# I Wave function for a helium atom

#### olgerm

Gold Member
Can you say whether I understood these things correctly?

to get condition on wavefunction $\Psi$ for a system that consists of 2 electrons(without taking spin into account) and helium nuclei I can solve schrödinger equation: $i*\frac{\partial \Psi}{\partial t}=-\frac{q_e*q_e}{\sqrt{(x_{electron1}-x_{electron2})^2+(y_{electron1}-y_{electron2})^2-(z_{electron1}-z_{electron2})^2}}-\frac{q_e*q_p}{\sqrt{(x_{electron1}-x_{proton})^2+(y_{electron1}-y_{proton})^2+(z_{electron1}-z_{proton})^2}}-\frac{q_e*q_p}{\sqrt{(x_{electron2}-x_{proton})^2+(y_{electron2}-y_{proton})^2+(z_{electron2}-z_{proton})^2}}-\\ \frac{1}{2*m_{electron}}* (\frac{\partial^2 \Psi}{\partial x_{electron1}^2}+\frac{\partial^2 \Psi}{\partial y_{electron1}^2}+\frac{\partial^2 \Psi}{\partial z_{electron1}^2})- \frac{1}{2*m_{electron}}* (\frac{\partial^2 \Psi}{\partial x_{electron2}^2}+\frac{\partial^2 \Psi}{\partial y_{electron2}^2}+\frac{\partial^2 \Psi}{\partial z_{electron2}^2})- \frac{1}{2*m_{proton}}* (\frac{\partial^2 \Psi}{\partial x_{proton}^2}+\frac{\partial^2 \Psi}{\partial y_{proton}^2}+\frac{\partial^2 \Psi}{\partial z_{proton}^2}))$

probability of particles being in corresponding positions is $P(t,x_{electron1},y_{electron1},z_{electron1},x_{electron2},y_{electron2},z_{electron2},x_{proton},y_{proton},z_{proton})=|\Psi(t,x_{electron1},y_{electron1},z_{electron1},x_{electron2},y_{electron2},z_{electron2},x_{proton},y_{proton},z_{proton})|^2$

to form helium atom wavefunction must be stationary aka not change in time. By solving time-independent schrdinger equation I find wavefunctions that satisfy normal schrödinger equation and so not change in time ($\frac{\partial |\Psi|}{\partial t}=0$).
$\frac{1}{2*m_{electron}}* (\frac{\partial^2 \Psi}{\partial x_{electron1}^2}+\frac{\partial^2 \Psi}{\partial y_{electron1}^2}+\frac{\partial^2 \Psi}{\partial z_{electron1}^2})- \frac{1}{2*m_{electron}}* (\frac{\partial^2 \Psi}{\partial x_{electron2}^2}+\frac{\partial^2 \Psi}{\partial y_{electron2}^2}+\frac{\partial^2 \Psi}{\partial z_{electron2}^2})- \frac{1}{2*m_{proton}}* (\frac{\partial^2 \Psi}{\partial x_{proton}^2}+\frac{\partial^2 \Psi}{\partial y_{proton}^2}+\frac{\partial^2 \Psi}{\partial z_{proton}^2}) -\frac{q_e*q_e}{\sqrt{(x_{electron1}-x_{electron2})^2+(y_{electron1}-y_{electron2})^2-(z_{electron1}-z_{electron2})^2}}-\frac{q_e*q_p}{\sqrt{(x_{electron1}-x_{proton})^2+(y_{electron1}-y_{proton})^2+(z_{electron1}-z_{proton})^2}}-\frac{q_e*q_p}{\sqrt{(x_{electron2}-x_{proton})^2+(y_{electron2}-y_{proton})^2+(z_{electron2}-z_{proton})^2}})=E*\Psi(r)$

to have hydrogen atom I have 1 electron and 1 proton and wavefunction must be stationary:
$-\frac{1}{2*m_{electron}}* (\frac{\partial^2 \Psi}{\partial x_{electron1}^2}+\frac{\partial^2 \Psi}{\partial y_{electron1}^2}+\frac{\partial^2 \Psi}{\partial z_{electron1}^2})- \frac{1}{2*m_{proton}}* (\frac{\partial^2 \Psi}{\partial x_{proton}^2}+\frac{\partial^2 \Psi}{\partial y_{proton}^2}+\frac{\partial^2 \Psi}{\partial z_{proton}^2}) -\frac{q_e*q_p}{\sqrt{(x_{electron1}-x_{proton})^2+(y_{electron1}-y_{proton})^2+(z_{electron1}-z_{proton})^2}}=E*\Psi(r)$

and the solution is: $\psi _{n,l,m}(r,\theta ,\phi )= (\frac{2^{l+1}*r^l*m_e^{l+3/2}*m_p^{l+3/2}*q_e^{2*l+3}*((n-l -1)!)^{1/2}} {n^{1+l}*(m_e+m_p)^{l+3/2}*((n+l )!)^{1/2}})*e^{-\frac{r*m_e*m_p*q_e^2}{n*(m_e+m_p)}}*L_{n-l -1}^{2l +1}(\frac{2*r*m_e*m_p*q_e^2}{n*(m_e+m_p)})*Y_l^m(\theta ,\phi )$(got it from Wikipedia)

since $P(A,B,C,D,E,F) \neq \int_{-\infty}^{\infty}(dx_1*\int_{-\infty}^{\infty}(dx_2*\int_{-\infty}^{\infty}(dx_3*P(A,B,C,x_1,x_2,x_3))))*\int_{-\infty}^{\infty}(dx_1*\int_{-\infty}^{\infty}(dx_2*\int_{-\infty}^{\infty}(dx_3*P(x_1,x_2,x_3,D,E,F))))$ positions of electron and proton in hydrogen atom are entagled.

To get in y-direction inertia of proton I us inertia operator $-i*\frac{\partial}{\partial y}$ :
$-i*\frac{\partial}{\partial y_{proton}} \Psi(t,x_{electron1},y_{electron1},z_{electron1},x_{electron2},y_{electron2},z_{electron2},x_{proton},y_{proton},z_{proton})=\Psi(t,x_{electron1},y_{electron1},z_{electron1},x_{electron2},y_{electron2},z_{electron2},x_{proton},p_{y\ proton},z_{proton})$

I used units where $\hbar=1$ and $k_{Coulomb}=1$

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#### DrClaude

Mentor
There is something that is not completely correct about what you wrote for the hydrogen atom. In the Schrödinger equation you set up, the independent variables are the cartesian coordinates of the electron and the proton, while the solution is written in terms of relative coordinates, excluding the center-of-mass motion.

For helium, you should be doing the same, namely first rewrite the Hamiltonian in the center-of-mass frame, then look for solutions in terms of the relative coordinates.

Note also that the label "proton" is not appropriate for helium, but should rather be "nucleus."

• dextercioby

#### olgerm

Gold Member
In the Schrödinger equation you set up, the independent variables are the cartesian coordinates of the electron and the proton, while the solution is written in terms of relative coordinates, excluding the center-of-mass motion.

Note also that the label "proton" is not appropriate for helium, but should rather be "nucleus."
Ok, I understand nucleus part. But generally if I have wavefunction of cartesian coordinates then by applying inertia operator $-i*\frac{\partial}{\partial y}$ I get new wavefunction whose argument for instead of y-postion is y-direction inertia of proton like:
$-i*\frac{\partial}{\partial y_{proton}} \Psi(t,x_{electron1},y_{electron1},z_{electron1},x_{electron2},y_{electron2},z_{electron2},x_{proton},y_{proton},z_{proton})=\Psi(t,x_{electron1},y_{electron1},z_{electron1},x_{electron2},y_{electron2},z_{electron2},x_{proton},p_{y\ proton},z_{proton})$?

#### weirdoguy

Besides, using whole long words like "proton" for labels and star as a product symbol makes your math hard to read. Not every product sign is necessary, and if it is use \cdot command instead of *.

#### DrClaude

Mentor
Ok, I understand nucleus part. But generally if I have wavefunction of cartesian coordinates then by applying inertia operator $-i*\frac{\partial}{\partial y}$ I get new wavefunction whose argument for instead of y-postion is y-direction inertia of proton like:
$-i*\frac{\partial}{\partial y_{proton}} \Psi(t,x_{electron1},y_{electron1},z_{electron1},x_{electron2},y_{electron2},z_{electron2},x_{proton},y_{proton},z_{proton})=\Psi(t,x_{electron1},y_{electron1},z_{electron1},x_{electron2},y_{electron2},z_{electron2},x_{proton},p_{y\ proton},z_{proton})$?
Nope. You need to do a Fourier transform with respect to $y_{proton}$ for that.

since $P(A,B,C,D,E,F) \neq \int_{-\infty}^{\infty}(dx_1*\int_{-\infty}^{\infty}(dx_2*\int_{-\infty}^{\infty}(dx_3*P(A,B,C,d,e,f))))*\int_{-\infty}^{\infty}(dx_1*\int_{-\infty}^{\infty}(dx_2*\int_{-\infty}^{\infty}(dx_3*P(x_1,x_2,x_3,D,E,F))))$ positions of electron and proton in hydrogen atom are entagled.
No. If you cannot write $\Psi(x,y,z,X,Y,Z) = \psi(x,y,z) \phi(X,Y,Z)$ where the degrees of freedom $(x,y,z)$ and $(X,Y,Z)$ belong to the two different particles, then they are entangled. Note that not just spatial degrees of freedom need to be taken into account, but also spin.

• • dextercioby and olgerm

#### olgerm

Gold Member
satisfy normal schrödinger equation and so not change in time ($\frac{\partial |\Psi|}{\partial t}=0$)
Is it Coorect or should it be: $\frac{\partial \Psi}{\partial t}=0$?

No. If you cannot write $\Psi(x,y,z,X,Y,Z) = \psi(x,y,z) \phi(X,Y,Z)$ where the degrees of freedom $(x,y,z)$ and $(X,Y,Z)$ belong to the two different particles, then they are entangled.
Should it be then:
since $`\Psi(A,B,C,D,E,F) \neq \int_{-\infty}^{\infty}(dx_1*\int_{-\infty}^{\infty}(dx_2*\int_{-\infty}^{\infty}(dx_3*\Psi(A,B,C,x_1,x_2,x_3))))*\int_{-\infty}^{\infty}(dx_1*\int_{-\infty}^{\infty}(dx_2*\int_{-\infty}^{\infty}(dx_3*\Psi(x_1,x_2,x_3,D,E,F))))$ positions of electron and proton in hydrogen atom are entangled.

$\psi(x,y,z)=\int_{-\infty}^{\infty}(dx_1*\int_{-\infty}^{\infty}(dx_2*\int_{-\infty}^{\infty}(dx_3*P(A,B,C,x_1,x_2,x_3))))$
$\phi(X,Y,Z)=\int_{-\infty}^{\infty}(dx_1*\int_{-\infty}^{\infty}(dx_2*\int_{-\infty}^{\infty}(dx_3*P(x_1,x_2,x_3,D,E,F))))$

#### HomogenousCow

You really don't need all those star symbols, multiplication is implied by adjecancy.

• weirdoguy

#### olgerm

Gold Member
Would like to get answers on topic, not on notation.

• weirdoguy

#### DrClaude

Mentor
Is it Coorect or should it be: $\frac{\partial \Psi}{\partial t}=0$?
It is correct, although one would usually write it as
$$\frac{\partial |\Psi|^2}{\partial t}=0$$
Physical states are defined up to a complex phase factor, to even for a stationary state $\frac{\partial \Psi}{\partial t} \neq 0$.

$\psi(x,y,z)=\int_{-\infty}^{\infty}(dx_1*\int_{-\infty}^{\infty}(dx_2*\int_{-\infty}^{\infty}(dx_3*P(A,B,C,x_1,x_2,x_3))))$
$\phi(X,Y,Z)=\int_{-\infty}^{\infty}(dx_1*\int_{-\infty}^{\infty}(dx_2*\int_{-\infty}^{\infty}(dx_3*P(x_1,x_2,x_3,D,E,F))))$
This is not correct as you are introducing probabilities. The wave function itself is what you should be considering.

Would like to get answers on topic, not on notation.
The problem is that your choice of notation makes it hard to read.

• weirdoguy

#### olgerm

Gold Member
This is not correct as you are introducing probabilities. The wave function itself is what you should be considering.

Yes, I actually meant:
since $\Psi(A,B,C,D,E,F) \neq \int_{-\infty}^{\infty}(dx_1*\int_{-\infty}^{\infty}(dx_2*\int_{-\infty}^{\infty}(dx_3*\Psi(A,B,C,x_1,x_2,x_3))))*\int_{-\infty}^{\infty}(dx_1*\int_{-\infty}^{\infty}(dx_2*\int_{-\infty}^{\infty}(dx_3*\Psi(x_1,x_2,x_3,D,E,F))))$ positions of electron and proton in hydrogen atom are entangled.

$\psi(A,B,C)=\int_{-\infty}^{\infty}(dx_1*\int_{-\infty}^{\infty}(dx_2*\int_{-\infty}^{\infty}(dx_3*\Psi(A,B,C,x_1,x_2,x_3))))$
$\phi(D,E,F)=\int_{-\infty}^{\infty}(dx_1*\int_{-\infty}^{\infty}(dx_2*\int_{-\infty}^{\infty}(dx_3*\Psi(x_1,x_2,x_3,D,E,F))))$

#### Haborix

I have looked at this thread a couple of times and I'm still not sure what the goal is. As of this moment, I understand you are interested in the helium atom and the properties of its stationary states. Part of the focus on notation arises because the goal is not really clear. Could we get a clear and concise statement of the problem/goal?

• weirdoguy

#### A. Neumaier

Would like to get answers on topic, not on notation.
Your nonstandard notation is distracting and makes your formulas difficult to read. Both significantly reduces the likelihood of getting an answer to the wanted point.

• weirdoguy and vanhees71

#### olgerm

Gold Member
I do not see anything non-standard here: integrals, plus-signs, multipication-signs, deriatives, square roots, integrals, functions.
I have also chosen easily understandable variable names.

• weirdoguy

#### A. Neumaier

I do not see anything non-standard here: integrals, plus-signs, multipication-signs, deriatives, square roots, integrals, functions.
I have also chosen easily understandable variable names.
Almost everything is very nonstandard: Indices are usually numbers or letters, not names. In mathematical formulas, multiplication is never written as *, but as juxtaposition. Parentheses are needed only to separate terms containing operators with different priorities, and are usually not written otherwise. Vector notation is usually preferred over writing out components, unless the latter are really needed. Hardly anyone writes out a distance $|\mathbf x-\mathbf y|$ in terms of components and the square root - this is done only when defining the distance, never when using it.

Following these conventions makes your excessively lengthy formulas (even running over multiple lines) much shorter without any loss of information. The resulting formulas are much easier to comprehend, and they look more like the familiar ones, hence are easier to interpret.

• Nugatory and weirdoguy

#### olgerm

Gold Member
Is it now easier to read? And is it correct? It should have same meaning like in post #__ ,but I am not sure if this is clearly understandable for me.

since $\Psi(A,B,C,D,E,F) \neq \int_{-\infty}^{\infty}dx_1\int_{-\infty}^{\infty}dx_2\int_{-\infty}^{\infty}dx_3\Psi(A,B,C,x_1,x_2,x_3)*\int_{-\infty}^{\infty}dx_1\int_{-\infty}^{\infty}dx_2\int_{-\infty}^{\infty}dx_3*\Psi(x_1,x_2,x_3,D,E,F)$ positions of electron and proton in hydrogen atom are entangled.

$\psi(A,B,C)=\int_{-\infty}^{\infty}dx_1\int_{-\infty}^{\infty}dx_2\int_{-\infty}^{\infty}dx_3\Psi(A,B,C,x_1,x_2,x_3)$
$\phi(D,E,F)=\int_{-\infty}^{\infty}dx_1\int_{-\infty}^{\infty}dx_2\int_{-\infty}^{\infty}dx_3\Psi(x_1,x_2,x_3,D,E,F)$

#### A. Neumaier

Is it now easier to read? And is it correct? It should have same meaning like in post #__ ,but I am not sure if this is clearly understandable for me.

since $\Psi(A,B,C,D,E,F) \neq \int_{-\infty}^{\infty}dx_1\int_{-\infty}^{\infty}dx_2\int_{-\infty}^{\infty}dx_3\Psi(A,B,C,x_1,x_2,x_3)*\int_{-\infty}^{\infty}dx_1\int_{-\infty}^{\infty}dx_2\int_{-\infty}^{\infty}dx_3*\Psi(x_1,x_2,x_3,D,E,F)$ positions of electron and proton in hydrogen atom are entangled.

$\psi(A,B,C)=\int_{-\infty}^{\infty}dx_1\int_{-\infty}^{\infty}dx_2\int_{-\infty}^{\infty}dx_3\Psi(A,B,C,x_1,x_2,x_3)$
$\phi(D,E,F)=\int_{-\infty}^{\infty}dx_1\int_{-\infty}^{\infty}dx_2\int_{-\infty}^{\infty}dx_3\Psi(x_1,x_2,x_3,D,E,F)$
Much easier to read. Still two superfluous stars. If an integral runs from $-\infty$ to $+\infty$ one usually deletes the bounds. This makes your formulas even better to read. You were also asked in post #11 to rephrase your goals to make more clear what you really want. So please start from scratch and recast your goals including your further simplified formulas, so that we have everything in one place.

#### ftr

I am not sure what you are after, but this problem is typically part of quantum chemistry. you can see the series of lectures in the link below (if you haven't seen them yet) where the Helium atom is tackled in chapter 9. Mind you it doesn't add anything new to understanding basic QM generally IMO.

#### olgerm

Gold Member
Is it true, that if $\Psi(A,B,C,D,E,F) \neq \int_{-\infty}^{\infty}dx_1\int_{-\infty}^{\infty}dx_2\int_{-\infty}^{\infty}dx_3\Psi(A,B,C,x_1,x_2,x_3)*\int_{-\infty}^{\infty}dx_1\int_{-\infty}^{\infty}dx_2\int_{-\infty}^{\infty}dx_3*\Psi(x_1,x_2,x_3,D,E,F)$
aka $\Psi(x_{electron},y_{electron},z_{electron},x_{proton},y_{proton},z_{proton}) \neq \int_{-\infty}^{\infty}(dx_1*\int_{-\infty}^{\infty}(dx_2*\int_{-\infty}^{\infty}(dx_3*\Psi(x_{electron},y_{electron},z_{electron},x_1,x_2,x_3))))*\int_{-\infty}^{\infty}(dx_1*\int_{-\infty}^{\infty}(dx_2*\int_{-\infty}^{\infty}(dx_3*\Psi(x_1,x_2,x_3,z_{electron},x_{proton},y_{proton},z_{proton}))))$
positions of electron and proton in hydrogen atom are entangled?

$\Psi$ is wavefiunction $\Psi(x_{electron},y_{electron},z_{electron},x_{proton},y_{proton},z_{proton})$.

• weirdoguy

#### A. Neumaier

Is it true that if $\Psi(\mathbf{y}, \mathbf{z})\neq \int d\mathbf{x}\Psi(\mathbf{y}, \mathbf{x})\int d\mathbf{x}\Psi(\mathbf{x}, \mathbf{z})$, the position $\mathbf{y}$ of the electron and the position $\mathbf{z}$ of the proton in a hydrogen atom are entangled?
There were still many superfluous bounds, stars and indices that no one but you writes. I corrected the grammar and shortened your statement to what it should have been to be standard. Can you see the difference in intelligibility?

• olgerm

Gold Member