The Wigner function negative value domains and energy function poles of the harmonic oscillator

Abstract

For a quantum harmonic oscillator, an explicit expression that describes the energy distribution as a coordinate function is obtained. The presence of the energy function poles is shown for the quantum system in domains where the Wigner function has negative values.

Appendices

Appendix A

From equations (6), (8), (9), (10), (17) and (16) for a harmonic oscillator, it follows that

$$\frac{1}{{f_{1} }}\frac{\partial P}{{\partial x}} = \frac{1}{{f_{1} }}\int\limits_{ - \infty }^{ + \infty } {v^{2} \frac{{\partial f_{2,n} }}{\partial x}} dv = - \frac{1}{m}\frac{{\partial U_{1} }}{\partial x} = - \omega^{2} x,$$

where in the one-dimensional case (9) the notation of $P_{\mu \lambda }$ is changed to $P$

$$\frac{x}{{\sigma_{x}^{2} }}\int\limits_{ - \infty }^{ + \infty } {v^{2} F^{\prime}_{n} } \left( {\tilde{\varepsilon }} \right)dv = - \omega^{2} xf_{1,n} = - \omega^{2} x\int\limits_{ - \infty }^{ + \infty } {F_{n} } \left( {\tilde{\varepsilon }} \right)dv,$$

$$0 = \int\limits_{ - \infty }^{ + \infty } {\left( {v^{2} F^{\prime}_{n} \left( {\tilde{\varepsilon }} \right) + \sigma_{x}^{2} \omega^{2} F_{n} \left( {\tilde{\varepsilon }} \right)} \right)} dv = \int\limits_{ - \infty }^{ + \infty } {\left( {v^{2} \frac{{\partial S_{2,n} }}{{\partial \tilde{\varepsilon }}} + \sigma_{x}^{2} \omega^{2} } \right)F_{n} \left( {\tilde{\varepsilon }} \right)} dv = f_{1,n} \left\langle {v^{2} \frac{{\partial S_{2,n} }}{{\partial \tilde{\varepsilon }}} + \sigma_{x}^{2} \omega^{2} } \right\rangle ,$$

$$\left\langle {v^{2} \frac{{\partial S_{2,n} }}{{\partial \tilde{\varepsilon }}} + \sigma_{x}^{2} \omega^{2} } \right\rangle = 0,\left\langle {v^{2} \frac{{\partial S_{2,n} }}{{\partial \tilde{\varepsilon }}}} \right\rangle = - \sigma_{x}^{2} \omega^{2} ,$$

(A.1)

where $S_{2,n} = {\text{Ln}} f_{2,n}$ Rewriting condition (7) in the form $P_{\mu \lambda } = - \alpha^{2} f_{1,n} \frac{{\partial^{2} S_{1,n} }}{{\partial x^{\mu } \partial x^{\lambda } }}$, $S_{1,n} = {\text{Ln}} f_{1,n}$, we obtain

$$\frac{1}{{f_{1,n} }}\int\limits_{ - \infty }^{ + \infty } {v^{2} f_{2,n} \left( {x,v} \right)dv} = - \alpha^{2} \frac{{\partial^{2} S_{1,n} }}{{\partial x^{2} }},\left\langle {v^{2} } \right\rangle = - \alpha^{2} \frac{{\partial^{2} S_{1,n} }}{{\partial x^{2} }}.$$

(A.2)

Substitute the distribution functions (14) into expressions (A.1) and (A.2). Let us start with expression (A.1).

$$S_{2,n} = {\text{Ln}} F_{n} \left( {\tilde{\varepsilon }} \right) = {\text{Ln}} B_{n} - \tilde{\varepsilon } + {\text{Ln}} L_{n} \left( {2\tilde{\varepsilon }} \right),\,\,\,\frac{{\partial S_{2,n} }}{{\partial \tilde{\varepsilon }}} = - 1 + 2\frac{{L_{n}^{\prime } \left( {2\tilde{\varepsilon }} \right)}}{{L_{n} \left( {2\tilde{\varepsilon }} \right)}},$$

(A.3)

where $B_{n} = \frac{{\left( { - 1} \right)^{n} }}{{2\pi \sigma_{v} \sigma_{x} }}$ Averaging expression (A.3) using (A.1), we obtain

$$\sigma_{x}^{2} \omega^{2} = \frac{{B_{n} }}{{f_{1,n} }}\int\limits_{ - \infty }^{ + \infty } {v^{2} e^{{ - \tilde{\varepsilon }}} \left( {L_{n} \left( {2\tilde{\varepsilon }} \right) + 2L_{n - 1}^{\left( 1 \right)} \left( {2\tilde{\varepsilon }} \right)} \right)} \,dv,$$

(A.4)

where we take into account that $L_{n}^{\prime } = L^{\prime}_{n - 1} - L_{n - 1}$, $L_{s}^{{\left( {\mu + 1} \right)}} = \sum\limits_{k = 0}^{s} {L_{k}^{\left( \mu \right)} }$ Considering the expression $L_{n}^{\left( \mu \right)} \left( x \right) = L_{n}^{{\left( {\mu + 1} \right)}} \left( x \right) - L_{n - 1}^{{\left( {\mu + 1} \right)}} \left( x \right)$ which at $\mu = 0$ will be $L_{n}^{{}} \left( x \right) = L_{n}^{\left( 1 \right)} \left( x \right) - L_{n - 1}^{\left( 1 \right)} \left( x \right)$, expression (A.4) will take the form:

$$\frac{{\sigma_{v}^{2} }}{{B_{n} }}f_{1,n} = \int\limits_{ - \infty }^{ + \infty } {v^{2} e^{{ - \tilde{\varepsilon }}} \left( {L_{n}^{\left( 1 \right)} \left( {2\tilde{\varepsilon }} \right) + L_{n - 1}^{\left( 1 \right)} \left( {2\tilde{\varepsilon }} \right)} \right)} \,dv.$$

(A.5)

The generalized Laguerre polynomials satisfy the relations

$$\begin{gathered} L_{n}^{\left( 1 \right)} \left( {2\tilde{\varepsilon }} \right) = \sum\limits_{k = 0}^{n} {L_{k}^{{}} \left( {\frac{{v^{2} }}{{\sigma_{v}^{2} }}} \right)L_{n - k}^{{}} \left( {\frac{{x^{2} }}{{\sigma_{x}^{2} }}} \right),\,\,\,} \hfill \\ L_{n - 1}^{\left( 1 \right)} \left( {2\tilde{\varepsilon }} \right) = \sum\limits_{k = 0}^{n - 1} {L_{k}^{{}} \left( {\frac{{v^{2} }}{{\sigma_{v}^{2} }}} \right)L_{n - 1 - k}^{{}} \left( {\frac{{x^{2} }}{{\sigma_{x}^{2} }}} \right)} . \hfill \\ \end{gathered}$$

(A.6)

Substituting (A.6) into (A.4), we obtain

$$\left( { - 1} \right)^{n} \frac{\sqrt \pi }{{2^{n + 1} n!}}H_{n}^{2} \left( {\frac{x}{{\sqrt 2 \sigma_{x} }}} \right) = \sum\limits_{k = 1}^{n} {L_{n - k}^{{}} \left( {\frac{{x^{2} }}{{\sigma_{x}^{2} }}} \right)} \left( {J_{k} + J_{k - 1} } \right) + L_{n}^{{}} \left( {\frac{{x^{2} }}{{\sigma_{x}^{2} }}} \right)J_{0} ,$$

(A.7)

where

$$J_{k} = \int\limits_{ - \infty }^{ + \infty } {\tau^{2} e^{{ - \tau^{2} }} L_{k}^{{}} \left( {2\tau^{2} } \right)} \,d\tau ,\,\,\,\tau = \frac{v}{{\sqrt 2 \sigma_{v}^{{}} }}.$$

(A.8)

Expression (A.7) allows representing the square of the Hermite polynomials $H_{n}^{2}$ in terms of the Laguerre polynomials $L_{k}^{{}}$ Let us calculate the integral (A.8).

$$J_{k} = \frac{1}{2}\int\limits_{ - \infty }^{ + \infty } {e^{{ - \tau^{2} }} L_{k}^{{}} \left( {2\tau^{2} } \right)d\tau } - 2J_{k - 1} + 2\int\limits_{ - \infty }^{ + \infty } {\tau^{2} e^{{ - \tau^{2} }} L_{k - 1}^{\prime } \left( {2\tau^{2} } \right)d\tau } .$$

(A.9)

Considering that $\int\limits_{ - \infty }^{ + \infty } {e^{{ - \tau^{2} }} L_{k} \left( {2\tau^{2} } \right)} \,d\tau = \left( { - 1} \right)^{k} \frac{\sqrt \pi }{{2^{k} k!}}H_{k}^{2} \left( 0 \right)$, we get

$$J_{k} = \left( { - 1} \right)^{k} \frac{\sqrt \pi }{{2^{k + 1} k!}}H_{k}^{2} \left( 0 \right) - 2J_{k - 1} + 2\int\limits_{ - \infty }^{ + \infty } {\tau^{2} e^{{ - \tau^{2} }} L_{k - 1}^{\prime } \left( {2\tau^{2} } \right)d\tau } ,$$

and hence

$$J_{k - 1} = \left( { - 1} \right)^{k - 1} \frac{\sqrt \pi }{{2^{k} \left( {k - 1} \right)!}}H_{k - 1}^{2} \left( 0 \right) - 2J_{k - 2} + 2\int\limits_{ - \infty }^{ + \infty } {\tau^{2} e^{{ - \tau^{2} }} L_{k - 1}^{\prime } \left( {2\tau^{2} } \right)d\tau } .$$

(A.10)

Substituting (A.10) into expression (A.9), we obtain

$$J_{k} = \left( { - 1} \right)^{k} \frac{\sqrt \pi }{{2^{k + 1} k!}}H_{k}^{2} \left( 0 \right) + \left( { - 1} \right)^{k} \frac{\sqrt \pi }{{2^{k - 1} \left( {k - 1} \right)!}}H_{k - 1}^{2} \left( 0 \right) + 2J_{k - 2} - 2\int\limits_{ - \infty }^{ + \infty } {\tau^{2} e^{{ - \tau^{2} }} L_{k - 2}^{\prime } \left( {2\tau^{2} } \right)d\tau .}$$

(A.11)

Then, let us carry out a similar substitution procedure for the integrals $J_{k - 2}$ and $J_{k - 3}$:

$$\begin{gathered} J_{k} = \left( { - 1} \right)^{k} \frac{\sqrt \pi }{{2^{k + 1} k!}}H_{k}^{2} \left( 0 \right) + \left( { - 1} \right)^{k} \frac{\sqrt \pi }{{2^{k - 1} \left( {k - 1} \right)!}}H_{k - 1}^{2} \left( 0 \right) + \left( { - 1} \right)^{k} \frac{\sqrt \pi }{{2^{k - 2} \left( {k - 2} \right)!}}H_{k - 2}^{2} \left( 0 \right) - \hfill \\ - 2J_{k - 3} + 2\int\limits_{ - \infty }^{ + \infty } {\tau^{2} e^{{ - \tau^{2} }} L_{k - 3}^{\prime } \left( {2\tau^{2} } \right)d\tau } . \hfill \\ \end{gathered}$$

(A.12)

$$\begin{gathered} J_{k} = \left( { - 1} \right)^{k} \frac{\sqrt \pi }{{2^{k + 1} k!}}H_{k}^{2} \left( 0 \right) + \left( { - 1} \right)^{k} \frac{\sqrt \pi }{{2^{k - 1} \left( {k - 1} \right)!}}H_{k - 1}^{2} \left( 0 \right) + \left( { - 1} \right)^{k} \frac{\sqrt \pi }{{2^{k - 2} \left( {k - 2} \right)!}}H_{k - 2}^{2} \left( 0 \right) + \hfill \\ + \left( { - 1} \right)^{k} \frac{\sqrt \pi }{{2^{k - 3} \left( {k - 3} \right)!}}H_{k - 3}^{2} \left( 0 \right) + 2J_{k - 4} - 2\int\limits_{ - \infty }^{ + \infty } {\tau^{2} e^{{ - \tau^{2} }} L_{k - 4}^{\prime } \left( {2\tau^{2} } \right)d\tau } . \hfill \\ \end{gathered}$$

(A.13)

The expression for $J_{0}$ has the form

$$J_{0} = \int\limits_{ - \infty }^{ + \infty } {\tau^{2} e^{{ - \tau^{2} }} L_{0}^{{}} \left( {2\tau^{2} } \right)d\tau } = \frac{1}{{2\sqrt 2 \sigma_{v}^{3} }}\int\limits_{ - \infty }^{ + \infty } {v^{2} e^{{ - \frac{{v^{2} }}{{2\sigma_{v}^{2} }}}} dv} = \frac{{\sqrt {2\pi } }}{2\sqrt 2 } = \frac{\sqrt \pi }{2}.$$

(A.14)

Let us consider the even ($k = 2m$) and odd ($k = 2m + 1$) values for the expression $J_{k}$ Proceeding with the iterative procedure expression (A.13) for $J_{2m}$ takes the form:

$$J_{2m} = \frac{\sqrt \pi }{{2^{2m + 1} \left( {2m} \right)!}}H_{2m}^{2} \left( 0 \right) + \sqrt \pi \sum\limits_{s = 1}^{2m} {\frac{{H_{2m - s}^{2} \left( 0 \right)}}{{2^{2m - s} \left( {2m - s} \right)!}}} .$$

(A.15)

Similarly, for $J_{2m + 1}$ we obtain

$$J_{2m + 1} = - \frac{\sqrt \pi }{{2^{2m + 2} \left( {2m + 1} \right)!}}H_{2m + 1}^{2} \left( 0 \right) - \sqrt \pi \sum\limits_{q = 1}^{2m + 1} {\frac{{H_{2m + 1 - q}^{2} \left( 0 \right)}}{{2^{2m + 1 - q} \left( {2m + 1 - q} \right)!}}} .$$

(A.16)

Comparing (A.15) and (A.16), we obtain a general expression for $J_{k}$

$$J_{k} = \left( { - 1} \right)^{k} \sqrt \pi \left\{ {\frac{1}{{2^{k + 1} k!}}H_{k}^{2} \left( 0 \right) + \sum\limits_{s = 1}^{k} {\frac{{H_{k - s}^{2} \left( 0 \right)}}{{2^{k - s} \left( {k - s} \right)!}}} } \right\}.$$

(A.17)

To transform expression (A.7), let us calculate the sum $J_{k} + J_{k - 1}$ using (A.17)

$$J_{k} + J_{k - 1} = \left( { - 1} \right)^{k} \sqrt \pi \left\{ {\frac{{H_{k}^{2} \left( 0 \right) - 2kH_{k - 1}^{2} \left( 0 \right)}}{{2^{k + 1} k!}} + \sum\limits_{s = 1}^{k} {\frac{{H_{k - s}^{2} \left( 0 \right) - 2\left( {k - s} \right)H_{k - s - 1}^{2} \left( 0 \right)}}{{2^{k - s} \left( {k - s} \right)!}}} } \right\}.$$

(A.18)

Substituting (A.18) into (A.7), we get

$$\begin{gathered} \frac{{\left( { - 1} \right)^{n} }}{{2^{n + 1} n!}}H_{n}^{2} \left( {\frac{x}{{\sqrt 2 \sigma_{x} }}} \right) = \frac{1}{2}L_{n}^{{}} \left( {\frac{{x^{2} }}{{\sigma_{x}^{2} }}} \right) + \hfill \\ + \sum\limits_{k = 1}^{n} {\left( { - 1} \right)^{k} L_{n - k}^{{}} \left( {\frac{{x^{2} }}{{\sigma_{x}^{2} }}} \right)} \left\{ {\frac{{H_{k}^{2} \left( 0 \right) - 2kH_{k - 1}^{2} \left( 0 \right)}}{{2^{k + 1} k!}} + \sum\limits_{s = 1}^{k} {\frac{{H_{k - s}^{2} \left( 0 \right) - 2\left( {k - s} \right)H_{k - s - 1}^{2} \left( 0 \right)}}{{2^{k - s} \left( {k - s} \right)!}}} } \right\}. \hfill \\ \end{gathered}$$

(A.19)

Expression (A.19) can be rewritten in a compact form using the Heaviside function

$$\eta \left( s \right) = \left\{ \begin{gathered} 0,\,\,\,s = 0, \hfill \\ 1,\,\,\,s > 0. \hfill \\ \end{gathered} \right. \frac{1 + \eta \left( s \right)}{2} = \left\{ \begin{gathered} \frac{1}{2},\,\,\,s = 0, \hfill \\ 1,\,\,\,s > 0. \hfill \\ \end{gathered} \right.$$

(A.20)

Using (A.20), expression (A.19) takes the following form:

$$\frac{{\left( { - 1} \right)^{n} }}{{2^{n + 1} n!}}H_{n}^{2} \left( {\frac{x}{{\sqrt 2 \sigma_{x} }}} \right) = \sum\limits_{k = 0}^{n} {\overline{C}_{k} L_{n - k}^{{}} \left( {\frac{{x^{2} }}{{\sigma_{x}^{2} }}} \right)} ,$$

(A.21)

where

$$\overline{C}_{k} = \left( { - 1} \right)^{k} \sum\limits_{s = 0}^{k} {\frac{1 + \eta \left( s \right)}{2}\frac{{H_{k - s}^{2} \left( 0 \right) - 2\left( {k - s} \right)H_{k - s - 1}^{2} \left( 0 \right)}}{{2^{k - s} \left( {k - s} \right)!}}} .$$

Now let us consider expression (A.2). First of all, we calculate the expression $\frac{{\partial^{2} S_{1} }}{{\partial x^{2} }}$ in (A.2)

$$\frac{{\partial^{2} S_{1} }}{{\partial x^{2} }} = - \frac{1}{{\sigma_{x}^{2} }}\left( {1 - \frac{{H^{\prime\prime}_{n} }}{{H_{n}^{{}} }} + \left( {\frac{{H^{\prime}_{n} }}{{H_{n}^{{}} }}} \right)^{2} } \right).$$

(A.22)

And find $\left\langle {v^{2} } \right\rangle$

$$\frac{{\left( { - 1} \right)^{n} \sqrt \pi }}{{2^{n + 1} n!\sigma_{v}^{2} }}H_{n}^{2} \left( {\frac{x}{{\sqrt 2 \sigma_{x} }}} \right)\left\langle {v^{2} } \right\rangle = \sum\limits_{k = 0}^{n} {L_{n - k}^{{}} \left( {\frac{{x^{2} }}{{\sigma_{x}^{2} }}} \right)} \,J_{k} - \sum\limits_{k = 0}^{n - 1} {L_{n - 1 - k}^{{}} \left( {\frac{{x^{2} }}{{\sigma_{x}^{2} }}} \right)} \,J_{k} ,$$

(A.23)

since $\int\limits_{ - \infty }^{ + \infty } {v^{2} e^{{ - \frac{{v^{2} }}{{2\sigma_{v}^{2} }}}} L_{k}^{{}} \left( {\frac{{v^{2} }}{{\sigma_{v}^{2} }}} \right)dv = } 2\sqrt 2 \sigma_{v}^{3} J_{k}$ Substituting (A.17) into (A.23), we obtain

$$\frac{{\left( { - 1} \right)^{n} }}{{2^{n + 1} n!\sigma_{v}^{2} }}H_{n}^{2} \left( {\frac{x}{{\sqrt 2 \sigma_{x} }}} \right)\left\langle {v^{2} } \right\rangle = \sum\limits_{k = 0}^{n} {C_{k} L_{n - k}^{{}} \left( {\frac{{x^{2} }}{{\sigma_{x}^{2} }}} \right)} ,$$

(A.24)

where

$$C_{k} = \left( { - 1} \right)^{k} \sum\limits_{s = 0}^{k} {\frac{1 + \eta \left( s \right)}{2}\frac{{H_{k - s}^{2} \left( 0 \right) + 2\left( {k - s} \right)H_{k - s - 1}^{2} \left( 0 \right)}}{{2^{k - s} \left( {k - s} \right)!}}} .$$

Appendix B

Let us calculate

$$\begin{array}{l} {\left\langle {\left\langle {{v^2}} \right\rangle } \right\rangle _n} = \int\limits_{ - \infty }^{ + \infty } {{f_{1,n}}\left( x \right){{\left\langle {{v^2}} \right\rangle }_{v,n}}\left( x \right)dx} = \frac{{{\alpha ^2}}}{{\sigma _x^2}} - \frac{{{\alpha ^2}}}{{\sigma _x^2}}\int\limits_{ - \infty }^{ + \infty } {{f_{1,n}}\left( x \right)\frac{{{{H''}_n}}}{{H_n^{}}}dx} + \frac{{{\alpha ^2}}}{{\sigma _x^2}}\int\limits_{ - \infty }^{ + \infty } {{f_{1,n}}\left( x \right){{\left( {\frac{{{{H'}_n}}}{{H_n^{}}}} \right)}^2}dx} ,\\ {\left\langle {\left\langle {{v^2}} \right\rangle } \right\rangle _n} = \frac{{{\alpha ^2}}}{{\sigma _x^2}} - \frac{{{\alpha ^2}}}{{\sigma _x^2}}\frac{1}{{{2^n}n!}}\frac{{\sqrt 2 {\sigma _x}}}{{\sqrt {2\pi } \sigma _x^{}}}\int\limits_{ - \infty }^{ + \infty } {{e^{ - {y^2}}}H_n^{}\left( y \right){{H''}_n}\left( y \right)dy} + \\ + \frac{{{\alpha ^2}}}{{\sigma _x^2}}\frac{1}{{{2^n}n!}}\frac{{\sqrt 2 {\sigma _x}}}{{\sqrt {2\pi } \sigma _x^{}}}\int\limits_{ - \infty }^{ + \infty } {{e^{ - {y^2}}}{{H'}_n}\left( y \right){{H'}_n}\left( y \right)dy} , \end{array}$$

and taking into account the differentiation formula for the Hermite polynomials $H_{n}^{\prime } \left( y \right) = 2nH_{n - 1} \left( y \right)$ and the orthogonality condition, we obtain the following expression:

$$\left\langle {\left\langle {v^{2} } \right\rangle } \right\rangle_{n} = \frac{{\alpha^{2} }}{{\sigma_{x}^{2} }} + \frac{{\alpha^{2} }}{{\sigma_{x}^{2} }}\frac{1}{{2^{n} n!}}\frac{{4n^{2} }}{\sqrt \pi }\sqrt \pi 2^{n - 1} \left( {n - 1} \right)! = \frac{{\alpha^{2} }}{{\sigma_{x}^{2} }}\left( {2n + 1} \right),$$

$$\left\langle {\left\langle {v^{2} } \right\rangle } \right\rangle_{n} = \sigma_{v}^{2} \left( {2n + 1} \right),$$

(B.1)

where relation (1.4) is taken into account. By virtue of the symmetry of expressions (1.6) and (1.8), we can rewrite the expression $\left\langle {\left\langle {x^{2} } \right\rangle } \right\rangle_{n}$ in a similar way

$$\left\langle {\left\langle {x^{2} } \right\rangle } \right\rangle_{n} = \sigma_{x}^{2} \left( {2n + 1} \right).$$

(B.2)

Appendix C

Let us derive the quantum potential (6) for the function $f_{1,n}$ (1.5):

$$\sqrt {f_{1,n} } = \sqrt {c_{n} } e^{{ - \frac{{y^{2} }}{2}}} H_{n} \left( y \right){\text{sgn}} H_{n} ,$$

(C.1)

$$\left( {\sqrt {f_{1,n} } } \right)_{xx} = \frac{{\sqrt {c_{n} } }}{{\sigma^{2} }}e^{{ - \frac{{y^{2} }}{2}}} {\text{sgn}} H_{n} \left[ {2n\left( {n - 1} \right)H_{n - 2} - 2nyH_{n - 1} + \frac{{y^{2} - 1}}{2}H_{n} } \right],$$

(C.2)

where $c_{n}^{ - 1} = 2^{n} n!\sqrt {2\pi } \sigma_{x}$, $y = \frac{x}{{\sqrt 2 \sigma_{x}^{{}} }}$, $H_{n}^{\prime } = 2nH_{n - 1}$ Taking recurrent relationship $H_{n} - 2yH_{n - 1} + 2\left( {n - 1} \right)H_{n - 2} = 0$ into account, expression (C.2) will take the following form:

$$\left( {\sqrt {f_{1,n} } } \right)_{xx} = \frac{{\sqrt {c_{n} } }}{{2\sigma^{2} }}e^{{ - \frac{{y^{2} }}{2}}} H_{n} {\text{sgn}} H_{n} \left[ {y^{2} - \left( {2n + 1} \right)} \right].$$

(C.3)

Substituting (C.1) and (C.3) into (6), we get

$${\text{Q}}_{n} \left( x \right) = - \frac{{\hbar^{2} }}{2m}\frac{{\left( {\sqrt {f_{1,n} } } \right)_{xx} }}{{\sqrt {f_{1,n} } }} = - \frac{{\hbar^{2} }}{{4m\sigma_{x}^{2} }}\left[ {y^{2} - \left( {2n + 1} \right)} \right] = - \frac{{m\omega^{2} x^{2} }}{2} + \hbar \omega \left( {n + \frac{1}{2}} \right),$$

(C.4)

where relations (1.4) are taken into account.