The ground state and the second excited state of a charged particle of mass m in a one-dimensional harmonic oscillator potential are both occupied. What is the expectation value of the particle position x as a function of time? What happens to the expectation value if the potential is subject to a

Question

The ground state and the second excited state of a charged particle of mass m in a one-dimensional harmonic oscillator potential are both occupied. What is the expectation value of the particle position x as a function of time? What happens to the expectation value if the potential is subject to a constant electric field [latex]\mathbf{E}[/latex] in the x direction?

Accepted Answer

The ground state and the second excited state of a charged particle of mass m in a one-dimensional harmonic oscillator potential are both occupied. We wish to find the expectation value of the particle position x as a function of time. We start by noting that [latex]|0 angle[/latex] and [latex]|2 angle[/latex] have no overlap when operated on by the position operator [latex]{\hat{x}}_{ }[/latex] . Hence,

[latex]\langle{\hat{x}}(t) angle={\frac{\hbar}{2m\omega}}\langle a_{0}\langle0|+a_{2}\langle2|\left|{\hat{b}}+{\hat{b}}^{\dagger} ight|a_{0}|0 angle+a_{2}|2 angle angle[/latex]

[latex]\langle{\hat{x}}(t) angle=|a_{0}|^{2}\langle0|0 angle + |a_{2} |^{2}\langle2|2 angle=\langle x angle[/latex]

and we conclude that there is no time dependence for the position expectation operator.

We note that [latex]|0 angle=\phi_{0}e^{-i\hbar\omega t/2}\; \mathrm{and}\;|2 angle=\phi_{2}e^{-i{5}\hbar{w}/2}[/latex], where [latex]\phi_{\mathrm{{0}}}[/latex] and [latex]\phi_{\mathrm {{2}}}[/latex] are the spatial wave functions for the ground state and the second excited state respectively wave functions for the ground state and the second excited state respectively.
In the presence of a constant electric field [latex]\mathbf{E}[/latex] in the x direction, a particle of charge [latex]{\mathcal{e}}[/latex] in the harmonic potential experiences a constant x-directed force [latex] e\left |\,\mathbf {E} ight|[/latex], which shifts the equilibrium position of the oscillator by [latex]\Delta{x}[/latex].

The new Hamiltonian is [latex]{H}=\hat{p}^{2}/2m+\kappa\hat {x}^{ \prime}{}^{2}/2+e|\mathrm{{\bf{F}}}|x^{\prime},[/latex] and the amount of the shift is given by [latex]e|{\bf E}|=\kappa\,\Delta x \mathrm{,~where~}\kappa\,=m\omega^{2}[/latex] is the oscillator force constant and [latex]x^{\prime}=[/latex] [latex]x+ riangle x.[/latex] The total energy is

[latex]E_{\mathrm{total}}={\frac{1}{2}}\kappa(x+\Delta x)^{2}-e \vert\mathbf{E}\vert(x+\Delta x)={\frac{1}{2}}\kappa x^{2}+\kappa x\Delta x+{\frac{1}{2}}\kappa\Delta x^{2}-e\vert\mathbf{E}\vert x-e\vert\mathbf{E}\vert\Delta x[/latex]

[latex]E_{\mathrm{total}}={\frac{1}{2}}\kappa x^{2}+{\frac{1}{2}}\kappa\Delta x^{2}-e\vert\mathbf{E}\vert\Delta x[/latex]

[latex]\mathrm{since}\,e|\mathbf{E}|=\kappa\,\Delta x\,\mathrm {and}[/latex]

[latex]E_{\mathrm{total}}=\frac{1}{2}\kappa x^{2}+\frac{1}{2}\kappa\frac{e^{2}|{\bf E}|^{2}}{\kappa^{2}}-\frac{e^{2}|{\bf E}|^{2}}{\kappa}=\frac{1}{2}\kappa x^{2}+\frac{1}{2}\frac{e^{2}|{\bf E}|^{2}}{\kappa}-\frac{e^{2}|{\bf E}|^{2}}{\kappa}[/latex]

[latex]E_{\mathrm{total}}=\frac{1}{2}m\omega^{2}x^{2}-\frac{1}{2}\frac{e^{2}|\mathbf{E}|^{2}}{\kappa}[/latex]

[latex]\mathrm{since}\,\kappa=m\omega^{2}.[/latex]

The oscillator frequency remains the same, but all of the energy levels are uniformly reduced by a fixed amount

[latex]\frac{-e^{2}|\mathbf{E}|}{2m\omega^{2}}[/latex]

Question 6.9: The ground state and the second excited state of a charged p......

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