Chapter 5, Problem 5.38P

(a)

To determine

Prove that for integers $k$ and $k\text{'}$ between $1$ and $N-1$, $\frac{1}{N}{\displaystyle \sum _{j=1}^N{e}^{i2\pi \left(k-k\text{'}\right)j/N}}={\delta }_{k\text{'},k}$ and $\frac{1}{N}{\displaystyle \sum _{j=1}^N{e}^{i2\pi \left(k+k\text{'}\right)j/N}}={\delta }_{k\text{'},N-k}$.

Answer to Problem 5.38P

It has been proved that for integers $k$ and $k\text{'}$ between $1$ and $N-1$, $\frac{1}{N}{\displaystyle \sum _{j=1}^N{e}^{i2\pi \left(k-k\text{'}\right)j/N}}={\delta }_{k\text{'},k}$ and $\frac{1}{N}{\displaystyle \sum _{j=1}^N{e}^{i2\pi \left(k+k\text{'}\right)j/N}}={\delta }_{k\text{'},N-k}$.

Explanation of Solution

Consider

$\begin{array}{c}\frac{1}{N}{\displaystyle \sum _{j=1}^N{e}^{i2\pi rj/N}}=\frac{1}{N}{\displaystyle \sum _{j=1}^N{\left(e^{i2\pi r/N}\right)}^j}\\ =\frac{1}{N}\frac{e^{2i\pi r}-1}{1-e^{i2\pi r/N}}\end{array}$

Since, $e^{2i\pi }=1$. For any integer $r$ the numerator becomes zero. And the sum must also be zero unless the denominator also vanishes.

The denominator vanishes only when $r=mN$ where $m$ is a integer.

For $\frac{1}{N}{\displaystyle \sum _{j=1}^N{e}^{i2\pi \left(k-k\text{'}\right)j/N}}={\delta }_{k\text{'},k}$,  $k-k\text{'}=mN$ and $m=0$. Thus, $k=k\text{'}$.

For $\frac{1}{N}{\displaystyle \sum _{j=1}^N{e}^{i2\pi \left(k+k\text{'}\right)j/N}}={\delta }_{k\text{'},N-k}$, $k+k\text{'}=mN$ and $m=1$. Thus $k\text{'}=N-k$.

Conclusion:

It has been proved that for integers $k$ and $k\text{'}$ between $1$ and $N-1$, $\frac{1}{N}{\displaystyle \sum _{j=1}^N{e}^{i2\pi \left(k-k\text{'}\right)j/N}}={\delta }_{k\text{'},k}$ and $\frac{1}{N}{\displaystyle \sum _{j=1}^N{e}^{i2\pi \left(k+k\text{'}\right)j/N}}={\delta }_{k\text{'},N-k}$.

(b)

To determine

The commutation relations for the ladder operators, $\left[{\widehat{a}}_{k-},{\widehat{a}}_{k\text{'}+}\right]={\delta }_{k,k\text{'}}$ and $\left[{\widehat{a}}_{k-},{\widehat{a}}_{k\text{'}-}\right]=\left[{\widehat{a}}_{k+},{\widehat{a}}_{k\text{'}+}\right]=0$.

Answer to Problem 5.38P

The commutation relations for the ladder operators, $\left[{\widehat{a}}_{k-},{\widehat{a}}_{k\text{'}+}\right]={\delta }_{k,k\text{'}}$ and $\left[{\widehat{a}}_{k-},{\widehat{a}}_{k\text{'}-}\right]=\left[{\widehat{a}}_{k+},{\widehat{a}}_{k\text{'}+}\right]=0$.

Explanation of Solution

Given the ladder operator

${\widehat{a}}_{k\pm }=\frac{1}{\sqrt{N}}{\displaystyle \sum _{j=1}^N{e}^{\pm i2\pi jk/N}}\left[\sqrt{\frac{m{\omega }_k}{2\hslash }}{x}_j\mp \sqrt{\frac{\hslash }{2m{\omega }_k}}\frac{\partial }{\partial x_j}\right]$        (I)

Solving for $\left[{\widehat{a}}_{k-},{\widehat{a}}_{k\text{'}+}\right]$ using equation (I)

$\begin{array}{c}\left[{\widehat{a}}_{k-},{\widehat{a}}_{k\text{'}+}\right]={\widehat{a}}_{k-}{\widehat{a}}_{k\text{'}+}-{\widehat{a}}_{k\text{'}+}{\widehat{a}}_{k-}\\ =\left\{\frac{1}{\sqrt{N}}{\displaystyle \sum _j{e}^{-i2\pi jk/N}\left[\sqrt{\frac{m{\omega }_k}{2\hslash }}{x}_j+\sqrt{\frac{\hslash }{2m{\omega }_k}}\frac{\partial }{\partial x_j}\right]}\right\}\left\{\frac{1}{\sqrt{N}}{\displaystyle \sum _{j,j\text{'}}{e}^{i2\pi j\text{'}k\text{'}/N}\left[\sqrt{\frac{m{\omega }_{k\text{'}}}{2\hslash }}{x}_{j\text{'}}\_\sqrt{\frac{\hslash }{2m{\omega }_{k\text{'}}}}\frac{\partial }{\partial x_{j\text{'}}}\right]}\right\}\\ -\left\{\frac{1}{\sqrt{N}}{\displaystyle \sum _{j,j\text{'}}{e}^{i2\pi j\text{'}k\text{'}/N}\left[\sqrt{\frac{m{\omega }_{k\text{'}}}{2\hslash }}{x}_{j\text{'}}\_\sqrt{\frac{\hslash }{2m{\omega }_{k\text{'}}}}\frac{\partial }{\partial x_{j\text{'}}}\right]}\right\}\left\{\frac{1}{\sqrt{N}}{\displaystyle \sum _j{e}^{-i2\pi jk/N}\left[\sqrt{\frac{m{\omega }_k}{2\hslash }}{x}_j+\sqrt{\frac{\hslash }{2m{\omega }_k}}\frac{\partial }{\partial x_j}\right]}\right\}\\ =\frac{1}{N}{\displaystyle \sum _{j,j\text{'}}{e}^{-i2\pi jk/N}{e}^{i2\pi j\text{'}k\text{'}/N}\left[\sqrt{\frac{{\omega }_k}{{\omega }_{k\text{'}}}}\left[x_j,\frac{\partial }{\partial x_{j\text{'}}}\right]+\sqrt{\frac{{\omega }_{k\text{'}}}{{\omega }_k}}\left[\frac{\partial }{\partial x_j},x_j\right]\right]}\\ =\frac{1}{N}{\displaystyle \sum _{j,j\text{'}}{e}^{i2\pi \left(j\text{'}k\text{'}-jk\right)/N}\left[\sqrt{\frac{{\omega }_k}{{\omega }_{k\text{'}}}}\left[x_j,\frac{\partial }{\partial x_{j\text{'}}}\right]+\sqrt{\frac{{\omega }_{k\text{'}}}{{\omega }_k}}\left[\frac{\partial }{\partial x_j},x_j\right]\right]}\end{array}$

Dropping the two terms involving commutators of two coordinates or two derivatives. The remaining commutators are $-{\delta }_{jj\text{'}}$ and ${\delta }_{jj\text{'}}$. The above equation becomes

$\begin{array}{c}\left[{\widehat{a}}_{k-},{\widehat{a}}_{k\text{'}+}\right]=\frac{1}{N}{\displaystyle \sum _j{e}^{i2\pi j\left(k\text{'}-k\right)/N}\left[\sqrt{\frac{{\omega }_k}{{\omega }_{k\text{'}}}}+\sqrt{\frac{{\omega }_{k\text{'}}}{{\omega }_k}}\right]}\\ =\frac{1}{N}{\displaystyle \sum _j{e}^{i2\pi j\left(k\text{'}-k\right)/N}\left[1\right]}\\ =\frac{1}{N}{\displaystyle \sum _j{e}^{i2\pi j\left(k\text{'}-k\right)/N}}\\ \left[{\widehat{a}}_{k-},{\widehat{a}}_{k\text{'}+}\right]={\delta }_{k,k\text{'}}\end{array}$

Hence it is proved that $\left[{\widehat{a}}_{k-},{\widehat{a}}_{k\text{'}+}\right]={\delta }_{k,k\text{'}}$.

Similarly solving for $\left[{\widehat{a}}_{k-},{\widehat{a}}_{k\text{'}-}\right]$ using equation (I)

$\begin{array}{c}\left[{\widehat{a}}_{k-},{\widehat{a}}_{k\text{'}-}\right]={\widehat{a}}_{k-}{\widehat{a}}_{k\text{'}-}-{\widehat{a}}_{k\text{'}-}{\widehat{a}}_{k-}\\ =\left\{\frac{1}{\sqrt{N}}{\displaystyle \sum _j{e}^{-i2\pi jk/N}\left[\sqrt{\frac{m{\omega }_k}{2\hslash }}{x}_j+\sqrt{\frac{\hslash }{2m{\omega }_k}}\frac{\partial }{\partial x_j}\right]}\right\}\left\{\frac{1}{\sqrt{N}}{\displaystyle \sum _{j,j\text{'}}{e}^{-i2\pi j\text{'}k\text{'}/N}\left[\sqrt{\frac{m{\omega }_{k\text{'}}}{2\hslash }}{x}_{j\text{'}}+\sqrt{\frac{\hslash }{2m{\omega }_{k\text{'}}}}\frac{\partial }{\partial x_{j\text{'}}}\right]}\right\}\\ -\left\{\frac{1}{\sqrt{N}}{\displaystyle \sum _{j,j\text{'}}{e}^{-i2\pi j\text{'}k\text{'}/N}\left[\sqrt{\frac{m{\omega }_{k\text{'}}}{2\hslash }}{x}_{j\text{'}}+\sqrt{\frac{\hslash }{2m{\omega }_{k\text{'}}}}\frac{\partial }{\partial x_{j\text{'}}}\right]}\right\}\left\{\frac{1}{\sqrt{N}}{\displaystyle \sum _j{e}^{-i2\pi jk/N}\left[\sqrt{\frac{m{\omega }_k}{2\hslash }}{x}_j+\sqrt{\frac{\hslash }{2m{\omega }_k}}\frac{\partial }{\partial x_j}\right]}\right\}\\ =\frac{1}{N}{\displaystyle \sum _{j,j\text{'}}{e}^{-i2\pi jk/N}{e}^{-i2\pi j\text{'}k\text{'}/N}\left[-\sqrt{\frac{{\omega }_k}{{\omega }_{k\text{'}}}}\left[x_j,\frac{\partial }{\partial x_{j\text{'}}}\right]+\sqrt{\frac{{\omega }_{k\text{'}}}{{\omega }_k}}\left[\frac{\partial }{\partial x_j},x_j\right]\right]}\\ =\frac{1}{N}{\displaystyle \sum _{j,j\text{'}}{e}^{-i2\pi \left(j\text{'}k\text{'}+jk\right)/N}\left[-\sqrt{\frac{{\omega }_k}{{\omega }_{k\text{'}}}}\left[x_j,\frac{\partial }{\partial x_{j\text{'}}}\right]+\sqrt{\frac{{\omega }_{k\text{'}}}{{\omega }_k}}\left[\frac{\partial }{\partial x_j},x_j\right]\right]}\end{array}$

Dropping the two terms involving commutators of two coordinates or two derivatives. The remaining commutators are $-{\delta }_{jj\text{'}}$ and ${\delta }_{jj\text{'}}$. The above equation becomes

$\left[{\widehat{a}}_{k-},{\widehat{a}}_{k\text{'}-}\right]=\frac{1}{N}{\displaystyle \sum _j{e}^{-i2\pi j\left(k\text{'}+k\right)/N}\left[-\sqrt{\frac{{\omega }_k}{{\omega }_{k\text{'}}}}+\sqrt{\frac{{\omega }_{k\text{'}}}{{\omega }_k}}\right]}$

Similarly solving for $\left[{\widehat{a}}_{k+},{\widehat{a}}_{k\text{'}+}\right]$ using equation (I)

$\begin{array}{c}\left[{\widehat{a}}_{k+},{\widehat{a}}_{k\text{'}+}\right]={\widehat{a}}_{k+}{\widehat{a}}_{k\text{'}+}-{\widehat{a}}_{k\text{'}+}{\widehat{a}}_{k+}\\ =\left\{\frac{1}{\sqrt{N}}{\displaystyle \sum _j{e}^{i2\pi jk/N}\left[\sqrt{\frac{m{\omega }_k}{2\hslash }}{x}_j-\sqrt{\frac{\hslash }{2m{\omega }_k}}\frac{\partial }{\partial x_j}\right]}\right\}\left\{\frac{1}{\sqrt{N}}{\displaystyle \sum _{j,j\text{'}}{e}^{i2\pi j\text{'}k\text{'}/N}\left[\sqrt{\frac{m{\omega }_{k\text{'}}}{2\hslash }}{x}_{j\text{'}}-\sqrt{\frac{\hslash }{2m{\omega }_{k\text{'}}}}\frac{\partial }{\partial x_{j\text{'}}}\right]}\right\}\\ -\left\{\frac{1}{\sqrt{N}}{\displaystyle \sum _{j,j\text{'}}{e}^{i2\pi j\text{'}k\text{'}/N}\left[\sqrt{\frac{m{\omega }_{k\text{'}}}{2\hslash }}{x}_{j\text{'}}-\sqrt{\frac{\hslash }{2m{\omega }_{k\text{'}}}}\frac{\partial }{\partial x_{j\text{'}}}\right]}\right\}\left\{\frac{1}{\sqrt{N}}{\displaystyle \sum _j{e}^{i2\pi jk/N}\left[\sqrt{\frac{m{\omega }_k}{2\hslash }}{x}_j-\sqrt{\frac{\hslash }{2m{\omega }_k}}\frac{\partial }{\partial x_j}\right]}\right\}\\ =\frac{1}{N}{\displaystyle \sum _{j,j\text{'}}{e}^{i2\pi jk/N}{e}^{i2\pi j\text{'}k\text{'}/N}\left[\sqrt{\frac{{\omega }_k}{{\omega }_{k\text{'}}}}\left[x_j,\frac{\partial }{\partial x_{j\text{'}}}\right]+\sqrt{\frac{{\omega }_{k\text{'}}}{{\omega }_k}}\left[\frac{\partial }{\partial x_j},x_j\right]\right]}\\ =\frac{1}{N}{\displaystyle \sum _{j,j\text{'}}{e}^{i2\pi \left(j\text{'}k\text{'}+jk\right)/N}\left[-\sqrt{\frac{{\omega }_k}{{\omega }_{k\text{'}}}}\left[x_j,\frac{\partial }{\partial x_{j\text{'}}}\right]+\sqrt{\frac{{\omega }_{k\text{'}}}{{\omega }_k}}\left[\frac{\partial }{\partial x_j},x_j\right]\right]}\end{array}$

Dropping the two terms involving commutators of two coordinates or two derivatives. The remaining commutators are $-{\delta }_{jj\text{'}}$ and ${\delta }_{jj\text{'}}$. The above equation becomes

$\left[{\widehat{a}}_{k+},{\widehat{a}}_{k\text{'}+}\right]=\frac{1}{N}{\displaystyle \sum _j{e}^{i2\pi j\left(k\text{'}+k\right)/N}\left[-\sqrt{\frac{{\omega }_k}{{\omega }_{k\text{'}}}}+\sqrt{\frac{{\omega }_{k\text{'}}}{{\omega }_k}}\right]}$

Equating the two commutator,

$\left[{\widehat{a}}_{k-},{\widehat{a}}_{k\text{'}-}\right]=\left[{\widehat{a}}_{k+},{\widehat{a}}_{k\text{'}+}\right]=0$

Conclusion:

Hence it is proved that the commutation relations for the ladder operators, $\left[{\widehat{a}}_{k-},{\widehat{a}}_{k\text{'}+}\right]={\delta }_{k,k\text{'}}$ and $\left[{\widehat{a}}_{k-},{\widehat{a}}_{k\text{'}-}\right]=\left[{\widehat{a}}_{k+},{\widehat{a}}_{k\text{'}+}\right]=0$.

(c)

To determine

Show that $x_j=R+\frac{1}{\sqrt{N}}{\displaystyle \sum _{k=1}^{N-1}\sqrt{\frac{\hslash }{2m{\omega }_k}}\left({\widehat{a}}_{k-}+{\widehat{a}}_{N-k+}\right)e^{i2\pi jk/N}}$ and $\frac{\partial }{\partial x_j}=\frac{1}{N}\frac{\partial }{\partial R}+\frac{1}{\sqrt{N}}{\displaystyle \sum _{k=1}^{N-1}\sqrt{\frac{m{\omega }_k}{2\hslash }}\left({\widehat{a}}_{k-}-{\widehat{a}}_{N-k+}\right)e^{i2\pi jk/N}}$.

Answer to Problem 5.38P

It is proved that $x_j=R+\frac{1}{\sqrt{N}}{\displaystyle \sum _{k=1}^{N-1}\sqrt{\frac{\hslash }{2m{\omega }_k}}\left({\widehat{a}}_{k-}+{\widehat{a}}_{N-k+}\right)e^{i2\pi jk/N}}$ and $\frac{\partial }{\partial x_j}=\frac{1}{N}\frac{\partial }{\partial R}+\frac{1}{\sqrt{N}}{\displaystyle \sum _{k=1}^{N-1}\sqrt{\frac{m{\omega }_k}{2\hslash }}\left({\widehat{a}}_{k-}-{\widehat{a}}_{N-k+}\right)e^{i2\pi jk/N}}$.

Explanation of Solution

Solving for $\left[{\widehat{a}}_{k-}+{\widehat{a}}_{N-k+}\right]$ using equation (I)

$\begin{array}{c}\left[{\widehat{a}}_{k-}+{\widehat{a}}_{N-k+}\right]={\widehat{a}}_{k-}{\widehat{a}}_{N-k+}-{\widehat{a}}_{k-}{\widehat{a}}_{N-k+}\\ =\left\{\frac{1}{\sqrt{N}}{\displaystyle \sum _j{e}^{-i2\pi jk/N}\left[\sqrt{\frac{m{\omega }_k}{2\hslash }}{x}_j+\sqrt{\frac{\hslash }{2m{\omega }_k}}\frac{\partial }{\partial x_j}\right]}\right\}\\ +\left\{\frac{1}{\sqrt{N}}{\displaystyle \sum _{j,j\text{'}}{e}^{i2\pi j\left(N-k\right)/N}\left[\sqrt{\frac{m{\omega }_{N-k}}{2\hslash }}{x}_j-\sqrt{\frac{\hslash }{2m{\omega }_{N-k}}}\frac{\partial }{\partial x_j}\right]}\right\}\end{array}$

Since ${\omega }_{N-k}={\omega }_k$ and $e^{i2\pi j\left(N-k\right)/N}=e^{i2\pi j}{e}^{-i2\pi jk/N}=e^{-i2\pi jk/N}$. The above equation becomes,

$\left[{\widehat{a}}_{k-}+{\widehat{a}}_{N-k+}\right]=\frac{2}{\sqrt{N}}{\displaystyle \sum _{j=1}^N\sqrt{\frac{m{\omega }_k}{2\hslash }}{e}^{-i2\pi jk/N}{x}_j}$        (II)

Using the above equation to solve for $\frac{1}{\sqrt{N}}{\displaystyle \sum _{k=1}^{N-1}\sqrt{\frac{\hslash }{2m{\omega }_k}}\left({\widehat{a}}_{k-}-{\widehat{a}}_{N-k+}\right)e^{i2\pi jk/N}}$

$\begin{array}{c}\frac{1}{\sqrt{N}}{\displaystyle \sum _{k=1}^{N-1}\sqrt{\frac{\hslash }{2m{\omega }_k}}\left({\widehat{a}}_{k-}-{\widehat{a}}_{N-k+}\right)e^{i2\pi jk/N}=}\frac{1}{\sqrt{N}}{\displaystyle \sum _{k=1}^{N-1}\sqrt{\frac{\hslash }{2m{\omega }_k}}\frac{2}{\sqrt{N}}{\displaystyle \sum _{j\text{'}=1}^N\sqrt{\frac{m{\omega }_k}{2\hslash }}{e}^{-i2\pi j\text{'}k/N}{x}_{j\text{'}}{e}^{i2\pi jk/N}}}\\ =\frac{1}{N}{\displaystyle \sum _{j\text{'}=1}^N{\displaystyle \sum _{k=1}^{N-1}{e}^{-i2\pi j\text{'}k/N}{x}_{j\text{'}}{e}^{i2\pi jk/N}}}\\ =\frac{1}{N}{\displaystyle \sum _{j\text{'}=1}^N{\displaystyle \sum _{k=1}^{N-1}{e}^{i2\pi \left(j-j\text{'}\right)k/N}{x}_{j\text{'}}}}\\ =\frac{1}{N}{\displaystyle \sum _{j\text{'}=1}^N{\displaystyle \sum _{k=1}^{N-1}{e}^{i2\pi \left(j-j\text{'}\right)k/N}-e^{i2\pi \left(j-j\text{'}\right)}{x}_{j\text{'}}}}\end{array}$

Further solving,

$\begin{array}{c}={\displaystyle \sum _{j\text{'}=1}^N\frac{1}{N}{\displaystyle \sum _{k=1}^N{e}^{i2\pi \left(j-j\text{'}\right)k/N}{x}_{j\text{'}}}}-\frac{1}{N}{\displaystyle \sum _{j\text{'}=1}^N{x}_{j\text{'}}}\\ ={\displaystyle \sum _{j\text{'}=1}^N{\delta }_{jj\text{'}}{x}_{j\text{'}}-R}\\ =x_{j\text{'}}-R\end{array}$

Since,

 $\begin{array}{c}\frac{1}{\sqrt{N}}{\displaystyle \sum _{k=1}^{N-1}\sqrt{\frac{\hslash }{2m{\omega }_k}}\left({\widehat{a}}_{k-}-{\widehat{a}}_{N-k+}\right)e^{i2\pi jk/N}=x_{j\text{'}}-R}\\ x_{j\text{'}}=R+\frac{1}{\sqrt{N}}{\displaystyle \sum _{k=1}^{N-1}\sqrt{\frac{\hslash }{2m{\omega }_k}}\left({\widehat{a}}_{k-}-{\widehat{a}}_{N-k+}\right)e^{i2\pi jk/N}}\end{array}$

Hence the first relation is proved, $x_j=R+\frac{1}{\sqrt{N}}{\displaystyle \sum _{k=1}^{N-1}\sqrt{\frac{\hslash }{2m{\omega }_k}}\left({\widehat{a}}_{k-}+{\widehat{a}}_{N-k+}\right)e^{i2\pi jk/N}}$.

Solving for $\left[{\widehat{a}}_{k-}-{\widehat{a}}_{N-k+}\right]$ using equation (I)

$\begin{array}{c}\left[{\widehat{a}}_{k-}-{\widehat{a}}_{N-k+}\right]=\left\{\frac{1}{\sqrt{N}}{\displaystyle \sum _j{e}^{-i2\pi jk/N}\left[\sqrt{\frac{m{\omega }_k}{2\hslash }}{x}_j+\sqrt{\frac{\hslash }{2m{\omega }_k}}\frac{\partial }{\partial x_j}\right]}\right\}\\ -\left\{\frac{1}{\sqrt{N}}{\displaystyle \sum _{j,j\text{'}}{e}^{i2\pi j\left(N-k\right)/N}\left[\sqrt{\frac{m{\omega }_{N-k}}{2\hslash }}{x}_j-\sqrt{\frac{\hslash }{2m{\omega }_{N-k}}}\frac{\partial }{\partial x_j}\right]}\right\}\end{array}$

Since ${\omega }_{N-k}={\omega }_k$ and $e^{i2\pi j\left(N-k\right)/N}=e^{i2\pi j}{e}^{-i2\pi jk/N}=e^{-i2\pi jk/N}$. The above equation becomes,

$\left[{\widehat{a}}_{k-}-{\widehat{a}}_{N-k+}\right]=\frac{2}{\sqrt{N}}{\displaystyle \sum _{j=1}^N\sqrt{\frac{\hslash }{2m{\omega }_k}}{e}^{-i2\pi jk/N}\frac{\partial }{\partial x_j}}$        (III)

Using the above equation to solve for $\frac{1}{\sqrt{N}}{\displaystyle \sum _{k=1}^{N-1}\sqrt{\frac{\hslash }{2m{\omega }_k}}\left({\widehat{a}}_{k-}-{\widehat{a}}_{N-k+}\right)e^{i2\pi jk/N}}$

$\begin{array}{c}\frac{1}{\sqrt{N}}{\displaystyle \sum _{k=1}^{N-1}\sqrt{\frac{m{\omega }_k}{2\hslash }}\left({\widehat{a}}_{k-}-{\widehat{a}}_{N-k+}\right)e^{i2\pi jk/N}}=\frac{1}{\sqrt{N}}{\displaystyle \sum _{k=1}^{N-1}\sqrt{\frac{m{\omega }_k}{2\hslash }}\frac{2}{\sqrt{N}}{\displaystyle \sum _{j\text{'}=1}^N\sqrt{\frac{m{\omega }_k}{2\hslash }}{e}^{-i2\pi j\text{'}k/N}\frac{\partial }{\partial x_{j\text{'}}}{e}^{i2\pi jk/N}}}\\ =\frac{1}{N}{\displaystyle \sum _{j\text{'}=1}^N{\displaystyle \sum _{k=1}^{N-1}{e}^{-i2\pi j\text{'}k/N}\frac{\partial }{\partial x_{j\text{'}}}{e}^{i2\pi jk/N}}}\\ =\frac{1}{N}{\displaystyle \sum _{j\text{'}=1}^N{\displaystyle \sum _{k=1}^{N-1}{e}^{i2\pi \left(j-j\text{'}\right)k/N}\frac{\partial }{\partial x_{j\text{'}}}}}\\ =\frac{1}{N}{\displaystyle \sum _{j\text{'}=1}^N{\displaystyle \sum _{k=1}^{N-1}{e}^{i2\pi \left(j-j\text{'}\right)k/N}-e^{i2\pi \left(j-j\text{'}\right)}\frac{\partial }{\partial x_{j\text{'}}}}}\end{array}$

Further solving,

$\begin{array}{c}={\displaystyle \sum _{j\text{'}=1}^N\frac{1}{N}{\displaystyle \sum _{k=1}^N{e}^{i2\pi \left(j-j\text{'}\right)k/N}\frac{\partial }{\partial x_{j\text{'}}}}}-\frac{1}{N}{\displaystyle \sum _{j\text{'}=1}^N\frac{\partial }{\partial x_{j\text{'}}}}\\ ={\displaystyle \sum _{j\text{'}=1}^N{\delta }_{jj\text{'}}\frac{\partial }{\partial x_{j\text{'}}}}-\frac{1}{N}{\displaystyle \sum _{j\text{'}=1}^N\frac{\partial }{\partial x_{j\text{'}}}}\\ =\frac{\partial }{\partial x_{j\text{'}}}-\frac{1}{N}\frac{\partial }{\partial R}\end{array}$

Since,

$\begin{array}{c}\frac{1}{\sqrt{N}}{\displaystyle \sum _{k=1}^{N-1}\sqrt{\frac{\hslash }{2m{\omega }_k}}\left({\widehat{a}}_{k-}-{\widehat{a}}_{N-k+}\right)e^{i2\pi jk/N}}=\frac{\partial }{\partial x_{j\text{'}}}-\frac{1}{N}\frac{\partial }{\partial R}\\ \frac{\partial }{\partial x_j}=\frac{1}{N}\frac{\partial }{\partial R}+\frac{1}{\sqrt{N}}{\displaystyle \sum _{k=1}^{N-1}\sqrt{\frac{m{\omega }_k}{2\hslash }}\left({\widehat{a}}_{k-}-{\widehat{a}}_{N-k+}\right)e^{i2\pi jk/N}}\end{array}$

Hence it is proved that $\frac{\partial }{\partial x_j}=\frac{1}{N}\frac{\partial }{\partial R}+\frac{1}{\sqrt{N}}{\displaystyle \sum _{k=1}^{N-1}\sqrt{\frac{m{\omega }_k}{2\hslash }}\left({\widehat{a}}_{k-}-{\widehat{a}}_{N-k+}\right)e^{i2\pi jk/N}}$.

Conclusion:

It is proved that $x_j=R+\frac{1}{\sqrt{N}}{\displaystyle \sum _{k=1}^{N-1}\sqrt{\frac{\hslash }{2m{\omega }_k}}\left({\widehat{a}}_{k-}+{\widehat{a}}_{N-k+}\right)e^{i2\pi jk/N}}$ and $\frac{\partial }{\partial x_j}=\frac{1}{N}\frac{\partial }{\partial R}+\frac{1}{\sqrt{N}}{\displaystyle \sum _{k=1}^{N-1}\sqrt{\frac{m{\omega }_k}{2\hslash }}\left({\widehat{a}}_{k-}-{\widehat{a}}_{N-k+}\right)e^{i2\pi jk/N}}$.

(d)

To determine

Show that $\widehat{H}=-\frac{{\hslash }^2}{2\left(Nm\right)}\frac{{\partial }^2}{\partial R^2}+{\displaystyle \sum _{k=1}^{N-1}\hslash {\omega }_k\left({\widehat{a}}_{k+}{\widehat{a}}_{k-}+\frac{1}{2}\right)}$.

Answer to Problem 5.38P

It has been proved that $\widehat{H}=-\frac{{\hslash }^2}{2\left(Nm\right)}\frac{{\partial }^2}{\partial R^2}+{\displaystyle \sum _{k=1}^{N-1}\hslash {\omega }_k\left({\widehat{a}}_{k+}{\widehat{a}}_{k-}+\frac{1}{2}\right)}$.

Explanation of Solution

Solving for $x_{j+1}-x_j$

$x_{j+1}-x_j=\frac{1}{\sqrt{N}}{\displaystyle \sum _{k=1}^{N-1}\sqrt{\frac{\hslash }{2m{\omega }_k}}\left({\widehat{a}}_{k-}+{\widehat{a}}_{N-k+}\right)e^{i2\pi jk/N}\left(e^{i2\pi k/N}-1\right)}$

Squaring on both sides,

${\left(x_{j+1}-x_j\right)}^2=\frac{1}{N}{\displaystyle \sum _{k,k\text{'}}^{N-1}\frac{\hslash }{2m\sqrt{{\omega }_k{\omega }_{k\text{'}}}}\left({\widehat{a}}_{k-}+{\widehat{a}}_{N-k+}\right)\left({\widehat{a}}_{k\text{'}-}+{\widehat{a}}_{N-k\text{'}+}\right)e^{i2\pi j\left(k+k\text{'}\right)/N}\left(e^{i2\pi k/N}-1\right)\left(e^{i2\pi k\text{'}/N}-1\right)}$        (IV)

Solving for $\left(e^{i2\pi k/N}-1\right)\left(e^{i2\pi \left(N-k\right)/N}-1\right)$ separately,

$\begin{array}{c}\left(e^{i2\pi k/N}-1\right)\left(e^{i2\pi \left(N-k\right)/N}-1\right)=e^{i\pi k/N}\left(e^{i\pi k/N}-e^{-i\pi k/N}\right)e^{-i\pi k/N}\left(e^{-i\pi k/N}-e^{i\pi k/N}\right)\\ =2i\sin \left(\frac{\pi k}{N}\right)\times \left(-2i\sin \left(\frac{\pi k}{N}\right)\right)\\ ={\left[2\sin \left(\frac{\pi k}{N}\right)\right]}^2\\ ={\left(\frac{{\omega }_k}{\omega }\right)}^2\end{array}$        (V)

Substitute equation (IV) and (V) in ${\displaystyle \sum _{j=1}^N\frac{1}{2}m{\omega }^2}{\left(x_{j+1}-x_j\right)}^2$

$\begin{array}{c}{\displaystyle \sum _{j=1}^N\frac{1}{2}m{\omega }^2}{\left(x_{j+1}-x_j\right)}^2={\displaystyle \sum _{k,k\text{'}}^{N-1}\frac{\hslash {\omega }^2}{4\sqrt{{\omega }_k{\omega }_{k\text{'}}}}\left({\widehat{a}}_{k-}+{\widehat{a}}_{N-k+}\right)\left({\widehat{a}}_{k\text{'}-}+{\widehat{a}}_{N-k\text{'}+}\right){\delta }_{k\text{'},N-k}\left(e^{i2\pi k/N}-1\right)\left(e^{i2\pi k\text{'}/N}-1\right)}\\ ={\displaystyle \sum _{k=1}^{N-1}\frac{\hslash {\omega }^2}{4{\omega }_k}\left({\widehat{a}}_{k-}+{\widehat{a}}_{N-k+}\right)\left({\widehat{a}}_{k\text{'}-}+{\widehat{a}}_{N-k\text{'}+}\right)\frac{{\omega }_k^2}{{\omega }^2}}\\ ={\displaystyle \sum _{k=1}^{N-1}\frac{\hslash {\omega }_k}{4}\left({\widehat{a}}_{k-}{\widehat{a}}_{k+}+{\widehat{a}}_{N-k+}{\widehat{a}}_{k+}+{\widehat{a}}_{k-}{\widehat{a}}_{N-k-}+{\widehat{a}}_{N-k+}{\widehat{a}}_{N-k-}\right)}\end{array}$        (VI)

Solving for $-\frac{{\hslash }^2}{2m}\frac{{\partial }^2}{\partial x_j^2}$

$\begin{array}{c}-\frac{{\hslash }^2}{2m}\frac{{\partial }^2}{\partial x_j^2}=\frac{1}{N}{\displaystyle \sum _{k,k\text{'}}^{N-1}\left(-\frac{{\hslash }^2}{2m}\right)\frac{m}{2\hslash }\sqrt{{\omega }_k{\omega }_{k\text{'}}}\left({\widehat{a}}_{N-k+}-{\widehat{a}}_{k-}\right)\left({\widehat{a}}_{k\text{'}-}-{\widehat{a}}_{N-k\text{'}+}\right)e^{i2\pi j\left(k+k\text{'}\right)/N}}\\ =-\frac{{\hslash }^2}{2m}\frac{1}{N^{3/2}}\frac{\partial }{\partial R}{\displaystyle \sum _{k=1}^{N-1}\sqrt{\frac{m{\omega }_k}{2\hslash }}\left({\widehat{a}}_{k\text{'}-}-{\widehat{a}}_{N-k\text{'}+}\right)e^{i2\pi jk/N}}+{\displaystyle \sum _{k=1}^{N-1}\sqrt{\frac{m{\omega }_k}{2\hslash }}\left({\widehat{a}}_{k\text{'}-}-{\widehat{a}}_{N-k\text{'}+}\right)e^{i2\pi jk/N}\frac{\partial }{\partial R}}\\ -\frac{{\hslash }^2}{2m}\frac{1}{N^2}\frac{{\partial }^2}{\partial R^2}\end{array}$

The middle term in the above equation vanish when it is summed over $j\left({\displaystyle \sum _{j=1}^N{e}^{i2\pi jk/N}}=N{\delta }_{k,0}=0\right)$. The above equation becomes

$\begin{array}{l}-\frac{{\hslash }^2}{2m}\frac{{\partial }^2}{\partial x_j^2}={\displaystyle \sum _{k=1}^{N-1}\left(-\frac{\hslash {\omega }_k}{4}\right)\left({\widehat{a}}_{k\text{'}-}-{\widehat{a}}_{N-k\text{'}+}\right)\left({\widehat{a}}_{N-k\text{'}-}-{\widehat{a}}_{k\text{'}+}\right)}-\frac{{\hslash }^2}{2Nm}\frac{{\partial }^2}{\partial R^2}\\ ={\displaystyle \sum _{k=1}^{N-1}\frac{\hslash {\omega }_k}{4}\left({\widehat{a}}_{k-}{\widehat{a}}_{k+}+{\widehat{a}}_{N-k+}{\widehat{a}}_{k+}+{\widehat{a}}_{k-}{\widehat{a}}_{N-k-}+{\widehat{a}}_{N-k+}{\widehat{a}}_{N-k-}\right)}-\frac{{\hslash }^2}{2Nm}\frac{{\partial }^2}{\partial R^2}\end{array}$

Adding the results

$\begin{array}{c}\widehat{H}=-\frac{{\hslash }^2}{2\left(Nm\right)}\frac{{\partial }^2}{\partial R^2}+{\displaystyle \sum _{k=1}^{N-1}\frac{\hslash {\omega }_k}{2}\left({\widehat{a}}_{k-}{\widehat{a}}_{k+}+{\widehat{a}}_{N-k+}{\widehat{a}}_{N-k-}\right)}\\ =-\frac{{\hslash }^2}{2\left(Nm\right)}\frac{{\partial }^2}{\partial R^2}+{\displaystyle \sum _{k=1}^{N-1}\frac{\hslash {\omega }_k}{2}\left({\widehat{a}}_{k-}{\widehat{a}}_{k+}+{\widehat{a}}_{k+}{a}_{k-}\right)}\\ \widehat{H}=-\frac{{\hslash }^2}{2\left(Nm\right)}\frac{{\partial }^2}{\partial R^2}+{\displaystyle \sum _{k=1}^{N-1}\hslash {\omega }_k\left({\widehat{a}}_{k+}{\widehat{a}}_{k-}+\frac{1}{2}\right)}\end{array}$

Conclusion:

It has been proved that $\widehat{H}=-\frac{{\hslash }^2}{2\left(Nm\right)}\frac{{\partial }^2}{\partial R^2}+{\displaystyle \sum _{k=1}^{N-1}\hslash {\omega }_k\left({\widehat{a}}_{k+}{\widehat{a}}_{k-}+\frac{1}{2}\right)}$.