Solving 4 simultaneous homogeneous equations

Question

$frac{-hbar^2}{2m}(a^2b_0-4ab_1+6b_2)-Ze^2b_1-Eb_0 = 0$
$frac{-hbar^2}{2m}(a^2b_1-6ab_2)-Ze^2b_2-Eb_1= 0$
$frac{-hbar^2}{2m}(-2ab_0+2b_1)-Ze^2b_0 = 0$
$frac{-hbar^2}{2m}(a^2b_2)-Eb_2 = 0$
here I need to find the values of $a,b_0,b_1,b_2$
E = $-displaystylefrac{Z^2e^2}{18a_0}$
$a_0 = displaystylefrac{hbar^2}{me^2}$
$Z,e,m$ 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
$a = displaystylefrac{Z}{3a_0}$
$b_0 = 27$
$b_1 = -displaystylefrac{18Z}{a_0}$
$b_2 = displaystylefrac{2Z^2}{a_0^2}$

arthur · Answer

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^2*e^2/(18*a0);
sa0 : a0 =  h^2/(m*e^2);

sa : a = Z/(3*a0);
sb0 : b0 = 27;
sb1 : b1 = 18*Z/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/(2*m) *(a^2*b0 - 4*a *b1 + 6*b2) - Z*e^2*b1 - E*b0$
tex(%)$
E2 : -h^2/(2*m) *(a^2*b1 - 6*a*b2) - Z*e^2*b2 - E*b1$
tex(%)$
E3 : -h^2/(2*m) *(-2*a*b0 + 2*b1) - Z*e^2*b0$
tex(%)$
E4 : -h^2/(2*m) *(a^2*b2) - E*b2$
tex(%)$

sE : E = -Z^2*e^2/(18*a0)$
tex(%)$
sa0 : a0 =  h^2/(m*e^2)$
tex(%)$

sa : a = Z/(3*a0)$
tex(%)$
sb0 : b0 = 27$
tex(%)$
sb1 : b1 = -18*Z/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(%);

$$ {it b_1}=-{{{it b_0},left(e^2,m,Z-a,h^2right)}over{
 h^2}}  tag{13}$$
$${it b_2}=-{{{it b_1},left(2,m,E+a^2,h^2right)}over{
 2,left(e^2,m,Z-3,a,h^2right)}}   tag{14}$$
$$ {it b_2}={{{it b_0},left(2,m,E+a^2,h^2right),left(
 e^2,m,Z-a,h^2right)}over{2,h^2,left(e^2,m,Z-3,a,h^2
 right)}}   tag{15}$$
$b_0$ 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(%);

$$a={{Z}over{3,{it a_0}}} tag{21}$$
$${it b_1}=-{{{it b_0},left(e^2,m,Z-{{h^2,Z}over{3,
 {it a_0}}}right)}over{h^2}}   tag{22}$$
$$ {it b_1}=-{{2,{it b_0},e^2,m,Z}over{3,h^2}}   tag{23}$$
$$ {it b_2}=-{{left(3,{it a_0},{it b_0},e^2,m-{it b_0}
 ,h^2right),Z^2}over{54,{it a_0}^2,h^2}}   tag{24}$$
$$ {it b_2}=-{{{it b_0},e^4,m^2,Z^2}over{27,h^4}}    tag{25}$$
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?.

Solving 4 simultaneous homogeneous equations

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