With the postulates of Quantum mechanics, we can proceed to start analyzing simple problems such as Harmonic oscillator. The reason we used to chose Harmonic oscillator because it is the one of the simplest classical problem for which we analysed the complete solution.
Note: One dimensional harmonic oscillator is considered.
We always used to start with the
Time Independent Schrodinger equation, where the potential is $ \frac{1}{2}kx^2 = \frac{1}{2} m\omega^2 x^2 $ , where $ \omega = \sqrt{\frac{k}{m}} $,
$$ \frac{-h^2}{2m} \frac{\partial^2 \psi(x)}{\partial{x^2}} + \frac{1}{2} m \omega^2 x^2 \psi(x) = E \psi(x) \,\,\,...eq.(1)$$
There are two methods to solve this differential equation and first we will look at the usual "Frobenius series method".
Before that, we should make some alterations so that the given equation will be more simplified.
Eq.(1) can be written as, $$ \frac{\partial^2\psi(x)}{\partial{x^2}} + \frac{-2m^2\omega^2x^2}{2\hbar^2}\psi(x) = \frac{-2mE}{\hbar^2}\psi(x)$$ $\rightarrow$ $$ \frac{\partial^2\psi(x)}{\partial{x^2}} + \frac{2mE}{\hbar^2}\psi(x) - \frac{m^2\omega^2x^2}{\hbar^2}\psi(x) = 0 \,\,\,...eq.(2) $$
To imply infinite series solutions (that includes all powers of x), we would like non-dimensionalize the above equation. Let us introduce a new dimensionless quantity $\rho = \alpha x$ where $\alpha$ should have the dimension that is inverse of the length. Using this relation wave function could be written in terms of "$\rho$" and using the chain rule, $$\frac{\partial\psi}{\partial{x}} = \frac{\partial\psi}{\partial\rho} \frac{\partial \rho}{\partial{x}} $$
that gives, $$ \frac{\partial\psi}{\partial{x}} = \frac{\partial\psi}{\partial\rho}\, \alpha $$ Again using chain rule for second derivative,
$\rightarrow$ $$ \frac{\partial^2\psi}{\partial{x^2}} = \frac{\partial}{\partial\rho} \left(\frac{\partial\psi}{\partial{x}}\right) \frac{\partial\rho}{\partial{x}} = \frac{\partial}{\partial\rho} \left(\alpha \frac{\partial\psi}{\partial\rho}\right) \alpha $$ $\rightarrow$ $$ \frac{\partial^2\psi}{\partial{x^2}} = \alpha^2 \frac{\partial^2\psi}{\partial\rho^2} \,\,\,...eq.(3)$$
Applying it in eq.(2), we get, $$ \alpha^2 \frac{\partial^2\psi}{\partial\rho^2} + \frac{2mE}{\hbar^2}\psi - \frac{m^2 \omega^2 \rho^2}{\hbar^2 \alpha^2}\psi = 0 $$
$\rightarrow$ $$ \frac{\partial^2\psi}{\partial\rho^2} + \frac{2mE} {\hbar^2 \alpha^2} \psi - \frac{m^2\omega^2\rho^2}{\hbar^2 \alpha^4}\psi = 0 \,\,\,...eq.(4)$$
From dimensional analysis we can see that the only combination of the constants "$ m,\, \omega,\,, \hbar\,$ to give the dimension of inverse of length is $ \sqrt{\frac{m\omega}{\hbar}} = \alpha $ . In other way, we we would like the make the coefficient equal to unity so that it would have no dimensions. Substituting for $\alpha $ value, we get, $$\frac{\partial^2\psi}{\partial\rho^2} + \frac{2mE\hbar}{\hbar^2 m\omega } \psi - \rho^2\psi = 0 $$ $\rightarrow$ $$ \frac{\partial^2\psi} {\partial\rho^2} + \frac{2E}{\hbar\omega}\psi - \rho^2\psi = 0 $$
Calling, $ \frac{2E}{\hbar\omega} = \lambda $ we arrive at our non dimensionalized equation, $$ \frac{\partial^2\psi}{\partial^2\rho} + (\lambda - \rho^2)\psi = 0 \,\,\,...eq.(5)$$
where $\lambda$ is also a dimensionless quantity.
Once we get, eq.(5), we would like to find the solution. But a close looking at the equation gives, for large values of $\rho $ the equation reduces to , $$ \frac{\partial^2\psi} {\partial\rho^2} = (\rho^2 - \lambda) \psi \approx \rho^2 \psi \,\,\,...eq.(6) $$
Eq.(6) has the approximate solution that is given by, $$ \psi(\rho) = A e^{\frac{-\rho^2}{2}} + B e^{\frac{\rho^2}{2}} $$ Since the wave function should be normalizable, the solution will have only, $$ \psi(\rho) = A e^{\frac{-\rho^2}{2}} = u(\rho) e^{\frac{-\rho^2}{2}}$$
Applying this as a trial solution in eq.(6) with corresponding derivatives gives us, $$ \left(\frac{\partial^2u}{\partial\rho^2} - 2\rho \frac{\partial{u}}{\partial\rho} + (\rho^2 - 1)u \right) e^{\frac{-\rho^2}{2}} + (\lambda - \rho^2) e^{\frac{-\rho^2}{2}} = 0 $$
$\rightarrow$ $$ \left[\frac{\partial^2u}{\partial\rho^2} - 2\rho\frac{\partial{u}}{\partial\rho} + (\lambda - 1) u\right] e^{\frac{-\rho^2}{2}} = 0 \,\,\,...eq.(7)$$
Thus we need to find $ u(\rho) $ such that, $$\frac{\partial^2u(\rho)}{\partial\rho^2} - 2\rho\frac{\partial{u(\rho)}}{\partial\rho} + (\lambda - 1) u(\rho) = 0 \,\,\,...eq.(8)$$
Eq.(8) is the known Hermite equation for which we know the solutions are Hermite polynomials.
For the solution of Hermite polynomials, see Hermite Polynomials derivation
Thus our general solution is, $$ \psi_n(\rho) = A_n H_n(\rho) e^{\frac{-\rho^2}{2}} \,\,\,...eq.(9)$$where the condition is that $ \lambda = 2n + 1 $ which implies , $$ E = (n + \frac{1}{2}) \hbar \omega \,\,\,...eq.(10)\\~\\ n = 0,1,2,...$$
To determine the constant $ A_n $ we normalize the wave function as,
$$ \int_{-\infty}^{\infty} |\psi(\rho)|^2 \, d\rho = 1 $$ since $ \rho = \alpha x $ it is as same as integrating from $ x = (-\infty,+\infty) $
We can choose A_n such that it satifies the following condition,
$\rightarrow$ $$ \frac{A_n^2}{\alpha} \,\int_{-\infty}^{\infty} H_n^2(\rho) e^{-\rho^2} \, d\rho = 1 \,\,\,...eq.(11)$$
But using the generating function of Hermite polynomials, $$ e^{-x^2 + 2x\rho} = \sum_{n=0}^{\infty}\frac{ H_n(\rho)}{n!} x^n \,\,\,...eq.(12)$$
$\rightarrow$ $$ \int _{-\infty}^{\infty} e^{-x^2 + 2x\rho} e^{-y^2+2y\rho} e^{-\rho^2} \, d\rho = \sum_{n=0}^{\infty} \sum_{m=0}^{\infty} \frac{x^n y^m}{n!m!} \int_{-\infty}^{\infty} H_n(\rho) H_m(\rho) e^{-\rho^2} \,d\rho \,\,..eq.(13)$$
$\rightarrow$
But the left hand side can be integrated to give,
$\rightarrow$ $$ \int_{-\infty}^{\infty} e^{-\rho^2+2(x+y)\rho-(x^2+y^2)} \,d\rho = \sqrt{\frac{\pi}{1}} e^{(x+y)^2-(x^2+y^2)} = \sqrt{\pi}\, e^{2xy} \,\,\,...eq.(14)$$
where we used the formula, $$ \int_{-\infty}^{\infty} e^{-ax^2 +bx+ c } \, dx = \sqrt{\frac{\pi}{a}} e^{\left(\frac{b^2}{4a}+c\right)} $$
where the comparison gives that, a = 1 , b = 2(x+y) , $ c = -(x^2+y^2)$
Again rewriting eq.(14) in series form,
$$ \sqrt {\pi} \, e^{2xy} = \sqrt{\pi}\, \sum_{n=0}^{\infty} \frac{(2xy)^n}{n!} $$ Using this, eq.(13) becomes, $$\sum_{n=0}^{\infty} \sum_{m=0}^{\infty} \frac{x^n y^m}{n!m!} \int_{-\infty}^{\infty} H_n(\rho) H_m(\rho) e^{-\rho^2} \,d\rho = \sqrt{\pi}\, \sum_{n=0}^{\infty} \frac{(2xy)^n}{n!} $$
By making m =n,
$$\sum_{n=0}^{\infty} \sum_{n=0}^{\infty} \frac{x^n y^n}{n!n!} \int_{-\infty}^{\infty} H_n(\rho) H_n(\rho) e^{-\rho^2} \,d\rho =\sqrt{\pi}\, \sum_{n=0}^{\infty} \frac{(2xy)^n}{n!} $$
Equal powers of the series are equated to give,
$$\int_{-\infty}^{\infty} H_n^2(\rho) e^{-\rho^2} \,d\rho = \sqrt{\pi}\,2^n\, n! \,\,\,...eq.(15)$$ and $$\int_{-\infty}^{\infty} H_n^2(\rho) \,e^{-\rho^2} \,d\rho = 0 \\~\\ for\,\, m\neq n$$ Substituting eq.(15) in eq.(11) gives, $$ \frac{A_n^2}{\alpha}\int_{-\infty}^{\infty} H_n^2(\rho) \,e^{-\rho^2} \,d\rho = \frac{A_n^2}{\alpha} \,\sqrt{\pi}\, 2^n\, n! = 1$$ which finally gives the value of the constant $$ A_n = \sqrt{\frac{\alpha}{\sqrt{\pi} \, 2^n \, n!}} $$
That's it! We arrived our final solution, i.e. the normalized stationary states of the quantum harmonic oscillator is,
$$ \psi_n(\rho) = \sqrt{\frac{\alpha}{\sqrt{\pi} \, 2^n \, n! }} \,H_n(\rho) \,e^{\frac{-\rho^2}{2}} \,\,\,...eq.(16)$$
Note: One dimensional harmonic oscillator is considered.
We always used to start with the
Time Independent Schrodinger equation, where the potential is $ \frac{1}{2}kx^2 = \frac{1}{2} m\omega^2 x^2 $ , where $ \omega = \sqrt{\frac{k}{m}} $,
$$ \frac{-h^2}{2m} \frac{\partial^2 \psi(x)}{\partial{x^2}} + \frac{1}{2} m \omega^2 x^2 \psi(x) = E \psi(x) \,\,\,...eq.(1)$$
There are two methods to solve this differential equation and first we will look at the usual "Frobenius series method".
Before that, we should make some alterations so that the given equation will be more simplified.
Eq.(1) can be written as, $$ \frac{\partial^2\psi(x)}{\partial{x^2}} + \frac{-2m^2\omega^2x^2}{2\hbar^2}\psi(x) = \frac{-2mE}{\hbar^2}\psi(x)$$ $\rightarrow$ $$ \frac{\partial^2\psi(x)}{\partial{x^2}} + \frac{2mE}{\hbar^2}\psi(x) - \frac{m^2\omega^2x^2}{\hbar^2}\psi(x) = 0 \,\,\,...eq.(2) $$
To imply infinite series solutions (that includes all powers of x), we would like non-dimensionalize the above equation. Let us introduce a new dimensionless quantity $\rho = \alpha x$ where $\alpha$ should have the dimension that is inverse of the length. Using this relation wave function could be written in terms of "$\rho$" and using the chain rule, $$\frac{\partial\psi}{\partial{x}} = \frac{\partial\psi}{\partial\rho} \frac{\partial \rho}{\partial{x}} $$
that gives, $$ \frac{\partial\psi}{\partial{x}} = \frac{\partial\psi}{\partial\rho}\, \alpha $$ Again using chain rule for second derivative,
$\rightarrow$ $$ \frac{\partial^2\psi}{\partial{x^2}} = \frac{\partial}{\partial\rho} \left(\frac{\partial\psi}{\partial{x}}\right) \frac{\partial\rho}{\partial{x}} = \frac{\partial}{\partial\rho} \left(\alpha \frac{\partial\psi}{\partial\rho}\right) \alpha $$ $\rightarrow$ $$ \frac{\partial^2\psi}{\partial{x^2}} = \alpha^2 \frac{\partial^2\psi}{\partial\rho^2} \,\,\,...eq.(3)$$
Applying it in eq.(2), we get, $$ \alpha^2 \frac{\partial^2\psi}{\partial\rho^2} + \frac{2mE}{\hbar^2}\psi - \frac{m^2 \omega^2 \rho^2}{\hbar^2 \alpha^2}\psi = 0 $$
$\rightarrow$ $$ \frac{\partial^2\psi}{\partial\rho^2} + \frac{2mE} {\hbar^2 \alpha^2} \psi - \frac{m^2\omega^2\rho^2}{\hbar^2 \alpha^4}\psi = 0 \,\,\,...eq.(4)$$
From dimensional analysis we can see that the only combination of the constants "$ m,\, \omega,\,, \hbar\,$ to give the dimension of inverse of length is $ \sqrt{\frac{m\omega}{\hbar}} = \alpha $ . In other way, we we would like the make the coefficient equal to unity so that it would have no dimensions. Substituting for $\alpha $ value, we get, $$\frac{\partial^2\psi}{\partial\rho^2} + \frac{2mE\hbar}{\hbar^2 m\omega } \psi - \rho^2\psi = 0 $$ $\rightarrow$ $$ \frac{\partial^2\psi} {\partial\rho^2} + \frac{2E}{\hbar\omega}\psi - \rho^2\psi = 0 $$
Calling, $ \frac{2E}{\hbar\omega} = \lambda $ we arrive at our non dimensionalized equation, $$ \frac{\partial^2\psi}{\partial^2\rho} + (\lambda - \rho^2)\psi = 0 \,\,\,...eq.(5)$$
where $\lambda$ is also a dimensionless quantity.
Once we get, eq.(5), we would like to find the solution. But a close looking at the equation gives, for large values of $\rho $ the equation reduces to , $$ \frac{\partial^2\psi} {\partial\rho^2} = (\rho^2 - \lambda) \psi \approx \rho^2 \psi \,\,\,...eq.(6) $$
Eq.(6) has the approximate solution that is given by, $$ \psi(\rho) = A e^{\frac{-\rho^2}{2}} + B e^{\frac{\rho^2}{2}} $$ Since the wave function should be normalizable, the solution will have only, $$ \psi(\rho) = A e^{\frac{-\rho^2}{2}} = u(\rho) e^{\frac{-\rho^2}{2}}$$
Applying this as a trial solution in eq.(6) with corresponding derivatives gives us, $$ \left(\frac{\partial^2u}{\partial\rho^2} - 2\rho \frac{\partial{u}}{\partial\rho} + (\rho^2 - 1)u \right) e^{\frac{-\rho^2}{2}} + (\lambda - \rho^2) e^{\frac{-\rho^2}{2}} = 0 $$
$\rightarrow$ $$ \left[\frac{\partial^2u}{\partial\rho^2} - 2\rho\frac{\partial{u}}{\partial\rho} + (\lambda - 1) u\right] e^{\frac{-\rho^2}{2}} = 0 \,\,\,...eq.(7)$$
Thus we need to find $ u(\rho) $ such that, $$\frac{\partial^2u(\rho)}{\partial\rho^2} - 2\rho\frac{\partial{u(\rho)}}{\partial\rho} + (\lambda - 1) u(\rho) = 0 \,\,\,...eq.(8)$$
Eq.(8) is the known Hermite equation for which we know the solutions are Hermite polynomials.
For the solution of Hermite polynomials, see Hermite Polynomials derivation
Thus our general solution is, $$ \psi_n(\rho) = A_n H_n(\rho) e^{\frac{-\rho^2}{2}} \,\,\,...eq.(9)$$where the condition is that $ \lambda = 2n + 1 $ which implies , $$ E = (n + \frac{1}{2}) \hbar \omega \,\,\,...eq.(10)\\~\\ n = 0,1,2,...$$
To determine the constant $ A_n $ we normalize the wave function as,
$$ \int_{-\infty}^{\infty} |\psi(\rho)|^2 \, d\rho = 1 $$ since $ \rho = \alpha x $ it is as same as integrating from $ x = (-\infty,+\infty) $
We can choose A_n such that it satifies the following condition,
$\rightarrow$ $$ \frac{A_n^2}{\alpha} \,\int_{-\infty}^{\infty} H_n^2(\rho) e^{-\rho^2} \, d\rho = 1 \,\,\,...eq.(11)$$
But using the generating function of Hermite polynomials, $$ e^{-x^2 + 2x\rho} = \sum_{n=0}^{\infty}\frac{ H_n(\rho)}{n!} x^n \,\,\,...eq.(12)$$
$\rightarrow$ $$ \int _{-\infty}^{\infty} e^{-x^2 + 2x\rho} e^{-y^2+2y\rho} e^{-\rho^2} \, d\rho = \sum_{n=0}^{\infty} \sum_{m=0}^{\infty} \frac{x^n y^m}{n!m!} \int_{-\infty}^{\infty} H_n(\rho) H_m(\rho) e^{-\rho^2} \,d\rho \,\,..eq.(13)$$
$\rightarrow$
But the left hand side can be integrated to give,
$\rightarrow$ $$ \int_{-\infty}^{\infty} e^{-\rho^2+2(x+y)\rho-(x^2+y^2)} \,d\rho = \sqrt{\frac{\pi}{1}} e^{(x+y)^2-(x^2+y^2)} = \sqrt{\pi}\, e^{2xy} \,\,\,...eq.(14)$$
where we used the formula, $$ \int_{-\infty}^{\infty} e^{-ax^2 +bx+ c } \, dx = \sqrt{\frac{\pi}{a}} e^{\left(\frac{b^2}{4a}+c\right)} $$
where the comparison gives that, a = 1 , b = 2(x+y) , $ c = -(x^2+y^2)$
Again rewriting eq.(14) in series form,
$$ \sqrt {\pi} \, e^{2xy} = \sqrt{\pi}\, \sum_{n=0}^{\infty} \frac{(2xy)^n}{n!} $$ Using this, eq.(13) becomes, $$\sum_{n=0}^{\infty} \sum_{m=0}^{\infty} \frac{x^n y^m}{n!m!} \int_{-\infty}^{\infty} H_n(\rho) H_m(\rho) e^{-\rho^2} \,d\rho = \sqrt{\pi}\, \sum_{n=0}^{\infty} \frac{(2xy)^n}{n!} $$
By making m =n,
$$\sum_{n=0}^{\infty} \sum_{n=0}^{\infty} \frac{x^n y^n}{n!n!} \int_{-\infty}^{\infty} H_n(\rho) H_n(\rho) e^{-\rho^2} \,d\rho =\sqrt{\pi}\, \sum_{n=0}^{\infty} \frac{(2xy)^n}{n!} $$
Equal powers of the series are equated to give,
$$\int_{-\infty}^{\infty} H_n^2(\rho) e^{-\rho^2} \,d\rho = \sqrt{\pi}\,2^n\, n! \,\,\,...eq.(15)$$ and $$\int_{-\infty}^{\infty} H_n^2(\rho) \,e^{-\rho^2} \,d\rho = 0 \\~\\ for\,\, m\neq n$$ Substituting eq.(15) in eq.(11) gives, $$ \frac{A_n^2}{\alpha}\int_{-\infty}^{\infty} H_n^2(\rho) \,e^{-\rho^2} \,d\rho = \frac{A_n^2}{\alpha} \,\sqrt{\pi}\, 2^n\, n! = 1$$ which finally gives the value of the constant $$ A_n = \sqrt{\frac{\alpha}{\sqrt{\pi} \, 2^n \, n!}} $$
That's it! We arrived our final solution, i.e. the normalized stationary states of the quantum harmonic oscillator is,
$$ \psi_n(\rho) = \sqrt{\frac{\alpha}{\sqrt{\pi} \, 2^n \, n! }} \,H_n(\rho) \,e^{\frac{-\rho^2}{2}} \,\,\,...eq.(16)$$
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