Excited states of the Gaussian two-electron quantum dot

Abstract

We consider the $1s2s\; {^{1,3}\! S}$ states of the two-electron three-dimensional quantum dot with a Gaussian one-body potential, $-V_0\exp (-\lambda r^2)$. For a single electron, a simple scaling relation allows the reduction into a one-parameter problem in terms of $\frac{V_0}{\lambda }$. However, for the two-electron system, the interelectronic repulsion term, $\frac{1}{r_{12}}$, frustrates this simple scaling transformation, so we face a genuine two-parameter system. We pay particular attention to the location and nature of the critical well-depths, at which the binding energy of the second electron vanishes. Several observations are noteworthy: For all $\lambda$, the triplet critical well-depth is lower than that in the singly excited singlet state. Hence, there exists a finite range of well-depths for which the triplet is bound and the singlet is not, a feature that can possibly be applied in some device. Above its critical well-depth, the triplet state energy is always lower than that of the singly excited singlet. Both well-depths are considerably higher than the critical well-depth in the ground state. The expectation value of the interelectronic repulsion is always lower in the triplet, like the harmonic quantum dot but unlike He-like atoms, the two-particle Debye (Yukawa) atom, or the confined He atom. In the infinite well-depth ($V_0$) limit, keeping the well-width $\left( \frac{1}{\lambda }\right)$ constant, the energies and other expectation values of the bound states of the two-electron Gaussian quantum dot approach those of a non-interacting harmonic two-electron system.

Graphic abstract

Appendices

Appendix A: Rescaling of the one-particle Gaussian energies

Some authors use Hamiltonians of the form

$$\begin{aligned} {\mathcal {H}}_A=-\frac{\hbar ^2}{2m^*}\nabla _{\rho }^2-V_A\exp (-\lambda _A\rho ^2) \, , \end{aligned}$$

where $m^*=m_0\alpha$ is an appropriate effective mass.

Writing $\rho =\sqrt{\eta } r$, we obtain

$$\begin{aligned} {\mathcal {H}}_A=-\frac{\hbar ^2}{2\eta m_0\alpha }\nabla _r^2-V_A\exp (-\lambda _A\eta r^2) \, . \end{aligned}$$

Multiplication by $\eta \alpha$ yields

$$\begin{aligned} {\mathcal {H}}=\eta \alpha {\mathcal {H}}_A=-\frac{\hbar ^2}{2 m_0}\nabla _r^2-\eta \alpha V_A\exp (-\lambda _A\eta r^2) \, . \end{aligned}$$

Choosing $V_0=\eta \alpha V_A$ and $\lambda =\lambda _A\eta$, we obtain

$$\begin{aligned} {\mathcal {H}}=-\frac{\hbar ^2}{2 m_0}\nabla _r^2-V_0\exp (-\lambda r^2) \, . \end{aligned}$$

The eigenvalues of ${\mathcal {H}}$ are obtained from those of ${\mathcal {H}}_A$ by multiplication by $\eta \alpha$.

Now, $\eta =\frac{\lambda }{\lambda _A}$. Hence, $V_0=\alpha \frac{\lambda }{\lambda _A} V_A$. To convert the eigenvalues of ${\mathcal {H}}_A$ to those of ${\mathcal {H}}$, we write

$$\begin{aligned} E=\eta \alpha E_A=\frac{\alpha }{\lambda _A} E_A \, . \end{aligned}$$

This relation is implicit in [11], and a similar argument is presented in [16].

Comparing our framework ($m=1$, $\lambda =\frac{1}{2}$ to that of Lai [5], etc. ($m^*=0.5$, $\lambda _A=1$, $V_A=400$), we should take $\eta =\frac{1}{4}$. That yields $V_0=\frac{1}{4}V_A=100$. Once we get the energy that corresponds to this $V_0$ in our framework, we have to divide it by 4 to get the energy obtained by Lai [5] for $\lambda =1$, $V_A=400$.

The eigenfunctions of ${\mathcal {H}}$ are obtained from those of ${\mathcal {H}}_A$ as follows

$$\begin{aligned} \psi (r)=\eta ^{\frac{3}{4}}\psi _A(\sqrt{\eta }r) \, , \end{aligned}$$

where the factor $\eta ^{\frac{3}{4}}$ is needed for normalization. It follows that the expectation value of $\exp (-\lambda r^2)$ satisfies

$$\begin{aligned}&\int _0^{\infty }\psi (r)\exp (-\lambda r^2)\psi (r)4\pi r^2 dr\\&\quad = \int _0^{\infty } \psi _A(r)\exp (-\lambda _A r^2)\psi _A(r)4\pi r^2 dr \,. \end{aligned}$$

Similarly,

$$\begin{aligned}&\int _0^{\infty }\psi (r)\Big (r^2\exp (-\lambda r^2)\Big )\psi (r)4\pi r^2 dr\\&\quad =\frac{\lambda _A}{\lambda }\int _0^{\infty }\psi _A(r)\Big (r^2\exp (-\lambda _A r^2)\Big )\psi _A(r)4\pi r^2 dr \, . \end{aligned}$$

Appendix B: The virial and the Hellmann–Feynman theorems for one electron in a Gaussian potential

B.1: The virial theorem

Consider the commutator

$$\begin{aligned} \left[ r\frac{\partial }{\partial r},{\mathcal {H}}\right] =\frac{\hbar ^2}{m_0}\nabla _r^2+2\lambda V_0 r^2\exp (-\lambda r^2) \, . \end{aligned}$$

Evaluating the expectation value with respect to an eigenfunction of ${\mathcal {H}}$ we obtain

$$\begin{aligned} \frac{\hbar ^2}{m_0}\langle \psi |\nabla _r^2|\psi \rangle +2\lambda V_0 \langle \psi |r^2\exp (-\lambda r^2)|\psi \rangle =0 \,. \end{aligned}$$

This can be written in the form

$$\begin{aligned} \lambda V_0 \langle \psi |r^2\exp (-\lambda r^2)|\psi \rangle =-\frac{\hbar ^2}{2m_0}\langle \psi |\nabla _r^2|\psi \rangle \, ,\end{aligned}$$

(B.1)

where the expression on the right is the expectation value of the kinetic energy.

B.2: The Hellmann–Feynman Theorem

We consider the Hamiltonian

$$\begin{aligned} {\mathcal {H}}=-\frac{\hbar ^2}{2m_0}\nabla _{r}^2-V_0\exp (-\lambda r^2) \, . \end{aligned}$$

Let $\psi$ be an eigenfunction with the eigenvalue E.

The Hellmann–Feynman theorem with respect to $V_0$ yields

$$\begin{aligned} \frac{\partial E}{\partial V_0}=-\left\langle \psi |\exp (-\lambda r^2)|\psi \right\rangle \, .\end{aligned}$$

(B.2)

Another variant of the Hellmann–Feynman theorem is

$$\begin{aligned} \frac{\partial E}{\partial \lambda }=V_0\left\langle \psi |r^2\exp (-\lambda r^2)|\psi \right\rangle \, . \end{aligned}$$

We encountered this expectation value in the context of the virial theorem.

We obtain the (somewhat surprising) result

$$\begin{aligned} \lambda \frac{\partial E}{\partial \lambda }=-\frac{\hbar ^2}{2m_0}\langle \psi |\nabla _r^2|\psi \rangle \, .\end{aligned}$$

(B.3)

Substituting equations (B.2) and (B.3) in equation (B.1), we obtain

$$\begin{aligned} E-\lambda \frac{\partial E}{\partial \lambda }-V_0\frac{\partial E}{\partial V_0}=0 \, , \end{aligned}$$

which can be written in the equivalent form

$$\begin{aligned} \lambda \frac{\partial }{\partial \lambda }\left( \frac{E}{\lambda }\right) =-V_0\frac{\partial }{\partial V_0}\left( \frac{E}{\lambda }\right) \, . \end{aligned}$$

Defining ${\mathcal {E}}=\frac{E}{\lambda }$ we obtain

$$\begin{aligned} \lambda \frac{\partial {\mathcal {E}}}{\partial \lambda }=-V_0\frac{\partial {\mathcal {E}}}{\partial V_0} \, , \end{aligned}$$

which holds if ${\mathcal {E}}(V_0,\lambda )=f\left( \frac{\lambda }{V_0}\right)$. This is fully consistent with the rescaling of the one-particle Gaussian energies, derived in Appendix A.

Appendix C: The virial and the Hellmann–Feynman theorems for two interacting electrons in a Gaussian potential

C.1: The virial theorem for the two-electron system

Consider the Hamiltonian

$$\begin{aligned} {\mathcal {H}}=-\frac{\hbar ^2}{2 m_0}\left( \nabla _1^2+\nabla _2^2\right) -V_0\Big (\exp (-\lambda r_1^2)\!+\!(\exp (-\lambda r_2^2)\Big )+\frac{1}{r_{12}}. \end{aligned}$$

We evaluate the commutator

$$\begin{aligned}&\left[ r_1\frac{\partial }{\partial r_1}+r_2\frac{\partial }{\partial r_2},{\mathcal {H}}\right] =\frac{\hbar ^2}{m_0}\Big (\nabla _1^2\\&\quad +\nabla _2^2\Big )+2\lambda V_0 \Big (r_1^2\exp (-\lambda r_1^2)+r_2^2\exp (-\lambda r_2^2)\Big )-\frac{1}{r_{12}} \, . \end{aligned}$$

Hence,

$$\begin{aligned}&\frac{\hbar ^2}{m_0}\left\langle \nabla _1^2+\nabla _2^2\right\rangle +2\lambda V_0 \left\langle r_1^2\exp (-\lambda r_1^2)\right. \\&\quad \left. +r_2^2\exp (-\lambda r_2^2)\right\rangle -\left\langle \frac{1}{r_{12}}\right\rangle =0 \, . \end{aligned}$$

C.2: The Hellmann–Feynman theorem for the two-electron system

Here,

$$\begin{aligned} \frac{\partial E}{\partial V_0}=-\left\langle \Psi \Big {|}\exp (-\lambda r_1^2)+\exp (-\lambda r_2^2)\Big {|}\Psi \right\rangle \, , \end{aligned}$$

(C.1)

and

$$\begin{aligned} \frac{\partial E}{\partial \lambda }=V_0\left\langle \Psi \Big {|}r_1^2\exp (-\lambda r_1^2)+r_2^2\exp (-\lambda r_2^2)\Big {|}\Psi \right\rangle .\qquad \end{aligned}$$

(C.2)

From the first relation, we obtain

$$\begin{aligned} E=-\frac{\hbar ^2}{2m_0}\left\langle \nabla _1^2+\nabla _2^2\right\rangle +V_0\frac{\partial E}{\partial V_0}+\left\langle \frac{1}{r_{12}}\right\rangle \, , \end{aligned}$$

and from the second relation, substituting in the virial theorem we obtain

$$\begin{aligned} \lambda \frac{\partial E}{\partial \lambda }=-\frac{\hbar ^2}{2m_0}\left\langle \nabla _1^2+\nabla _2^2\right\rangle +\frac{1}{2}\left\langle \frac{1}{r_{12}}\right\rangle \, . \end{aligned}$$

Combining these relations, we obtain

$$\begin{aligned} \left\langle \frac{1}{r_{12}}\right\rangle =2\left( E-\lambda \frac{\partial E}{\partial \lambda }-V_0\frac{\partial E}{\partial V_0}\right) \,. \end{aligned}$$

(C.3)

Appendix D: Interelectronic repulsion in the harmonic oscillator framework

We consider the harmonic oscillator radial wave functions

$$\begin{aligned} \phi _{1s}(r)= & {} \frac{1}{\pi ^\frac{3}{4}}\exp (-\frac{r^2}{2}) \, , \nonumber \\ \phi _{2s}(r)= & {} \sqrt{\frac{3}{2}}\frac{1}{\pi ^{\frac{3}{4}}}(1-\frac{2}{3}r^2)\exp (-\frac{r^2}{2}) \, , \nonumber \\ \phi _{2p}(r)= & {} \sqrt{\frac{2}{3}}\frac{1}{\pi ^\frac{3}{4}}r\exp (-\frac{r^2}{2}) \, . \end{aligned}$$

(D.1)

The interelectronic repulsion in the $1s^2$ state is

$$\begin{aligned} \left\langle \frac{1}{r_{12}}\right\rangle = 2\int _0^{\infty }dr_2 4\pi r_2^2 \int _{r_2}^{\infty }dr_1 4\pi r_1^2\frac{1}{r_1}\phi _{1s}^2(r_1)\phi _{1s}^2(r_2) \, . \end{aligned}$$

For the $1s2s\; ^3S$ state

$$\begin{aligned} \left\langle \frac{1}{r_{12}}\right\rangle = J_{1s2s}-K_{1s2s} \, , \end{aligned}$$

where

$$\begin{aligned} J_{1s2s}= & {} \int _0^{\infty }dr_2 4\pi r_2^2\int _{r_2}^{\infty }dr_1 4\pi r_1^2\frac{1}{r_1}\phi _{1s}^2(r_1)\phi _{2s}^2(r_2)\\&+\int _0^{\infty }dr_2 4\pi r_2^2\int _0^{r_2}dr_1 4\pi r_1^2\frac{1}{r_2}\phi _{1s}^2(r_1)\phi _{2s}^2(r_2) , \end{aligned}$$

and

$$\begin{aligned}&K_{1s2s}=2\int _0^{\infty }dr_2 4\pi r_2^2\int _0^{r_2}dr_1 \\&\quad \times 4\pi r_1^2\frac{1}{r_2}\phi _{1s}(r_1)\phi _{2s}(r_1)\phi _{1s}(r_2)\phi _{2s}(r_2) \, . \end{aligned}$$

The corresponding singlet is a linear combination of the two states 1s2s and $2p^2$ that are degenerate in the non-interacting limit. Hence, we consider the $2\times 2$ matrix with elements

$$\begin{aligned} H_{11}=J_{1s2s}+K_{1s2s} \, , \\ H_{22}=F_0+10 F_2 \end{aligned}$$

where

$$\begin{aligned} F_0= & {} \int _0^{\infty }dr_2 4\pi r_2^2\int _{r_2}^{\infty }dr_1 4\pi r_1^2\frac{1}{r_1}\phi _{2p}^2(r_1)\phi _{2p}^2(r_2)\\&+\int _0^{\infty }dr_2 4\pi r_2^2\int _0^{r_2}dr_1 4\pi r_1^2\frac{1}{r_2}\phi _{2p}^2(r_1)\phi _{2p}^2(r_2)\, , \\ F_2= & {} \frac{1}{25}\left( \int _0^{\infty }dr_2 4\pi r_2^2\int _{r_2}^{\infty }dr_1 4\pi r_1^2\frac{r_2^2}{r_1^3}\phi _{2p}^2(r_1)\phi _{2p}^2(r_2) \right. \\&\left. +\int _0^{\infty }dr_2 4\pi r_2^2\int _0^{r_2}dr_1 4\pi r_1^2\frac{r_1^2}{r_2^3}\phi _{2p}^2(r_1)\phi _{2p}^2(r_2)\right) \, , \end{aligned}$$

and

$$\begin{aligned} H_{12}= & {} H_{21}=-\sqrt{\frac{2}{3}}\int _0^{\infty }dr_2 4\pi r_2^2\int _0^{r_2}dr_1\\&4\pi r_1^2\frac{r_1}{r_2^2}\phi _{1s}(r_1)\phi _{2s}(r_2)\phi _{2p}(r_1)\phi _{2p}(r_2) \\&+\int _0^{\infty }dr_2 4\pi r_2^2\int _{r_2}^{\infty }dr_1\\&4\pi r_1^2\frac{r_2}{r_1^2}\phi _{1s}(r_1)\phi _{2s}((r_2)\phi _{2p}(r_1)\phi _{2p}(r_2) \, . \end{aligned}$$

Evaluating these integrals, we obtain

$$\begin{aligned} \left\langle 1s^2\; ^1S\bigg {|}\frac{1}{r_{12}}\bigg {|} 1s^2\; ^1S\right\rangle =\sqrt{\frac{2}{\pi }}\approx 0.797885 \, , \\ \left\langle 1s2s\; ^3S\bigg {|}\frac{1}{r_{12}}\bigg {|} 1s2s\; ^3S\right\rangle =\frac{2}{3}\sqrt{\frac{2}{\pi }} \approx 0.531923 \, . \end{aligned}$$

For the excited singlet state

$$\begin{aligned} \left\langle 1s2s\; ^1S\bigg {|}\frac{1}{r_{12}}\bigg {|} 1s2s\; ^1S\right\rangle =\frac{11}{12}\sqrt{\frac{2}{\pi }} \, , \\ \left\langle 1s2s\; ^1S\bigg {|}\frac{1}{r_{12}}\bigg {|} 2p^2\; ^1S\right\rangle =\frac{1}{12}\sqrt{\frac{2}{\pi }} \, , \\ \left\langle 2p^2\, ^1S\bigg {|}\frac{1}{r_{12}}\bigg {|} 2p^2\, ^1S\right\rangle =\frac{11}{12}\sqrt{\frac{2}{\pi }} \, . \end{aligned}$$

The equality of the two diagonal matrix elements was not expected. Diagonalizing this $2\times 2$ matrix, the lower eigenvalue is

$$\begin{aligned} \left\langle 1s2s+2p^2\, ^1S\bigg {|}\frac{1}{r_{12}}\bigg {|} 1s2s+2p^2\, ^1S\right\rangle =\frac{5}{6}\sqrt{\frac{2}{\pi }} \approx 0.664904 \, . \end{aligned}$$