# Schrödinger equation for 2 particles

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1. Jun 21, 2015

### olgerm

U(x,y,z,t)*ψ(x,y,z,t)-(ħ/(2*m))*(d2ψ(x,y,z,t)/dx2+d2ψ(x,y,z,t)/dy2+d2ψ(x,y,z,t)/dz2)=ħ*i*dψ(x,y,z,t)/dt
qproton=-qe

Schrödinger equation for electron in hydrogen atom (if we consider proton as point charge which is moving at a constant speed vproton=(vp;x;vp;y;vp;z).) is:
Ue(x,y,z,t)*ψe(x,y,z,t)-(ħ/(2*me))*(d2ψe(x,y,z,t)/dx2+d2ψe(x,y,z,t)/dy2+d2ψe(x,y,z,t)/dz2)=ħ*i*dψe(x,y,z,t)/dt
Ue(x,y,z,t)=qe*qproton/(rdistance from electron to proton0*4*π)=-qe2/(((xp+t*vp;x-xe-t*ve;x)2+(yp+t*vp;y-ye-t*ve;y)2+(zp+t*vp;z-ze-t*ve;z)2)(1/2)0*4*π)
⇒-qe2/(((xp+t*vp;x-xe-t*ve;x)2+(yp+t*vp;y-ye-t*ve;y)2+(zp+t*vp;z-ze-t*ve;z)2)(1/2)0*4*π)*ψ(x,y,z,t)-(ħ/(2*me))*(d2ψ(x,y,z,t)/dx2+d2ψ(x,y,z,t)/dy2+d2ψ(x,y,z,t)/dz2)=ħ*i*dψ(x,y,z,t)/dt
by solving it we can get electron wave function ψe(x,y,z,t) in hydrogen atom. Am I right?

But if we consider proton as wave like we did with electron:
Ue(x,y,z,t)*ψe(x,y,z,t)-(ħ/(2*me))*(d2ψe(x,y,z,t)/dx2+d2ψe(x,y,z,t)/dy2+d2ψe(x,y,z,t)/dz2)=ħ*i*dψe(x,y,z,t)/dt
Up(x,y,z,t)*ψp(x,y,z,t)-(ħ/(2*mp))*(d2ψp(x,y,z,t)/dx2+d2ψp(x,y,z,t)/dy2+d2ψp(x,y,z,t)/dz2)=ħ*i*dψp(x,y,z,t)/dt
its obvious that their potential energys are equal to each other[Ue(x,y,z,t)=Up(x,y,z,t)], but to what potential energy function equals [U(x,y,z,t)=?]?

I know it is good approximation to consider proton as point charge .I am asking this to understand Schrödinger equation better.

Last edited: Jun 21, 2015
2. Jun 21, 2015

### blue_leaf77

I don't quite understand with some of the notations you used there, seems like U is the electrostatic potential between proton and electron. In that case I don't think it can be written as a function of time. Second the derivative of the wavefunction on the left side should be of second order, you probably need to check again the form of a general Schroedinger equation. Finally it seems like to me you want to consider the massive proton as mobile along with the electron, it's a two body problem. The common way to solve this is to change the coordinate so that the variables become the relative distance between electron-proton and the coordinate of the center of mass.

Last edited: Jun 21, 2015
3. Jun 21, 2015

### Khashishi

For a two body problem like the hydrogen atom, you can reduce it into a one body problem by solving it in the center of mass coordinates using the reduced mass. But maybe you know that and you are trying this method as an exercise.

You need to write a single wavefunction as a function of both the electron and proton positions $\psi(\mathbf{r}_e,\mathbf{r}_p,t)$. The potential energy also is a function of both coordinates $V(\mathbf{r}_e,\mathbf{r}_p,t)$.

4. Jun 21, 2015

### olgerm

Yes, I want to solve this as 2-"body" problem.

Should Uproton be function of both electron and proton coordinates [Up(xp,yp,zp,xe,ye,ze,t)] or just proton coordinates [Up(xp,yp,zp,t)],if I am solving this as to 2-"body problem"?

what Upotential energy of proton(...) equals to?

5. Jun 21, 2015

### blue_leaf77

Remember, the electrostatic potential energy is the work needed to bring a charged body from infinity to a distance r from another charged body fixed in some place. Would this work be different whether what you bring is the electron while the proton is fixed in place or you bring the proton instead and the electron fixed?
Just warning: it seems like you tend to want to write the potential felt by proton and that felt by electron separately. Coulomb interaction is a mutual interaction, if there is only one charged particle in space, it won't feel any Coulomb field.

6. Jun 21, 2015

### olgerm

It is work needed to bring a charged body from a distance r to infinity.
I do remember: ∫rdr*(q1*q2/(r2*4*π*ε0))=q1*q2/(r*4*π*ε0)
If these are point charges then r=((xe-xp)2+(ye-yp)2+(ze-zp)2)(1/2).
But what r equals to if both are wave functions?

Last edited: Jun 21, 2015
7. Jun 21, 2015

### blue_leaf77

It equals to the expression you gave there. Indeed, in QM we do not say that a particle like electrons or protons to be point like, in fact we have never known how they exactly look like otherwise we violate the Heisenberg uncertainty principle. The coordinates of proton and electron appear as variables in Schroedinger equation, so you don't need to worry too much about them because the wavefunction to be solved should be a function of these coordinates. In fact, a complete specification of x,y,z and t will only give you the probability to find the electron at coordinates (x,y,z) at time t.

8. Jun 21, 2015

### olgerm

Ue(x,y,z,t)*ψe(x,y,z,t)-(ħ/(2*me))*(d2ψe(x,y,z,t)/dx2+d2ψe(x,y,z,t)/dy2+d2ψe(x,y,z,t)/dz2)=ħ*i*dψe(x,y,z,t)/dt
Up(x,y,z,t)*ψp(x,y,z,t)-(ħ/(2*mp))*(d2ψp(x,y,z,t)/dx2+d2ψp(x,y,z,t)/dy2+d2ψp(x,y,z,t)/dz2)=ħ*i*dψp(x,y,z,t)/dt
Ue=Up=q1*q2/(((xe-xp)2+(ye-yp)2+(ze-zp)2)(1/2)*4*π*ε0)

q1*q2/(((xe-xp)2+(ye-yp)2+(ze-zp)2)(1/2)*4*π*ε0)*ψe(xe,ye,ze,t)-(ħ/(2*me))*(d2ψe(xe,ye,ze,t)/dx2+d2ψe(xe,ye,ze,t)/dy2+d2ψe(xe,ye,ze,t)/dz2)=ħ*i*dψe(xe,ye,ze,t)/dt
q1*q2/(((xe-xp)2+(ye-yp)2+(ze-zp)2)(1/2)*4*π*ε0)*ψp(xp,yp,zp,t)-(ħ/(2*mp))*(d2ψp(xp,yp,zp,t)/dx2+d2ψp(xp,yp,zp,t)/dy2+d2ψp(xp,yp,zp,t)/dz2)=ħ*i*dψp(xp,yp,zp,t)/dt

Is it correct?
To get wave function I just had to solve this functional equation with respect ψe(xe,ye,ze,t) and ψp(xp,yp,zp,t)?

9. Jun 21, 2015

### blue_leaf77

The wavefunction cannot be calculated that way. First of all the correct Schroedinger equation for proton-electron system is
$$-\frac{\hbar^2}{2m_p} \nabla_p^2 \psi(\mathbf{r_p,r_e},t) -\frac{\hbar^2}{2m_e} \nabla_e^2 \psi(\mathbf{r_p,r_e},t) - \frac{e^2}{r_{12}} \psi(\mathbf{r_p,r_e},t) = i\hbar \frac{\partial} {\partial t} \psi(\mathbf{r_p,r_e},t)$$
where $\mathbf{r_p}$ and $\mathbf{r_e}$ are the coordinates of the proton and electron respectively. The presence of subscript "e" in the nabla operator like $\nabla_e$ indicates that the differentiation is taken w.r.t to the electron coordinates, likewise for "p" subscript for proton. $r_{12} = |\mathbf{r_e} - \mathbf{r_p}|$.
That kind of differential equation cannot be separated in terms of $\mathbf{r_e}$ and $\mathbf{r_p}$ due to the presence of the mutual Coulomb interaction. However that equation can be separated if you change the variable to the relative distance between proton and electron $\mathbf{r} = \mathbf{r_e} - \mathbf{r_p}$ and the coordinate of the center of mass of the electron-proton system $\mathbf{R} = \frac{m_e \mathbf{r_e} + m_p \mathbf{r_p}}{m_e + m_p}$, so that in the end the wavefunction is separated into the relative motion part and center of mass motion part.

10. Jun 22, 2015

### olgerm

What you mean cant be calculated that way? Does it give wrong answer if I solve it for Ψ(x,y,z,t)?
Or that, it is just very hard to solve such functional equation?

Is it analogues to classical 2-body problem where we can write

$x_1(t) = \iint \frac {x_1(t)*(m_1*m_2*G+q_1*q_2/(4*π*ε_0))}{((x_1(t)-x_2(t))^2+(y_1(t)-y_2(t))^2+(z_1(t)-z_2(t))^2)^{3/2}*m_1} \, dt^2$

$y_1(t) = \iint \frac {y_1(t)*(m_1*m_2*G+q_1*q_2/(4*π*ε_0))}{((x_1(t)-x_2(t))^2+(y_1(t)-y_2(t))^2+(z_1(t)-z_2(t))^2)^{3/2}*m_1} \, dt^2$

$z_1(t) = \iint \frac {z_1(t)*(m_1*m_2*G+q_1*q_2/(4*π*ε_0))}{((x_1(t)-x_2(t))^2+(y_1(t)-y_2(t))^2+(z_1(t)-z_2(t))^2)^{3/2}*m_1} \, dt^2$

$x_2(t) = \iint \frac {x_1(t)*(m_1*m_2*G+q_1*q_2/(4*π*ε_0))}{((x_1(t)-x_2(t))^2+(y_1(t)-y_2(t))^2+(z_1(t)-z_2(t))^2)^{3/2}*m_2} \, dt^2$

$y_2(t) = \iint \frac {y_1(t)*(m_1*m_2*G+q_1*q_2/(4*π*ε_0))}{((x_1(t)-x_2(t))^2+(y_1(t)-y_2(t))^2+(z_1(t)-z_2(t))^2)^{3/2}*m_2} \, dt^2$

$z_2(t) = \iint \frac {z_1(t)*(m_1*m_2*G+q_1*q_2/(4*π*ε_0))}{((x_1(t)-x_2(t))^2+(y_1(t)-y_2(t))^2+(z_1(t)-z_2(t))^2)^{3/2}*m_2} \, dt^2$

or take canter of mass to (0;0;0) point and solve 2-body problem by using angular momentum and energy conservation law?
In classical 2-body problem both ways give same result ,but first one is just very hard to solve. Is it the same in QM 2-"body" problem where we can write
or reduce it into 1-"body" problem by using center of mass?
And both give the same solution just the first one is very hard to solve?

Last edited: Jun 22, 2015
11. Jun 22, 2015

### blue_leaf77

The basic Hamiltonian is this
and if this equation
were to be true, this means you have somehow decouple the above basic Schroedinger equation into two. But how will you do this, the former just cannot lead to the second equations. What you did is just making two copies of Hamiltonian and letting it act to the wavefunctions of electron and of proton, assuming they have been separated from each other.
The difference between QM and CM is that in QM quantities such as position and momentum (time excluded) are operators (in addition to the introduction of wavefunction), in contrast to CM where they are just ordinary quantity. In CM I know you can write those equations either from the Lagrangian or Newton's second law, but look what is the consequence of solving such equations. Upon solving them you will obviously get x(t), y(t), z(t) and so on deterministically at each time, this is not supported in QM as you may know that the variance of position and momentum is not independent.
If you want to look for a closer analogy with CM, you may want to look up Heisenberg picture of QM. For one body problem where no time-dependent potential present, you can find equation like
$$m\frac{d}{dt} \langle x \rangle = -\langle \nabla V(\mathbf{r}) \rangle$$.
that's for one moving body problem, for two bodies I guess you can also get similar expression to those found in CM.

12. Jun 22, 2015

### olgerm

Solution of QM 2-"body" problem is one wave function Ψ(xe,ye,ze,xp,yp,zp,t)?
And probability of electron being at coordinates(xe,ye,ze) and proton on coordinates (xp,yp,zp) at time t is p=|Ψ(xe,ye,ze,xp,yp,zp,t)|2?
Is always QM N-"body" problem solution one wave function which arguments are coordinates of all particles and time?

$U_{System Potential Energy}(r_1,r_2,r_3,...,r_n,t)-\sum_{n=1}^N((\frac{d^2Ψ(r_1,r_2,r_3,...,r_n,t)}{dx_n^2}+\frac{d^2Ψ(r_1,r_2,r_3,...,r_n,t)}{dy_n^2}+\frac{d^2Ψ(r_1,r_2,r_3,...,r_n,t)}{dz_n^2})*\frac{ħ^2}{m_n})=i*ħ \frac{dΨ(r_1,r_2,r_3,...,r_n,t)}{dt}$

If it is true then my first try to solve it as 2 wave functions: one for proton and one electron makes no sense.

Last edited: Jun 22, 2015
13. Jun 22, 2015

### blue_leaf77

Yes, because a single system in QM is described by single Schroedinger equation, therefore there can only be one solution/one wavefunction to this equation. If there are more than one particle, we simply need to add the corresponding term into the Hamiltonian and the wavefunction should obviously be a function of all degree of freedom in the system.

Yes.

That's the so called Schroedinger equation. Everything in Schroedinger picture QM must depart from the Schroedinger equation, that's your firm starting point. Whether you can latter separate the Schroedinger equation in question or not, that's up to the complexity of the maths.

Last edited: Jun 22, 2015