- $\frac{-\hbar^2}{2m}(a^2b_0-4ab_1+6b_2)-Ze^2b_1-Eb_0 = 0$
- $\frac{-\hbar^2}{2m}(a^2b_1-6ab_2)-Ze^2b_2-Eb_1= 0$
- $\frac{-\hbar^2}{2m}(-2ab_0+2b_1)-Ze^2b_0 = 0$
- $\frac{-\hbar^2}{2m}(a^2b_2)-Eb_2 = 0$
here I need to find the values of $a,b_0,b_1,b_2$
E = $-\displaystyle\frac{Z^2e^2}{18a_0}$
$a_0 = \displaystyle\frac{\hbar^2}{me^2}$
$Z,e,m$ are constants(atomic number,charge of electron, mass of electron)
$\hbar$ is also a constant
I got relations between $b_0$ and $b_1$, $b_0$ and $b_2$. But the value of $b_0$ is becoming zero which makes the value of $b_1$ and $b_2$ zero. Is there a way to solve this?
The solutions given are
$a = \displaystyle\frac{Z}{3a_0}$
$b_0 = 27$
$b_1 = -\displaystyle\frac{18Z}{a_0}$
$b_2 = \displaystyle\frac{2Z^2}{a_0^2}$
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K Sreerag
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Welcome to MSE.
Test the consistency of the equations by substituting the given solutions into the first equation.
Using maxima to do the algebra. Equations $(1)$ to $(4)$ and $(11)$ are equal to zero.
E1 : -h/(2*m) *(a^2*b0 - 4*a *b1 + 6*b2) - Z*e^2*b1 - E*b0;
E2 : -h/(2*m) *(a^2*b1 - 6*a*b2) - Z*e^2*b2 - E*b1;
E3 : -h/(2*m) *(-2*a*b0 + 2*b1) - Z*e^2*b0;
E4 : -h/(2*m) *(a^2*b2) - E*b2;
sE : E = -Z^2e^2/(18a0);
sa0 : a0 = h^2/(m*e^2);
sa : a = Z/(3a0);
sb0 : b0 = 27;
sb1 : b1 = 18Z/a0;
sb2 : b2 = 2*Z^2/a0^2;
AllIntoEqn1 : subst(sa0,subst(sE,subst(sa,subst(sb2,subst(sb1,subst(sb0,E1))))));
SolveForh : solve(AllIntoEqn1,[h]);
$$-{\it b_1}\,e^2\,Z-{\it b_0}\,E-{{\left(6\,{\it b_2}-4\,a\, {\it b_1}+a^2\,{\it b_0}\right)\,h}\over{2\,m}} \tag{1}$$
$$-{\it b_2}\,e^2\,Z-{\it b_1}\,E-{{\left(a^2\,{\it b_1}-6\,a\, {\it b_2}\right)\,h}\over{2\,m}} \tag{2}$$
$$-{\it b_0}\,e^2\,Z-{{\left(2\,{\it b_1}-2\,a\,{\it b_0}\right)\,h }\over{2\,m}} \tag{3}$$
$$-{\it b_2}\,E-{{a^2\,{\it b_2}\,h}\over{2\,m}} \tag{4}$$
$$E=-{{e^2\,Z^2}\over{18\,{\it a_0}}} \tag{5}$$
$${\it a_0}={{h^2}\over{e^2\,m}} \tag{6}$$
$$a={{Z}\over{3\,{\it a_0}}} \tag{7}$$
$${\it b_0}=27 \tag{8}$$
$${\it b_1}=-{{18\,Z}\over{{\it a_0}}} \tag{9}$$
$${\it b_2}={{2\,Z^2}\over{{\it a_0}^2}} \tag{10}$$
Substituting equations $(5)$ to $(10)$ into $(1)$ , (AllIntoEqn1 command) results in equation $(11)$.
$${{39\,e^4\,m\,Z^2}\over{2\,h^2}}-{{39\,e^4\,m\,Z^2}\over{2\,h^3}} \tag{11}$$
Solving equation $(11)$ for h gives equation $(12)$.
$$h=1 \tag{12}$$
This gives an incorrect value for $\hbar$.
There seems to be something wrong with the equations.
The powers of h in the denominators of $(11)$ don't match.
Section $2$ : @Luka fix ${\hbar}^2$
/* h^2 */
E1 : -h^2/(2m) (a^2b0 - 4a b1 + 6b2) - Ze^2b1 - Eb0$
tex(%)$
E2 : -h^2/(2m) (a^2b1 - 6ab2) - Ze^2b2 - Eb1$
tex(%)$
E3 : -h^2/(2m) (-2ab0 + 2b1) - Ze^2b0$
tex(%)$
E4 : -h^2/(2m) (a^2b2) - Eb2$
tex(%)$
sE : E = -Z^2e^2/(18a0)$
tex(%)$
sa0 : a0 = h^2/(m*e^2)$
tex(%)$
sa : a = Z/(3a0)$
tex(%)$
sb0 : b0 = 27$
tex(%)$
sb1 : b1 = -18Z/a0$
tex(%)$
sb2 : b2 = 2*Z^2/a0^2$
tex(%)$
AllIntoEqn1 : subst(sa0,subst(sE,subst(sa,subst(sb2,subst(sb1,subst(sb0,E1))))))$
tex(%)$
$$-{\it b_1}\,e^2\,Z-{\it b_0}\,E-{{\left(6\,{\it b_2}-4\,a\, {\it b_1}+a^2\,{\it b_0}\right)\,h^2}\over{2\,m}} \tag{1b}$$
$$-{\it b_2}\,e^2\,Z-{\it b_1}\,E-{{\left(a^2\,{\it b_1}-6\,a\, {\it b_2}\right)\,h^2}\over{2\,m}} \tag{2b}$$
$$-{\it b_0}\,e^2\,Z-{{\left(2\,{\it b_1}-2\,a\,{\it b_0}\right)\,h^2 }\over{2\,m}} \tag{3b}$$
$$-{\it b_2}\,E-{{a^2\,{\it b_2}\,h^2}\over{2\,m}} \tag{4b}$$
$$E=-{{e^2\,Z^2}\over{18\,{\it a_0}}} \tag{5b}$$
$${\it a_0}={{h^2}\over{e^2\,m}} \tag{6b}$$
$$a={{Z}\over{3\,{\it a_0}}} \tag{7b}$$
$${\it b_0}=27 \tag{8b}$$
$${\it b_1}=-{{18\,Z}\over{{\it a_0}}} \tag{9b}$$
$${\it b_2}={{2\,Z^2}\over{{\it a_0}^2}} \tag{10b}$$
$$0 \tag{11b}$$
The given solutions are consistent with equation $(1)$.
Solve $b_1$ and $b_2$ in terms of $b_0$:
sb1 : facsum(solve(E3,b1));
tex(%);
sb2 : facsum(solve(E2,b2));
tex(%);
sb2 : facsum(subst(sb1,solve(E2,b2)));
tex(%);
$$ {\it b_1}=-{{{\it b_0}\,\left(e^2\,m\,Z-a\,h^2\right)}\over{ h^2}} \tag{13}$$
$${\it b_2}=-{{{\it b_1}\,\left(2\,m\,E+a^2\,h^2\right)}\over{ 2\,\left(e^2\,m\,Z-3\,a\,h^2\right)}} \tag{14}$$
$$ {\it b_2}={{{\it b_0}\,\left(2\,m\,E+a^2\,h^2\right)\,\left( e^2\,m\,Z-a\,h^2\right)}\over{2\,h^2\,\left(e^2\,m\,Z-3\,a\,h^2 \right)}} \tag{15}$$
$b_0$ comes out as a linear multiple in the equations of $b_1$ , $b_2$ and thus equations $(1)$ to $(3)$ indicating it is arbitrary.
$\boxed{b_0 \: \text{is arbitrary that's why it was coming out as }0.}$
Solving for a.
sa2 : solve(E4,a^2)$
tex(%);
subst(sE,sa2)$
tex(%);
subst(sa0,subst(sE,sa2))$
tex(%);
solve(subst(sa0,subst(sE,sa2)),a)$
tex(%);
From equation $(4)$
$$ a^2=-{{2\,m\,E}\over{h^2}} \tag{16}$$
Substitute E from $(5)$
$$ a^2={{e^2\,m\,Z^2}\over{9\,{\it a_0}\,h^2}} \tag{17}$$
Substitute $a_0$ from $(6)$
$$ a^2={{e^4\,m^2\,Z^2}\over{9\,h^4}} \tag{18}$$
Solve for a.
$$ a=-{{e^2\,m\,Z}\over{3\,h^2}} , a={{e^2\,m\,Z}\over{3\,h^2}} \tag{19}$$
From equation $(6)$ substitute $a_0$ take the positive value:
$$ a={{Z}\over{3\,a_0}} \tag{20}$$
Solving for $b_1$ and $b_2$:
ra : a = Z/(3*a0)$
tex(%);
subst(ra,sb1)$
tex(%);
rb1 : subst(sa0,subst(ra,sb1))$
tex(%);
ratsimp(subst(sE,subst(ra,sb2)))$
tex(%);
subst(sa0,ratsimp(subst(sE,subst(ra,sb2))))$
tex(%);
$$a={{Z}\over{3\,{\it a_0}}} \tag{21}$$
$${\it b_1}=-{{{\it b_0}\,\left(e^2\,m\,Z-{{h^2\,Z}\over{3\, {\it a_0}}}\right)}\over{h^2}} \tag{22}$$
$$ {\it b_1}=-{{2\,{\it b_0}\,e^2\,m\,Z}\over{3\,h^2}} \tag{23}$$
$$ {\it b_2}=-{{\left(3\,{\it a_0}\,{\it b_0}\,e^2\,m-{\it b_0} \,h^2\right)\,Z^2}\over{54\,{\it a_0}^2\,h^2}} \tag{24}$$
$$ {\it b_2}=-{{{\it b_0}\,e^4\,m^2\,Z^2}\over{27\,h^4}} \tag{25}$$
Using equation $(6)$ for $a_0$
$$ {\it b_1}=-{{2\,{\it b_0}\,Z}\over{3\,a_0}} \tag{26}$$
$$ {\it b_2}=-{{{\it b_0}\,Z^2}\over{27\,{a_0}^2}} \tag{27}$$
$b_2$ doesn't match? All the algebra is automated?.
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$h$ and $c$ are commonly set to 1, so that is not necessarily wrong, but in (11) the powers should be equal, as you've said. – Luka Jul 24 '20 at 12:05
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$\frac{-\hbar^2}{2m}(a^2b_0-4ab_1+6b_2)-Ze^2b_1-Eb_0 = 0$ instead of $\frac{-\hbar}{2m}(a^2b_0-4ab_1+6b_2)-Ze^2b_1-Eb_0 = 0$ maybe? – Luka Jul 24 '20 at 12:06
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@Luka But I still don't understand why is b0 arbitrary. When can we take a variable as arbitrary? – K Sreerag Jul 27 '20 at 12:35
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@KSreerag This would probably be better answered by a mathematician. So you have 4 equations and 4 variables that you would like to determine. And you would like to get a value for each one in terms of things like $a_0, E, Z, etc.$ in other words a definite value for each variable. If not all variables are expressible like that, in your example you get values like $b_1=b_0\times ...$, $b_2=b_0\times ...$, $b_3=b_0\times ...$, we say that variables can be parameterized by $b_0$. $b_0$ is a parameter you set, so that it suits your model or theory. – Luka Jul 27 '20 at 13:01
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@KSreerag From a physical standpoint: The free parameter, $b_0$ in your case, tells you something about your model, like the strength of a coupling, a spring constant, etc. . In the case you've provided above it's probably the strenght or coupling of the electric field (electron-nucleus maybe?), since you have terms like $Ze^2b_0$. And all other equations or couplings depend on how your $b_0$ is set, thats why it turns up in all of the equations for $b_1,b_2,...$ . – Luka Jul 27 '20 at 13:05
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@KSreerag You could set $b_0=1$ run a simulation and see how your sistem behaves or what solutions you get, then you could try $b_0=2$ and again calculate a solution and so on... Then you see how your sistem behaves based on the paramater $b_0$ for example and use this to get the best parameter $b_0$ value. – Luka Jul 27 '20 at 13:09
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@KSreerag - Wave equations often have arbitrary amplitude e.g. $\displaystyle \frac{d^2x}{dt^2} = -w^2x$ is solved by $x = Ae^{\pm iwt}$ where $A$ is an arbitrary amplitude. $b_0$ plays that role in the above equations. – Jul 27 '20 at 13:29
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https://wikimedia.org/api/rest_v1/media/math/render/svg/9f57865f4e7f8414e2c7b1a2ae06a30eb3b0d013
– Luka Jul 24 '20 at 12:08