Question

Find the matrix elements $<n\left|x\right|n\text{'}>$a­­­­nd $\mathbf{<}\mathbf{n}\mathbf{\left|}\mathit{p}\mathbf{\right|}\mathbf{n}\mathbf{\text{'}}\mathbf{>}$in the (orthonormal) basis of stationary states for the harmonic oscillator (Equation 2.67). You already calculated the "diagonal" elements $\left(n=n\text{'}\right)$ in Problem 2.12; use the same technique for the general case. Construct the corresponding (infinite) matrices, X and P . Show that$\left(1/2m\right){\mathbf{P}}^{\mathbf{2}}\mathbf{+}\left({\mathrm{m\omega }}^2/2\right){\mathbf{X}}^{\mathbf{2}}\mathbf{=}\mathbf{H}$is diagonal, in this basis. Are its diagonal elements what you would expect? Partial answer:

$\mathbf{<}\mathbf{n}\mathbf{\left|}\mathbf{x}\mathbf{\right|}\mathbf{n}\mathbf{\text{'}}\mathbf{>}\mathbf{=}\sqrt{\frac{\sout{\mathbf{h}}}{\mathbf{2}\mathbf{m}\mathbf{\omega }}}\left(\sqrt{n\text{'}}{\delta }_{n,n\text{'}-1}+\sqrt{n}{\delta }_{n,n\text{'}-1}\right)$

Short answer

Required answers are

$$\begin{gathered} X=\sqrt{\frac{\sout{h}}{2m\omega }}\left(\begin{array}{cccccc}0& \sqrt{1}& 0& 0& 0& 0\\ \sqrt{1}& 0& \sqrt{2}& 0& 0& 0\\ 0& \sqrt{2}& 0& \sqrt{3}& 0& 0\\ 0& 0& \sqrt{3}& 0& \sqrt{4}& 0\\ 0& 0& 0& \sqrt{4}& 0& \sqrt{5}\end{array}⋮\right) \\ P=i\sqrt{\frac{m\sout{h\omega }}{2}}\left(\begin{array}{cccccc}0& -\sqrt{1}& 0& 0& 0& 0\\ \sqrt{1}& 0& -\sqrt{2}& 0& 0& 0\\ 0& \sqrt{2}& 0& -\sqrt{3}& 0& 0\\ 0& 0& \sqrt{3}& 0& -\sqrt{4}& 0\\ 0& 0& 0& \sqrt{4}& 0& -\sqrt{5}\end{array}⋮\right) \\ H=\frac{\sout{h\omega }}{2}\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 3& 0& 0\\ 0& 0& 5& 0\\ 0& 0& 0& 7\end{array}\right) \end{gathered}$$

Step-by-step solution

Diagonal elements of the matrices X and  P.

In a harmonic oscillator, the rising and dropping operators are used to compute $<x>$, and$\mathbf{<}\mathbf{p}\mathbf{>}$, they are both zero for all motionless states. These measures are the diagonal elements of the matricesX andP. That is:

localid="1658140622863" ${<X>}_{\mathbf{n}\mathbf{n}}\mathbf{=}\mathbf{<}\mathbf{n}\mathbf{\left|}\mathbf{x}\mathbf{\right|}\mathbf{n}\mathbf{>}$

From equation 2.69 and equation, 2.66 we have:

$$\begin{gathered} x=\sqrt{\frac{\sout{h}}{2m\omega }}\left(a_{+}+a_{-}\right)p \\ =\sqrt{\frac{\sout{h}m\omega }{2}}\left(a_{+}-a_{-}\right)a_{+}\overline{)n>} \\ =\sqrt{n+1}\overline{)n+1>a_{-}\overline{)n>}} \\ x=\sqrt{n}\overline{)n-1} \end{gathered}$$

The general matrix elements for the operator  x

The general matrix elements for the operator x can then be calculated as:

Let $n\text{'}=n-1$in the first term, so we get:

role="math" localid="1658141146213"

Construct the infinite matrices

Use (1) and (2) to construct the infinite matrices to get (noting that n and n' run from zero to infinity):

$X=\sqrt{\frac{\sout{h}}{2m\omega }}\left(\begin{array}{cccccc}0& \sqrt{1}& 0& 0& 0& 0\\ \sqrt{1}& 0& \sqrt{2}& 0& 0& 0\\ 0& \sqrt{2}& 0& \sqrt{3}& 0& 0\\ 0& 0& \sqrt{3}& 0& \sqrt{4}& 0\\ 0& 0& 0& \sqrt{4}& 0& \sqrt{5}\end{array}⋮\right)$

and:

$P=i\sqrt{\frac{m\sout{h\omega }}{2}}\left(\begin{array}{cccccc}0& -\sqrt{1}& 0& 0& 0& 0\\ \sqrt{1}& 0& -\sqrt{2}& 0& 0& 0\\ 0& \sqrt{2}& 0& -\sqrt{3}& 0& 0\\ 0& 0& \sqrt{3}& 0& -\sqrt{4}& 0\\ 0& 0& 0& \sqrt{4}& 0& -\sqrt{5}\end{array}⋮\right)$

Show that the matrix is diagonal

To show that $\left(1/2m\right)P^2+\left(m{\omega }^2/2\right)X^2=\mathrm{H}$is diagonal. Square the matrices above:

role="math" localid="1658141619383" $$\begin{gathered} X^2=\frac{\sout{h}}{2m\omega }\left(\begin{array}{cccccc}1& 0& \sqrt{1.2}& 0& 0& 0\\ 0& 3& 0& \sqrt{2.3}& 0& 0\\ \sqrt{1.2}& 0& 5& 0& \sqrt{3.4}& 0\\ 0& \sqrt{2.3}& 0& 7& 0& \sqrt{4.5}\end{array}\right) \\ {P}^2=\frac{m\sout{h\omega }}{2}\left(\begin{array}{cccccc}-1& 0& \sqrt{1.2}& 0& 0& 0\\ 0& -3& 0& \sqrt{2.3}& 0& 0\\ \sqrt{1.2}& 0& -5& 0& \sqrt{3.4}& 0\\ 0& \sqrt{2.3}& 0& -7& 0& \sqrt{4.5}\end{array}\right) \end{gathered}$$

Substitute in the Hamiltonian equation, and we get,

role="math" localid="1658141818347" $$\begin{gathered} H=-\frac{\sout{h\omega }}{4}\left(\begin{array}{cccccc}-1& 0& \sqrt{1.2}& 0& 0& 0\\ 0& -3& 0& \sqrt{2.3}& 0& 0\\ \sqrt{1.2}& 0& -5& 0& \sqrt{3.4}& 0\\ 0& \sqrt{2.3}& 0& -7& 0& \sqrt{4.5}\end{array}\right) \\ +\frac{\sout{h}\omega }{4}\left(\begin{array}{cccccc}1& 0& \sqrt{1.2}& 0& 0& 0\\ 0& 3& 0& \sqrt{2.3}& 0& 0\\ \sqrt{1.2}& 0& 5& 0& \sqrt{3.4}& 0\\ 0& \sqrt{2.3}& 0& 7& 0& \sqrt{4.5}\end{array}\right) \end{gathered}$$

We get,

$H=\frac{\sout{h}\omega }{2}\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 3& 0& 0\\ 0& 0& 5& 0\\ 0& 0& 0& 7\end{array}\right)$

So, it is diagonal with the diagonal element is given by:

$H_{nn}=\left(n+\frac{1}{2}\right)\sout{h}\omega$