the final equation, the energy eigenvalue equation,
$$-\frac{\hbar^2}{2 \mu r} \partial_{r}^2 [r^2 R_{nl}(r)] + \frac{\hbar^2 l(l+1)}{2 \mu r^2} R_{nl} + V(r) R_{nl}=E_{nl} R_{nl}.$$
This finishes the proof that indeed the time-independent Schrödinger equation separates in spherical coordinates.
 
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  • #22
@vanhees71 Thank you for the answer, I know Noether's theorem. Maybe what make me confused is that I know this: if there is an Hamiltonian which can be written as the sum of different Hamiltonians and each term of the sum commutes with the others it's possible to demonstrate that the total eigenvalues are the sum of single eigenvalues and the total eigenfunction is equal to the product of the single eigenfunctions (see 3d harmonic oscillator in cartesian coordinates). In this problem the total eigenfunction is equal to a product of eigenfunctions so I thought we were using the same method but, after reading your answer, I think we are using something little different in fact, for example, when we search the eigenvalues of Hydrogen atom (central Hamiltonian problem) we find 'em imposing the wavefunction to be ##L-squared## we do not write the Hamiltonian as the sum of single Hamiltonians and then add up the eigenvalues of single Hamiltonians and again with the 3D harmonic oscillator if we solve it in cartesian coordinates we immediately come up with eigenvalues equal to the sum of single eigenvalues and eigenfunction as product of single eigenfunctions BUT if we study the 3d oscillator in spherical coordinates we solve the entire radial TISE. Does this makes sense?
 
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Related to Separable Hamiltonian for central potential

1. What is a separable Hamiltonian for central potential?

A separable Hamiltonian for central potential is a mathematical construct used in quantum mechanics to describe the motion of a particle in a central potential. It is separable because it can be broken down into two independent parts: the radial part and the angular part.

2. What is the significance of a separable Hamiltonian for central potential?

The significance of a separable Hamiltonian for central potential is that it allows for the simplification of the Schrödinger equation, making it easier to solve and analyze. It also provides insight into the symmetries and conserved quantities of the system.

3. How is a separable Hamiltonian for central potential derived?

A separable Hamiltonian for central potential is derived by using the separation of variables technique, where the wavefunction is expressed as a product of two functions: one depending on the radial coordinate and the other on the angular coordinates. These functions are then substituted into the Schrödinger equation, resulting in two separate equations for the radial and angular parts.

4. Can a separable Hamiltonian for central potential be used for any central potential?

Yes, a separable Hamiltonian for central potential can be used for any central potential, as long as it is spherically symmetric. This means that the potential energy only depends on the distance from the center and not on the direction.

5. What are the applications of a separable Hamiltonian for central potential?

A separable Hamiltonian for central potential has various applications in quantum mechanics, such as in the study of atomic and molecular systems, nuclear physics, and solid state physics. It is also used in the development of analytical and numerical techniques for solving the Schrödinger equation.

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