Ground state of the hydrogen atom via Dirac equation in a minimal-length scenario

Abstract

In this work we calculate the correction to the ground state energy of the hydrogen atom due to contributions arising from the presence of a minimal length. The minimal-length scenario is introduced by means of modifying the Dirac equation through a deformed Heisenberg algebra (Kempf algebra). With the introduction of the Coulomb potential in the new Dirac energy operator, we calculate the energy shift of the ground state of the hydrogen atom in first order of the parameter related to the minimal length via perturbation theory.

Appendices

Appendix A: Relativistic energy of the electron

From Eq. (18) we have a linear homogeneous system of equations for ϕ and χ,

$$\begin{aligned} &{ \begin{pmatrix} - E_{ML} + mc^{2} & c ( \boldsymbol {\sigma} \cdot\hat{\bf p} ) + \beta c ( \boldsymbol {\sigma} \cdot\hat{\bf p} )^{3} \\ c ( \boldsymbol {\sigma} \cdot\hat{\bf p} ) + \beta c ( \boldsymbol {\sigma} \cdot\hat{\bf p} )^{3} & - E_{ML} - mc^{2} \\ \end{pmatrix}} \\ &{\quad {}\times \begin{pmatrix} \phi\\ \chi\\ \end{pmatrix} = 0,} \end{aligned}$$

(A.1)

which has non-trivial solution only for

$$ \begin{vmatrix} - \bigl( E_{ML} - mc^{2} \bigr) & c ( \boldsymbol {\sigma} \cdot\hat{\bf p} ) \bigl( 1 + \beta p^{2} \bigr) \\ c ( \boldsymbol {\sigma} \cdot\hat{\bf p} ) \bigl( 1 + \beta p^{2} \bigr) & - \bigl( E_{ML} + mc^{2} \bigr) \\ \end{vmatrix} = 0. $$

(A.2)

Thus, after throwing away terms of order β ², the relativistic energy of the free electron in the regarded scenario of minimal length can be obtained from

$$ E^{2}_{ML} = p^{2}c^{2} + m^{2}c^{4} + 2\beta c^{2} p^{4}, $$

(A.3)

as was expected from $E^{2}_{ML} = c^{2}P^{2} + m^{2}c^{4}$.

Appendix B: Corrections in the first order of perturbation

In order to obtain

$$\begin{aligned} &{\langle\psi| ( \boldsymbol {\alpha} \cdot\hat{\bf p} )^{3} | \psi \rangle} \\ &{\quad {}= \int \bigl(\phi^{\dag}, \chi^{\dag} \bigr) \begin{pmatrix} 0 & ( \boldsymbol {\sigma} \cdot\hat{\bf p} )^{3} \\ ( \boldsymbol {\sigma} \cdot\hat{\bf p} )^{3} & 0 \\ \end{pmatrix} \begin{pmatrix} \phi\\ \chi\\ \end{pmatrix} \,d^{3}\boldsymbol {x},} \end{aligned}$$

(B.1)

where ϕ and χ are two-component eigenspinors of the state,

$$ \langle \boldsymbol {x} | \psi\rangle= \begin{pmatrix} \phi\\ \chi\\ \end{pmatrix} = \begin{pmatrix} F(r)Y^{j, m}_{j-1/2} (\theta, \phi ) \\ -if(r)Y^{j, m}_{j+1/2} (\theta, \phi ) \\ \end{pmatrix} , $$

(B.2)

$Y^{j, m}_{j \pm1/2} (\theta, \phi )$ are the common eigenspinor-function of $\hat{j}_{z}$ and $\hat{J}^{2}$ and

$$\begin{aligned} &{F(r) = x^{\gamma} e^{-ax} \sum_{\nu= 0}^{n^{\prime}} a_{\nu} x^{\nu},} \end{aligned}$$

(B.3)

$$\begin{aligned} &{f(r) = x^{\gamma} e^{-ax} \sum_{\nu= 0}^{n^{\prime}} b_{\nu} x^{\nu},} \end{aligned}$$

(B.4)

where

$$\begin{aligned} &{n^{\prime} = n - \biggl( j + \frac{1}{2} \biggr),} \end{aligned}$$

(B.5)

$$\begin{aligned} &{\gamma= -1 + \sqrt{ \biggl( j + \frac{1}{2} \biggr)^{2} - \alpha^{2}},} \end{aligned}$$

(B.6)

$$\begin{aligned} &{a = \sqrt{1 - \frac{E^2}{m^2 c^4}},} \end{aligned}$$

(B.7)

and

$$ x = \biggl( \frac{mc}{\hbar} \biggr) r, $$

(B.8)

we employ the following identity [28]:

$$ \boldsymbol {\sigma} \cdot\hat{\bf p} = \boldsymbol {\sigma} \cdot \boldsymbol {e}_{r} \biggl( -i\hbar\frac{\partial}{\partial r} + i\frac{\boldsymbol {\sigma} \cdot\hat {\bf L}}{r} \biggr), $$

(B.9)

with

$$ \boldsymbol {\sigma} \cdot \boldsymbol {e}_{r}Y^{j, m}_{j \pm1/2} = - Y^{j, m}_{j \pm1/2}. $$

(B.10)

After some algebra we get

$$\begin{aligned} &{( \boldsymbol {\sigma} \cdot\hat{\bf p} ) \phi_{1}} \\ &{\quad {} = i\hbar \biggl[ \frac{dF}{dr} - \biggl(j - \frac{1}{2} \biggr) \frac {F}{r} \biggr] Y^{j, m}_{j+\frac{1}{2}} ,} \end{aligned}$$

(B.11)

$$\begin{aligned} &{( \boldsymbol {\sigma} \cdot\hat{\bf p} ) \phi_{2}} \\ &{\quad {} = \hbar \biggl[ \frac{df}{dr} + \biggl(j + \frac{3}{2} \biggr) \frac{f}{r} \biggr] Y^{j, m}_{j-\frac{1}{2}} ,} \end{aligned}$$

(B.12)

and

$$\begin{aligned} &{( \boldsymbol {\sigma} \cdot\hat{\bf p} )^{2} \phi_{1}} \\ &{\quad {} = -\hbar^{2} \biggl[ \frac{d^{2}F}{dr^{2}} + 2 \frac{1}{r}\frac{dF}{dr}} \\ &{\qquad {} - \biggl(j - \frac{1}{2} \biggr) \biggl(j + \frac{1}{2} \biggr) \frac {F}{r^{2}} \biggr] Y^{j, m}_{j-\frac{1}{2}},} \end{aligned}$$

(B.13)

$$\begin{aligned} &{( \boldsymbol {\sigma} \cdot\hat{\bf p} )^{2} \phi_{2}} \\ &{\quad {} = i \hbar^{2} \biggl[ \frac{d^{2}f}{dr^{2}} + 2 \frac{1}{r}\frac{df}{dr}} \\ &{\qquad {} - \biggl(j + \frac{1}{2} \biggr) \biggl(j + \frac{3}{2} \biggr) \frac {f}{r^{2}} \biggr] Y^{j, m}_{j+\frac{1}{2}},} \end{aligned}$$

(B.14)

and

$$\begin{aligned} &{( \boldsymbol {\sigma} \cdot\hat{\bf p} )^{3} \phi_{1}} \\ &{\quad {} = - i\hbar ^{3} \biggl[ \frac{d^{3}F}{dr^{3}} - \biggl(j - \frac{5}{2} \biggr) \frac{1}{r}\frac{d^{2}F}{dr^{2}} \biggr] Y^{j, m}_{j+\frac{1}{2}}} \\ &{\qquad {}+ i\hbar^{3} \biggl[ \biggl(j + \frac{1}{2} \biggr) \biggl(j + \frac{3}{2} \biggr)\frac{1}{r^{2}}\frac{dF}{dr} \biggr] Y^{j, m}_{j+\frac{1}{2}}} \\ &{\qquad {}- i\hbar^{3} \biggl[ \biggl(j - \frac{1}{2} \biggr) \biggl(j + \frac{1}{2} \biggr) \biggl(j + \frac{3}{2} \biggr)\frac{F}{r^{3}} \biggr] Y^{j, m}_{j+\frac{1}{2}},} \end{aligned}$$

(B.15)

$$\begin{aligned} &{( \boldsymbol {\sigma} \cdot\hat{\bf p} )^{3} \phi_{2}} \\ &{\quad {} = - \hbar ^{3} \biggl[ \frac{d^{3}f}{dr^{3}} + \biggl(j + \frac{7}{2} \biggr) \frac{1}{r}\frac{d^{2}f}{dr^{2}} \biggr] Y^{j, m}_{j-\frac{1}{2}}} \\ &{\qquad {}+ \hbar^{3} \biggl[ \biggl(j - \frac{1}{2} \biggr) \biggl(j + \frac{1}{2} \biggr)\frac{1}{r^{2}}\frac{df}{dr} \biggr] Y^{j, m}_{j-\frac{1}{2}}} \\ &{\qquad {}+ \hbar^{3} \biggl[ \biggl(j - \frac{1}{2} \biggr) \biggl(j + \frac{1}{2} \biggr) \biggl(j + \frac{3}{2} \biggr) \frac{f}{r^{3}} \biggr] Y^{j, m}_{j-\frac{1}{2}}.} \end{aligned}$$

(B.16)

Finally,

$$\begin{aligned} &{\langle\psi| ( \boldsymbol {\alpha} \cdot\hat{\bf p} )^{3} | \psi \rangle} \\ &{\quad {}= - \hbar^{3}\int \biggl( F^{*}\frac{d^{3}f}{dr^{3}} - f^{*} \frac {d^{3}F}{dr^{3}} \biggr) r^2 \,dr} \\ &{\qquad {}- \hbar^{3} \biggl(j + \frac{7}{2} \biggr)\int F^{*} \frac {d^{2}f}{dr^{2}} r \,dr} \\ &{\qquad {}- \hbar^{3} \biggl(j - \frac{5}{2} \biggr) \int f^{*} \frac {d^{2}F}{dr^{2}} r \,dr} \\ &{\qquad {}+ \hbar^{3} \biggl(j - \frac{1}{2} \biggr) \biggl(j + \frac{1}{2} \biggr) \int F^{*} \frac{df}{dr} \,dr} \\ &{\qquad {}- \hbar^{3} \biggl(j + \frac{1}{2} \biggr) \biggl(j + \frac{3}{2} \biggr) \int f^{*} \frac{dF}{dr} \,dr} \\ &{\qquad {}+ \hbar^{3} \biggl(j - \frac{1}{2} \biggr) \biggl(j + \frac{1}{2} \biggr) \biggl(j + \frac{3}{2} \biggr)} \\ &{\qquad {}\times \int \bigl( F^{*}f + f^{*}F \bigr) \frac{dr}{r}.} \end{aligned}$$

(B.17)

2.1 B.1 Ground state energy

For the ground state we have $j = \frac{1}{2}$ and n′=0, then

$$\begin{aligned} &{F_{0}(r) = a_{0} x^{\gamma} e^{-ax},} \end{aligned}$$

(B.18)

$$\begin{aligned} &{f_{0}(r) = b_{0} x^{\gamma} e^{-ax},} \end{aligned}$$

(B.19)

where

$$\begin{aligned} &{\gamma= \epsilon- 1,} \end{aligned}$$

(B.20)

$$\begin{aligned} &{a_{0} = \biggl( \frac{2a}{b} \biggr)^{\gamma+ 1} \sqrt{ \frac{ (1 + \epsilon )}{\varGamma ( 2\gamma+3 ) }},} \end{aligned}$$

(B.21)

$$\begin{aligned} &{b_{0} = \sqrt{\frac{1 - \epsilon}{1 + \epsilon}}a_{0},} \end{aligned}$$

(B.22)

with $\epsilon= \sqrt{1- \alpha^{2}}$.

Hence

$$ \langle\psi_{0} | ( \boldsymbol {\alpha} \cdot\hat{\bf p} )^{3} | \psi_{0} \rangle= m^{3}c^{3} \frac{ ( 1- \epsilon^{2} )^{2}}{\epsilon (2 \epsilon- 1 )}, $$

(B.23)

where E ₀=mc ² ϵ is the energy of the |ψ ₀〉 ground state of the hydrogen atom obtained from the ordinary Dirac equation.