The two lowest eigenvalues of the harmonic oscillator in the presence of a Gaussian perturbation

Abstract

In this note, we consider a one-dimensional quantum mechanical particle constrained by a parabolic well perturbed by a Gaussian potential. As the related Birman–Schwinger operator is trace class, the Fredholm determinant can be exploited in order to compute the modified eigenenergies which differ from those of the harmonic oscillator due to the presence of the Gaussian perturbation. By taking advantage of Wang’s results on scalar products of four eigenfunctions of the harmonic oscillator, it is possible to evaluate quite accurately the two lowest lying eigenvalues as functions of the coupling constant $\lambda$.

Appendix: proofs of some of the previous results

In this appendix, we wish to provide the reader with the mathematical details of some results used throughout the article.

1.1 Calculation of the optimal range of admissible $\lambda$ ensuring the invertibility of $\lambda M_{1/2}^{(0)}$

Theorem A

The operator $\left[ 1-\lambda M_{1/2}^{(0)} \right] ^{-1}$

$$\begin{aligned} M_{1/2}^{(0)}=\sum _{n=1}^{\infty } \frac{e^{-(\cdot )^2/2}|\psi _n \rangle \langle \psi _n|e^{-(\cdot )^2/2}}{n}, \end{aligned}$$

exists for any positive $\lambda <\frac{1}{\sqrt{2} \ln 2} \approx 1.020$.

Proof

As $M_{1/2}^{(0)}>0$, its trace class norm is exactly its trace, so that

$$\begin{aligned} \text {tr} [M_{1/2}^{(0)}]=\sum _{m=0}^{\infty }\sum _{n=1}^{\infty } \frac{(\psi _m,e^{-(\cdot )^2/2}\psi _n)^2}{n}=\sum _{n=1}^{\infty } \frac{ \left| \left| e^{-(\cdot )^2/2}\psi _n \right| \right| _2^2}{n}, \end{aligned}$$

(A.1)

$\square$

as follows easily from the orthonormality of the eigenfunctions of the harmonic oscillator. Since

$$\begin{aligned} \left( \psi _n,e^{-(\cdot )^2}\psi _n \right) =\sqrt{\pi }\left( \psi _n, \psi _0^2 \psi _n \right) =\sqrt{\pi } \frac{\psi _{2n}^2(0)}{\sqrt{2}}, \end{aligned}$$

the rhs of (A.1) is equal to

$$\begin{aligned} \sqrt{2 \pi } \sum _{n=1}^{\infty } \frac{\psi _{2n}^2(0)}{2n}=\sqrt{2 \pi } \lim _{\epsilon \rightarrow 0_{+}} \left[ \sum _{n=0}^{\infty } \frac{\psi _{2n}^2(0)}{2n+\epsilon } - \frac{\psi _0^2(0)}{\epsilon } \right] . \end{aligned}$$

(A.2)

By mimicking what was done in [18,19,20], the rhs of (A.2) can be written as

$$\begin{aligned}&\sqrt{2 \pi } \lim _{\epsilon \rightarrow 0_{+}} \left[ \left( H_0-\frac{1}{2}+\epsilon \right) ^{-1}(0,0) - \frac{1}{\sqrt{\pi }\epsilon } \right] \nonumber \\&\quad =\sqrt{2 \pi } \lim _{\epsilon \rightarrow 0_{+}} \left[ \frac{1}{\sqrt{\pi }}\int _0^1 \frac{s^{\epsilon -1}}{(1-s^2)^{1/2}}\mathrm{d}s - \frac{1}{\sqrt{\pi }\epsilon } \right] . \end{aligned}$$

(A.3)

After expressing the second term inside the square brackets as an integral and simplifying the integrand, the latter limit becomes

$$\begin{aligned} \sqrt{2} \lim _{\epsilon \rightarrow 0_{+}} \int _0^1 \frac{s^{\epsilon -1}\left[ 1-(1-s^2)^{1/2}\right] }{(1-s^2)^{1/2}}\mathrm{d}s=\sqrt{2} \lim _{\epsilon \rightarrow 0_{+}} \int _0^1 \frac{s^{\epsilon +1}}{(1-s^2)^{1/2}\left[ 1+(1-s^2)^{1/2}\right] }\mathrm{d}s.\nonumber \\ \end{aligned}$$

(A.4)

As we may perform the limit inside the integral, the above rhs becomes

$$\begin{aligned} \sqrt{2} \int _0^1 \frac{s}{(1-s^2)^{1/2}\left[ 1+(1-s^2)^{1/2}\right] }\mathrm{d}s=-\sqrt{2} \int _0^1 \frac{\mathrm{d}(1-s^2)^{1/2}}{\mathrm{d}s}\frac{1}{\left[ 1+(1-s^2)^{1/2}\right] }\mathrm{d}s,\nonumber \\ \end{aligned}$$

(A.5)

which, after setting $y=(1-s^2)^{1/2}$, gets transformed into

$$\begin{aligned} \sqrt{2} \int _0^1 \frac{1}{1+y}dy=\sqrt{2} \left[ \ln (1+y) \right] _0^1=\sqrt{2} \ln 2. \end{aligned}$$

(A.6)

Therefore, $\left| \left| M_{1/2}^{(0)} \right| \right| _1=\sqrt{2} \ln 2$. Then, for any $\lambda <\frac{1}{\sqrt{2} \ln 2}$, we have $\lambda \left| \left| M_{1/2}^{(0)} \right| \right| _1 <1$, which ensures the existence of $\left[ 1-\lambda M_{1/2}^{(0)} \right] ^{-1}$.

Before closing this subsection, it might be worth noting that the same result could have been achieved by expressing the integral on the rhs of (A.3) as a ratio of values of the gamma function, as was done in [22, 24,25,26].

1.2 Proof of the expression for $\epsilon _0(\lambda )$ given in equation (3.12)

Theorem B

The following positive series converges and the sum is

$$\begin{aligned} \sum _{n=1}^{\infty } \frac{ \psi _{2n}^2(0)}{2^{2n+1}n}=\frac{\ln (8-4\sqrt{3})}{\sqrt{\pi }}. \end{aligned}$$

(A.7)

Proof

First of all, we notice that:

$$\begin{aligned}&\sum _{n=1}^{\infty } \frac{ \psi _{2n}^2(0)}{2^{2n+1}n} =\sum _{n=1}^{\infty } \frac{ \psi _{2n}^2(0)}{2^{2n+1}} \int _0^{1} x^{n-1}\mathrm{d}x \nonumber \\&\quad =\int _0^{1} \left[ \sum _{n=1}^{\infty } \frac{ \psi _{2n}^2(0)}{2^{2n+1}}\, x^{n-1} \right] \mathrm{d}x =\int _0^{1} \left[ \sum _{n=1}^{\infty } \psi _{2n}^2(0) \left( \frac{x}{4}\right) ^n \right] \frac{\mathrm{d}x}{2x}. \end{aligned}$$

(A.8)

$\square$

As follows in a rather straightforward manner from the definition of the normalised eigenfunctions of the harmonic oscillator (2.6) (see [16, 18, 27]),

$$\begin{aligned} \psi _{2n}(0)= \frac{H_{2n}(0)}{\sqrt{2^{2n} (2n)! \sqrt{\pi }}}, \end{aligned}$$

and as is well known (see [3, 28])

$$\begin{aligned} H_{2n}(0)= (-1)^n \frac{(2n)!}{n!}, \end{aligned}$$

so that the series inside the square brackets on the rhs of (A.8) becomes

$$\begin{aligned}&\sum _{n=1}^{\infty } \psi _{2n}^2(0) \left( \frac{x}{4}\right) ^n = \sum _{n=1}^{\infty } \frac{ (2n)!}{\sqrt{\pi }\, 2^{2n}\, (n!)^2} \left( \frac{x}{4}\right) ^n \nonumber \\&\quad = \frac{1}{\sqrt{\pi }} \sum _{n=1}^{\infty } \frac{ (2n)!}{(n!)^2} \left( \frac{x}{2^4} \right) ^n = \frac{1}{\sqrt{\pi }} \left( \frac{2}{\sqrt{4-x}}-1\right) . \end{aligned}$$

(A.9)

Going with this result to the rhs of (A.8) and integrating from 0 to 1, we get:

$$\begin{aligned} \sum _{n=1}^{\infty } \frac{ \psi _{2n}^2(0)}{2^{2n+1}n} = \frac{1}{2\sqrt{\pi }} \int _0^{1} \left( \frac{2}{\sqrt{4-x}}-1\right) \frac{\mathrm{d}x}{x}= \frac{\ln (8-4\sqrt{3})}{\sqrt{\pi }}, \end{aligned}$$

(A.10)

which completes the proof of our claim.

1.3 Sum of the series in equation (4.6)

Theorem C

The following positive series converges and the sum is

$$\begin{aligned} \sum _{n=1}^{\infty } \frac{(n+1)\psi _{2(n+1)}^2(0)}{2^{2(n+1)}n}=\frac{2\sqrt{3}-3\left( 1-\ln [8-4\sqrt{3}] \right) }{12\sqrt{\pi }}. \end{aligned}$$

(A.11)

Proof

The proof is similar to the previous one: Using the eigenfunctions (2.6)

$$\begin{aligned} \psi _{2(n+1)}(0)= \frac{H_{2(n+1)}(0)}{\sqrt{2^{2(n+1)} (2n+2)! \sqrt{\pi }}}, \end{aligned}$$

(A.12)

and the fact that

$$\begin{aligned} H_{2(n+1)}(0)= (-1)^{n+1} \frac{(2n+2)!}{(n+1)!}, \end{aligned}$$

$\square$

it is straightforward to deduce the following

$$\begin{aligned}&\sum _{n=1}^{\infty } \frac{(n+1)\psi _{2(n+1)}^2(0)}{2^{2(n+1)}n}= \sum _{n=1}^{\infty } \frac{\psi _{2(n+1)}^2(0)}{2^{2(n+1)}}\, (n+1) \int _0^{1} x^{n-1}\mathrm{d}x\\&\quad =\int _0^{1} \left[ \sum _{n=1}^{\infty } \frac{\psi _{2(n+1)}^2(0)}{2^{2(n+1)}}\, (n+1)\, x^{n} \right] \frac{\mathrm{d}x}{x} =\int _0^{1} \frac{\mathrm{d}}{\mathrm{d}x }\left[ \sum _{n=1}^{\infty } \frac{\psi _{2(n+1)}^2(0)}{2^{2(n+1)}}\, x^{n+1} \right] \frac{\mathrm{d}x}{x}\\&\quad =\int _0^{1} \frac{\mathrm{d}}{\mathrm{d}x }\left[ \sum _{n=1}^{\infty } \psi _{2(n+1)}^2(0)\, \left( \frac{x}{2^2}\right) ^{n+1} \right] \frac{\mathrm{d}x}{x}\\&\quad =\int _0^{1} \frac{\mathrm{d}}{\mathrm{d}x }\left[ \sum _{n=1}^{\infty } \frac{(2n+2)!}{2^{2(n+1)} (n+1)!^2 \sqrt{\pi }} \left( \frac{x}{2^2}\right) ^{n+1} \right] \frac{\mathrm{d}x}{x}\\&\quad =\frac{1}{\sqrt{\pi }} \int _0^{1} \frac{\mathrm{d}}{\mathrm{d}x }\left[ \sum _{n=1}^{\infty } \frac{(2n+2)!}{ (n+1)!^2 } \left( \frac{x}{2^4}\right) ^{n+1} \right] \frac{\mathrm{d}x}{x} =\frac{1}{\sqrt{\pi }} \int _0^{1} \frac{\mathrm{d}}{\mathrm{d}x }\left[ \frac{2}{\sqrt{4-x}}-1-\frac{x}{8} \right] \frac{\mathrm{d}x}{x}\\&\quad = \frac{1}{12\sqrt{\pi }} \left( 2\sqrt{3}-3\left[ 1-\ln (8-4\sqrt{3}) \right] \right) , \end{aligned}$$

which completes our proof.