Non degenerate perturbation example: Difference between revisions
LaurelWinter (talk | contribs) (New page: From David J. Griffiths "Introduction to Quantum Mechanics" Problem 6.3 "Two identical bosons are placed in an infinite square well. They interact weakly with one another, via the potenti...) |
LaurelWinter (talk | contribs) No edit summary |
||
| Line 7: | Line 7: | ||
b) Use first-order perturbation theory to estimate the effect of the particle-particle interaction on the energies of the ground state and the first excited state. | b) Use first-order perturbation theory to estimate the effect of the particle-particle interaction on the energies of the ground state and the first excited state. | ||
For the infinite square well the unperturbed energy is given by <math> E^{0}_n = \frac{n^{2}\pi^{2}\hbar^{2}}{2ma^2}</math>and the ground state wave function is given by <math>\Psi^{0}_n = \sqrt{\frac{2}{a}}sin(\frac{n\pi x}{a})</math>. | For the infinite square well the unperturbed energy is given by <math> E^{0}_n = \frac{n^{2}\pi^{2}\hbar^{2}}{2ma^2}</math>and the ground state wave function is given by <math>\Psi^{0}_n = \sqrt{\frac{2}{a}}sin\left(\frac{n\pi x}{a}\right)</math>. | ||
| Line 14: | Line 14: | ||
For the case of identical bosons with no interaction between them the ground state is given by the following: <math>\psi^0 _{11} (x) = \psi^0_1(x_1)\psi^0_1(x_2)</math> | For the case of identical bosons with no interaction between them the ground state is given by the following: <math>\psi^0 _{11} (x) = \psi^0_1(x_1)\psi^0_1(x_2)</math> | ||
<math>\psi^0 _{11} (x) = \sqrt{\frac{2}{a}}sin(\frac{\pi x_1}{a})\sqrt{\frac{2}{a}}sin(\frac{\pi x_2}{a})</math> | <math>\psi^0 _{11} (x) = \sqrt{\frac{2}{a}}sin\left(\frac{\pi x_1}{a}\right)\sqrt{\frac{2}{a}}sin\left(\frac{\pi x_2}{a}\right)</math> | ||
While the energy is given by <math>E^0_{1total} = E^0_1(x_1) + E^0_1(x_2)</math>. | While the energy is given by <math>E^0_{1total} = E^0_1(x_1) + E^0_1(x_2)</math>. | ||
| Line 20: | Line 20: | ||
<math>E^0_1 = 2E^0_1 = (2)\frac{\pi^2 \hbar^2}{2ma^2} = \frac{\pi^2 \hbar^2}{ma^2}</math> | <math>E^0_1 = 2E^0_1 = (2)\frac{\pi^2 \hbar^2}{2ma^2} = \frac{\pi^2 \hbar^2}{ma^2}</math> | ||
The first excited state is given by <math>\psi^0_{12}(x) = \frac{1}{\sqrt{2}}[ \psi^0_1(x_1)\psi^0_2(x_2) + \psi^0_1(x_2)\psi^0_2(x_1)]</math> | The first excited state is given by <math>\psi^0_{12}(x) = \frac{1}{\sqrt{2}}\left[ \psi^0_1(x_1)\psi^0_2(x_2) + \psi^0_1(x_2)\psi^0_2(x_1)\right]</math> | ||
<math>\psi^0_{12} = \frac{1}{\sqrt{2}}[ \sqrt{\frac{2}{a}}sin(\frac{\pi x_1}{a})\sqrt{\frac{2}{a}}sin(\frac{2\pi x_2}{a}) + \sqrt{\frac{2}{a}}sin(\frac{2\pi x_1}{a})\sqrt{\frac{2}{a}}sin(\frac{\pi x_2}{a})]</math> | <math>\psi^0_{12} = \frac{1}{\sqrt{2}}[ \sqrt{\frac{2}{a}}sin\left(\frac{\pi x_1}{a}\right)\sqrt{\frac{2}{a}}sin\left(\frac{2\pi x_2}{a}\right) + \sqrt{\frac{2}{a}}sin | ||
\left(\frac{2\pi x_1}{a}\right)\sqrt{\frac{2}{a}}sin\left(\frac{\pi x_2}{a}\right)]</math> | |||
<math>\psi^0_{12} = \frac{\sqrt{2}}{a}[sin(\frac{\pi x_1}{a})sin(\frac{2 \pi x_2}{a}) + sin(\frac{2\pi x_1}{a})sin(\frac{\pi x_2}{a})]</math> | <math>\psi^0_{12} = \frac{\sqrt{2}}{a}[sin\left(\frac{\pi x_1}{a}\right)sin\left(\frac{2 \pi x_2}{a}\right) + sin\left(\frac{2\pi x_1}{a}\right)sin\left(\frac{\pi x_2}{a}\right)]</math> | ||
While the energy is given by <math>E^0_2 = E^0_2(x_1) + E^0_2(x_2)</math> | |||
<math>E^0_2 = \frac{\pi ^2 \hbar^2}{2ma^2} + \frac{4\pi^2 \hbar^2}{2ma^2} = \frac{5 \pi^2 \hbar^2}{2ma^2}</math> | <math>E^0_2 = \frac{\pi ^2 \hbar^2}{2ma^2} + \frac{4\pi^2 \hbar^2}{2ma^2} = \frac{5 \pi^2 \hbar^2}{2ma^2}</math> | ||
| Line 38: | Line 39: | ||
The interaction potential given in the problem is the perturbed part of the Hamiltonian. Therefore the first order perturbation in integral form for the ground state is given by: | The interaction potential given in the problem is the perturbed part of the Hamiltonian. Therefore the first order perturbation in integral form for the ground state is given by: | ||
<math>E^1_1 = \int \psi^1_1 [-aV_0 \delta(x_1 - x_2) ] \psi ^1_1 dx</math>, where <math>\psi^1_1 = \frac{2}{a} sin(\frac{\pi x_1}{a}) sin(\frac{\pi x_2}{a})</math> | <math>E^1_1 = \int \psi^1_1 [-aV_0 \delta(x_1 - x_2) ] \psi ^1_1 dx</math>, where <math>\psi^1_1 = \frac{2}{a} sin\left(\frac{\pi x_1}{a}\right) sin\left(\frac{\pi x_2}{a}\right)</math> | ||
<math>E^1_1 = \int_0^a \int_0^a (\frac{2}{a})^2 sin^2(\frac{\pi x_1}{a}) sin^2(\frac{\pi x_2}{a}) [-a V_0 \delta(x_1 - x_2) ] dx_1 dx_2</math> | <math>E^1_1 = \int_0^a \int_0^a \left(\frac{2}{a}\right)^2 sin^2\left(\frac{\pi x_1}{a}\right) sin^2\left(\frac{\pi x_2}{a}\right) [-a V_0 \delta(x_1 - x_2) ] dx_1 dx_2</math> | ||
<math>E^1_1 = -aV_0 (\frac{2}{a})^2 \int_0^a sin^4 (\frac{ \pi x}{a}) dx</math> | <math>E^1_1 = -aV_0 \left(\frac{2}{a}\right)^2 \int_0^a sin^4 \left(\frac{ \pi x}{a}\right) dx</math> | ||
<math>E^1_1 = -aV_0 (\frac{2}{a})^2 \frac{a}{\pi} \int_0^\pi sin^4y dy</math> | Then do a substitution such that <math>y = \frac{\pi x}{a}</math> | ||
<math>E^1_1 = -aV_0 \left(\frac{2}{a}\right)^2 \frac{a}{\pi} \int_0^\pi sin^4y dy</math> | |||
The integral of <math>sin^4(y)</math> is given by <math>\frac{1}{32} (12y - 8 sin(2y) + sin(4y))</math>. Since <math>y= \pi</math> in this case both sine terms go to zero and the integrand yields <math>\frac{3 \pi}{8}</math>. | The integral of <math>sin^4(y)</math> is given by <math>\frac{1}{32} (12y - 8 sin(2y) + sin(4y))</math>. Since <math>y= \pi</math> in this case both sine terms go to zero and the integrand yields <math>\frac{3 \pi}{8}</math>. | ||
| Line 54: | Line 57: | ||
As for the first order perturbation to the energy for the first excited state the wavefunction is given as | As for the first order perturbation to the energy for the first excited state the wavefunction is given as | ||
<math>\psi^0_{12} = \frac{\sqrt{2}}{a}[sin(\frac{\pi x_1}{a})sin(\frac{2 \pi x_2}{a}) + sin(\frac{2\pi x_1}{a})sin(\frac{\pi x_2}{a})]</math> | <math>\psi^0_{12} = \frac{\sqrt{2}}{a}[sin\left(\frac{\pi x_1}{a}\right)sin\left(\frac{2 \pi x_2}{a}\right) + sin\left(\frac{2\pi x_1}{a}\right)sin\left(\frac{\pi x_2}{a}\right)]</math> | ||
Note that if this system was for fermions then this wavefunction would be zero since there would be a minus sign in place of the plus sign. | Note that if this system was for fermions then this wavefunction would be zero since there would be a minus sign in place of the plus sign. | ||
<math>E^1_2 = -aV_0 \int_0^a \int_0^a [\frac{\sqrt{2}}{a} (sin(\frac{\pi x_1}{a})sin(\frac{2 \pi x_2}{a}) + sin(\frac{2\pi x_1}{a})sin(\frac{\pi x_2}{a})]^2 \delta(x_1 - x_2) dx_1 dx_2</math> | <math>E^1_2 = -aV_0 \int_0^a \int_0^a [\frac{\sqrt{2}}{a} \left(sin\left(\frac{\pi x_1}{a}\right)sin\left(\frac{2 \pi x_2}{a}\right) + sin\left(\frac{2\pi x_1}{a}\right)sin\left(\frac{\pi x_2}{a}\right)\right)]^2 \delta(x_1 - x_2) dx_1 dx_2</math> | ||
<math>E^1_2 = -aV_0 (\frac{2}{a^2}) \int_0^a [sin(\frac{\pi x}{a})sin(\frac{2 \pi x}{a}) + sin(\frac{2\pi x}{a})sin(\frac{\pi x}{a})]^2 dx</math> | <math>E^1_2 = -aV_0 \left(\frac{2}{a^2}\right) \int_0^a [sin\left(\frac{\pi x}{a}\right)sin\left(\frac{2 \pi x}{a}\right) + sin\left(\frac{2\pi x}{a}\right)sin\left(\frac{\pi x}{a}\right)]^2 dx</math> | ||
<math>E^1_2 = -\frac{2V_0}{a} \int_0^a [4sin^2(\frac{\pi x}{a})sin^2(\frac{2 \pi x}{a})] dx</math> | <math>E^1_2 = -\frac{2V_0}{a} \int_0^a [4sin^2\left(\frac{\pi x}{a}\right)sin^2\left(\frac{2 \pi x}{a}\right)] dx</math> | ||
Use the same substitution as before <math>y = \frac{\pi x}{a}</math> | Use the same substitution as before <math>y = \frac{\pi x}{a}</math> | ||
<math>E^1_2 = -\frac{8V_0}{a} (\frac{a}{\pi}) \int_0^\pi [sin^2(y) sin^2 (2y) dy] </math> | <math>E^1_2 = -\frac{8V_0}{a} \left(\frac{a}{\pi}\right) \int_0^\pi [sin^2(y) sin^2 (2y) dy] </math> | ||
use the identity <math> sin(2y) = 2cos(y)sin(y)</math> such that <math>sin^2(2y) = 4 cos^2(y) sin^2(y)</math> | use the identity <math> sin(2y) = 2cos(y)sin(y)</math> such that <math>sin^2(2y) = 4 cos^2(y) sin^2(y)</math> | ||
| Line 74: | Line 77: | ||
The <math>sin^4(y)</math> integrand is the same as before and yields <math>\frac{3\pi}{8}</math>. The <math>sin^6(y)</math> integrand is <math>\frac{5y}{16} - \frac{15}{64} sin(2y) + \frac{3}{64} sin (4y) - \frac{1}{192} sin(6y)</math> where all the sine terms are zero so it just yields <math>\frac{5 \pi}{16}</math>. | The <math>sin^4(y)</math> integrand is the same as before and yields <math>\frac{3\pi}{8}</math>. The <math>sin^6(y)</math> integrand is <math>\frac{5y}{16} - \frac{15}{64} sin(2y) + \frac{3}{64} sin (4y) - \frac{1}{192} sin(6y)</math> where all the sine terms are zero so it just yields <math>\frac{5 \pi}{16}</math>. | ||
<math>E^1_2 = -\frac{32V_0}{\pi} (\frac{3 \pi}{8} - \frac{5 \pi}{16}) = -2V_0</math> | <math>E^1_2 = -\frac{32V_0}{\pi} \left(\frac{3 \pi}{8} - \frac{5 \pi}{16}\right) = -2V_0</math> | ||
Revision as of 22:10, 9 April 2010
From David J. Griffiths "Introduction to Quantum Mechanics" Problem 6.3
"Two identical bosons are placed in an infinite square well. They interact weakly with one another, via the potential Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V(x_1, x_2) = -aV_0 \delta(x_1 - x_2)} , where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V_0} is a constant with the dimensions of energy, and a is the width of the well).
a)First, ignoring the interaction between the particles, find the ground state and the first excited state -- both the wave function and the associated energies.
b) Use first-order perturbation theory to estimate the effect of the particle-particle interaction on the energies of the ground state and the first excited state.
For the infinite square well the unperturbed energy is given by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E^{0}_n = \frac{n^{2}\pi^{2}\hbar^{2}}{2ma^2}} and the ground state wave function is given by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Psi^{0}_n = \sqrt{\frac{2}{a}}sin\left(\frac{n\pi x}{a}\right)} .
a)
For the case of identical bosons with no interaction between them the ground state is given by the following: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi^0 _{11} (x) = \psi^0_1(x_1)\psi^0_1(x_2)}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi^0 _{11} (x) = \sqrt{\frac{2}{a}}sin\left(\frac{\pi x_1}{a}\right)\sqrt{\frac{2}{a}}sin\left(\frac{\pi x_2}{a}\right)}
While the energy is given by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E^0_{1total} = E^0_1(x_1) + E^0_1(x_2)} .
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E^0_1 = 2E^0_1 = (2)\frac{\pi^2 \hbar^2}{2ma^2} = \frac{\pi^2 \hbar^2}{ma^2}}
The first excited state is given by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi^0_{12}(x) = \frac{1}{\sqrt{2}}\left[ \psi^0_1(x_1)\psi^0_2(x_2) + \psi^0_1(x_2)\psi^0_2(x_1)\right]}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi^0_{12} = \frac{1}{\sqrt{2}}[ \sqrt{\frac{2}{a}}sin\left(\frac{\pi x_1}{a}\right)\sqrt{\frac{2}{a}}sin\left(\frac{2\pi x_2}{a}\right) + \sqrt{\frac{2}{a}}sin \left(\frac{2\pi x_1}{a}\right)\sqrt{\frac{2}{a}}sin\left(\frac{\pi x_2}{a}\right)]}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi^0_{12} = \frac{\sqrt{2}}{a}[sin\left(\frac{\pi x_1}{a}\right)sin\left(\frac{2 \pi x_2}{a}\right) + sin\left(\frac{2\pi x_1}{a}\right)sin\left(\frac{\pi x_2}{a}\right)]}
While the energy is given by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E^0_2 = E^0_2(x_1) + E^0_2(x_2)}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E^0_2 = \frac{\pi ^2 \hbar^2}{2ma^2} + \frac{4\pi^2 \hbar^2}{2ma^2} = \frac{5 \pi^2 \hbar^2}{2ma^2}}
b)
The first order perturbation, the energy correction is given by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E^1_n = \langle \psi^1_n | H' | \psi^1_n \rangle} for the ground state.
The interaction potential given in the problem is the perturbed part of the Hamiltonian. Therefore the first order perturbation in integral form for the ground state is given by:
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E^1_1 = \int \psi^1_1 [-aV_0 \delta(x_1 - x_2) ] \psi ^1_1 dx} , where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi^1_1 = \frac{2}{a} sin\left(\frac{\pi x_1}{a}\right) sin\left(\frac{\pi x_2}{a}\right)}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E^1_1 = \int_0^a \int_0^a \left(\frac{2}{a}\right)^2 sin^2\left(\frac{\pi x_1}{a}\right) sin^2\left(\frac{\pi x_2}{a}\right) [-a V_0 \delta(x_1 - x_2) ] dx_1 dx_2}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E^1_1 = -aV_0 \left(\frac{2}{a}\right)^2 \int_0^a sin^4 \left(\frac{ \pi x}{a}\right) dx}
Then do a substitution such that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y = \frac{\pi x}{a}}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E^1_1 = -aV_0 \left(\frac{2}{a}\right)^2 \frac{a}{\pi} \int_0^\pi sin^4y dy}
The integral of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle sin^4(y)} is given by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{32} (12y - 8 sin(2y) + sin(4y))} . Since Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y= \pi} in this case both sine terms go to zero and the integrand yields Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{3 \pi}{8}} .
Therefore the first order correction to the energy is Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E^1_1 = -\frac{3}{2} V_0 }
As for the first order perturbation to the energy for the first excited state the wavefunction is given as
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi^0_{12} = \frac{\sqrt{2}}{a}[sin\left(\frac{\pi x_1}{a}\right)sin\left(\frac{2 \pi x_2}{a}\right) + sin\left(\frac{2\pi x_1}{a}\right)sin\left(\frac{\pi x_2}{a}\right)]}
Note that if this system was for fermions then this wavefunction would be zero since there would be a minus sign in place of the plus sign.
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E^1_2 = -aV_0 \int_0^a \int_0^a [\frac{\sqrt{2}}{a} \left(sin\left(\frac{\pi x_1}{a}\right)sin\left(\frac{2 \pi x_2}{a}\right) + sin\left(\frac{2\pi x_1}{a}\right)sin\left(\frac{\pi x_2}{a}\right)\right)]^2 \delta(x_1 - x_2) dx_1 dx_2}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E^1_2 = -aV_0 \left(\frac{2}{a^2}\right) \int_0^a [sin\left(\frac{\pi x}{a}\right)sin\left(\frac{2 \pi x}{a}\right) + sin\left(\frac{2\pi x}{a}\right)sin\left(\frac{\pi x}{a}\right)]^2 dx}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E^1_2 = -\frac{2V_0}{a} \int_0^a [4sin^2\left(\frac{\pi x}{a}\right)sin^2\left(\frac{2 \pi x}{a}\right)] dx}
Use the same substitution as before Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y = \frac{\pi x}{a}}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E^1_2 = -\frac{8V_0}{a} \left(\frac{a}{\pi}\right) \int_0^\pi [sin^2(y) sin^2 (2y) dy] }
use the identity Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle sin(2y) = 2cos(y)sin(y)} such that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle sin^2(2y) = 4 cos^2(y) sin^2(y)}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E^1_2 = -\frac{8V_0}{ \pi} 4 \int_0^ \pi sin^2(y) sin^2(y) cos^2(y) = - \frac{32 V_0}{ \pi} \int_0^\pi (sin^4(y) - sin^6(y)) dy}
The Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle sin^4(y)} integrand is the same as before and yields Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{3\pi}{8}} . The Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle sin^6(y)} integrand is Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{5y}{16} - \frac{15}{64} sin(2y) + \frac{3}{64} sin (4y) - \frac{1}{192} sin(6y)} where all the sine terms are zero so it just yields Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{5 \pi}{16}} .
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E^1_2 = -\frac{32V_0}{\pi} \left(\frac{3 \pi}{8} - \frac{5 \pi}{16}\right) = -2V_0}