Solving 4 simultaneous homogeneous equations

Question

1. $\frac{-\hbar^2}{2m}(a^2b_0-4ab_1+6b_2)-Ze^2b_1-Eb_0 = 0$
2. $\frac{-\hbar^2}{2m}(a^2b_1-6ab_2)-Ze^2b_2-Eb_1= 0$
3. $\frac{-\hbar^2}{2m}(-2ab_0+2b_1)-Ze^2b_0 = 0$
4. $\frac{-\hbar^2}{2m}(a^2b_2)-Eb_2 = 0$
   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
   $a = \displaystyle\frac{Z}{3a_0}$
   $b_0 = 27$
   $b_1 = -\displaystyle\frac{18Z}{a_0}$
   $b_2 = \displaystyle\frac{2Z^2}{a_0^2}$

Answer 1

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.

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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(%);

$$ {\it b_1}=-{{{\it b_0}\,\left(e^2\,m\,Z-a\,h^2\right)}\over{ h^2}} \tag{13}$$

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

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

$b_0$ 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?.