Hydrogen Atom In Ground State

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The Hydrogen Atom

Michael Fowler, UVa

Factoring Out the Centre of Mass Motion

The hydrogen atom consists of two particles, the proton and the electron, interacting via the Coulomb potential $V ({\vec{r}}_{ane} - {\vec{r}}_{ii}) = e^{two} / r$ , where as usual $r = | {\vec{r}}_{1} - {\vec{r}}_{2} |$ . Writing the masses of the two particles as $m_{1}, 1000_{2}$ Schrödinger's equation for the atom is:

$(- \frac{ℏ^{2}}{2 m_{1}} {\vec{\nabla}}_{1}^{two} - \frac{ℏ^{2}}{ii m_{2}} {\vec{\nabla}}_{2}^{2} - \frac{e^{2}}{r}) ψ ({\vec{r}}_{1}, {\vec{r}}_{2}) = E ψ ({\vec{r}}_{1}, {\vec{r}}_{ii}) .$

But ${\vec{r}}_{1}, {\vec{r}}_{2}$ are non the almost natural position variables for describing this organisation: since the potential depends only on the relative position, a better choice is $\vec{r}, \vec{R}$ defined by:

$\vec{r} = {\vec{r}}_{ane} - {\vec{r}}_{2}, \vec{R} = \frac{{chiliad}_{1} {\vec{r}}_{one} + k_{two} {\vec{r}}_{2}}{m_{1} + m_{two}}$

so $\vec{R}$ is the centre of mass of the arrangement. It is convenient at the same time to announce the total mass by $1000 = m_{1} + m_{two},$ and the reduced mass by $m = \frac{1000_{1} {yard}_{two}}{m_{ane} + m_{2}} .$

Transforming in straightforward manner to the variables $\vec{r}, \vec{R}$ Schrödinger's equation becomes

$(- \frac{ℏ^{2}}{2 M} {\vec{\nabla}}_{R}^{2} - \frac{ℏ^{2}}{2 m} {\vec{\nabla}}_{r}^{two} - \frac{e^{2}}{r}) ψ (\vec{R}, \vec{r}) = E ψ (\vec{R}, \vec{r}) .$

Writing the wave function

$ψ (\vec{R}, \vec{r}) = Ψ (\vec{R}) ψ (\vec{r})$

nosotros tin split the equation into two:

$\begin{array}{l} (- \frac{ℏ^{2}}{2 M} {\vec{\nabla}}_{R}^{two}) Ψ (\vec{R}) = {Due east}_{R} Ψ (\vec{R}) \\ (- \frac{ℏ^{ii}}{two m} {\vec{\nabla}}_{r}^{2} + Five (\vec{r})) ψ (\vec{r}) = E_{r} ψ (\vec{r}) \end{array}$

and the total system energy is $East = E_{R} + E_{r} .$ Note that the movement of the center of mass is (of form) simply that of a free particle, having a lilliputian plane wave solution. From at present on, we shall simply be concerned with the relative move of the particles. Since the proton is far heavier than the electron, we volition nigh always ignore the difference between the electron mass and the reduced mass, but information technology should be noted that the deviation is hands detectable spectroscopically: for example, the lines shift if the proton is replaced by a deuteron (heavy hydrogen).

We're ready to write Schrödinger's equation for the hydrogen atom, dropping the $r$ suffixes in the second equation to a higher place, and writing out ${\vec{\nabla}}^{2}$ explicitly in spherical coordinates:

$\begin{array}{l} - \frac{ℏ^{2}}{2 m} (\frac{1}{r^{ii}} \frac{\partial}{\partial r} (r^{2} \frac{\partial ψ}{\partial r}) + \frac{1}{r^{2} \sin θ} \frac{\partial}{\partial θ} (\sin θ \frac{\partial ψ}{\partial θ}) + \frac{one}{r^{2} \sin^{two} θ} \frac{\partial^{2} ψ}{\partial φ^{two}}) - \frac{{east}^{ii}}{r} ψ \\ = E ψ . \end{array}$

Factoring Out the Angular Dependence: the Radial Equation

Since the potential is spherically symmetric, the Hamiltonian $H$ commutes with the athwart momentum operators $L^{two}, L_{z}$ so nosotros can construct a common set of eigenkets of the three operators $H, 50^{2}, L_{z} .$ The angular dependence of these eigenkets is therefore that of the $Y_{50}^{1000}$ 's, so the solutions must exist of the course

$ψ_{East fifty yard} (r, θ, ϕ) = R_{E l m} (r) Y_{l}^{m} (θ, ϕ) .$

Now, observe that in the Schrödinger equation above, the angular role of ${\vec{\nabla}}^{2}$ is exactly the differential operator $50^{ii} / 2 chiliad r^{ii}$ , so operating on $ψ_{E l m} (r, θ, ϕ) = R_{E 50 m} (r) Y_{l}^{m} (θ, ϕ)$ it will give $ℏ^{2} l (l + 1) / ii 1000 r^{two}$ .

The spherical harmonic $Y_{50}^{m}$ can then exist cancelled from the two sides of the equation leaving:

$\begin{array}{l} - \frac{ℏ^{ii}}{2 k} (\frac{1}{r^{2}} \frac{d}{d r} (r^{2} \frac{d}{d r}) - \frac{l (l + one)}{r^{2}}) R_{E fifty} (r) - \frac{{due east}^{two}}{r} R_{Eastward l} (r) \\ = Due east R_{Eastward l} (r) \end{array}$

information technology now being apparent that $R (r)$ cannot depend on $chiliad .$

The radial derivatives simplify if one factors out $ane / r$ from the function $R,$ writing

$R_{E l} (r) = \frac{u (r)}{r}$

and temporarily suppressing the $E$ and $50$ to reduce ataxia.

The equation becomes:

$- \frac{ℏ^{two}}{2 thousand} (\frac{d^{ii}}{d r^{2}} - \frac{50 (l + i)}{r^{2}}) u (r) - \frac{e^{two}}{r} u (r) = Eastward u (r) .$

Rearranging,

$- \frac{ℏ^{2}}{two g} \frac{d^{2} u (r)}{d r^{2}} + (\frac{ℏ^{ii}}{2 g} \frac{fifty (fifty + 1)}{r^{two}} - \frac{{east}^{2}}{r}) u (r) = E u (r) .$

Note that this is the same as the Schrödinger equation for a particle in ane dimension, restricted to $r > 0$ , in a potential (for $l \neq 0$ ) going to positive infinity at the origin, so negative and going to nil at big distances, so it always has a minimum for some positive $r .$

Nosotros are interested in jump states of the proton-electron system, so $E$ will be a negative quantity. At large separations, the wave equation simplifies to

$- \frac{ℏ^{2}}{2 m} \frac{d^{ii} u (r)}{d r^{2}} ≅ E u (r) (for large r)$

having estimate solutions $e^{κ r}, e^{- κ r} where κ = \sqrt{- 2 m E / ℏ^{two}} .$

The bound states we are looking for, of course, have exponentially decreasing wave functions at large distances.

Going to a Dimensionless Variable

To further simplify the equation, we introduce the dimensionless variable

$ρ = κ r, κ = \sqrt{- 2 k E / ℏ^{two}}$

giving

$\frac{d^{2} u (ρ)}{d ρ^{2}} = (1 - \frac{two ν}{ρ} + \frac{l (l + i)}{ρ^{2}}) u (ρ)$

where (for reasons which will become credible shortly) we have introduced $ν$ defined past

$ii ν = {due east}^{2} κ / E .$

Observe that in transforming from $r$ to the dimensionless variable $ρ$ the scaling factor $κ$ depends on energy, so will be different for unlike energy spring states!

Consider now the beliefs of the wave part near the origin. The dominant term for sufficiently small $ρ$ is the centrifugal ane, and so

$\frac{d^{2} u (ρ)}{d ρ^{2}} ≅ \frac{50 (50 + 1)}{ρ^{2}} u (ρ)$

for which the solutions are $u (ρ) \sim ρ^{- 50}, u (ρ) \sim ρ^{fifty + 1} .$ Since the moving ridge part cannot exist atypical, we cull the second.

Nosotros have established that the wave function decays as $e^{- κ r} = {due east}^{- ρ}$ at large distances, and goes as $ρ^{l + one}$ shut to the origin. Factoring out these ii asymptotic behaviors, define $w (ρ)$ past

$u (ρ) = e^{- ρ} ρ^{l + 1} w (ρ) .$

It is straightforward (if tedious!) to establish that $w (ρ)$ satisfies the differential equation:

$ρ \frac{d^{ii} w (ρ)}{d ρ^{two}} + 2 (fifty + 1 - ρ) \frac{d w (ρ)}{d ρ} + 2 (ν - (l + 1)) w (ρ) = 0.$

Putting in a trial series solution $w (ρ) = \sum_{g = 0}^{\infty} w_{k} ρ^{k}$ gives a recurrence relation between successive coefficients:

$\frac{w_{k + one}}{{west}_{chiliad}} = \frac{2 (k + l + 1 - ν)}{(m + 1) (one thousand + 2 (l + 1))} .$

For big values of $k,$ $w_{k + i} / {west}_{chiliad} \to ii / k$ , so $w_{k} \sim 2^{one thousand} / thou!$ and therefore $w (ρ) \sim {due east}^{2 ρ}$ . This means we have found the diverging radial wavefunction $u (ρ) \sim e^{ρ}$ , which is in fact the right behavior for general values of the energy.

To discover the spring states, nosotros must choose energies such that the series is non an infinite i. As long as the serial stops somewhere, the exponential subtract will eventually have over, and yield a finite (bound state) wave part. Just equally for the unproblematic harmonic oscillator, this tin only happen if for some $one thousand, w_{grand + i} = 0.$ Inspecting the ratio ${westward}_{k + ane} / w_{k}$ , evidently the condition for a leap country is that

$ν = n, an integer$

in which example the series for $westward (ρ)$ terminates at $k = north - l - 1.$ From now on, since we know that for the functions we're interested in $ν$ is an integer, we replace $ν$ past $n .$

To find the energies of these bound states, think $2 north = 2 ν = e^{2} κ / East$ and $κ = \sqrt{- 2 g Eastward / ℏ^{ii}}$ , so

$4 n^{2} = \frac{e^{4} κ_{northward}^{ii}}{{Eastward}_{north}^{ii}} = - \frac{{east}^{four}}{{Eastward}_{n}^{ii}} \frac{ii 1000 E_{north}}{ℏ^{ii}},$

and then

${East}_{north} = - \frac{m {due east}^{4}}{2 ℏ^{two}} \frac{1}{n^{2}} = - \frac{xiii.six}{{north}^{2}} ev = - \frac{i}{{north}^{2}} Rydberg .$

(This defines the Rydberg, a popular unit of energy in atomic physics.)

Remarkably, this is the very same series of spring state energies found by Bohr from his model! Of course, this had better be the case, since the series of energies Bohr found correctly deemed for the spectral lines emitted by hot hydrogen atoms. Find, though, that there are some important differences with the Bohr model: the energy here is determined entirely by $n,$ called the master quantum number, but, in contrast to Bohr'south model, $n$ is not the athwart momentum. The truthful ground state of the hydrogen atom, $n = 1,$ has nil athwart momentum: since $n = k + l + 1$ , $n = i$ means both $fifty = 0$ and $k = 0.$ The ground state wave function is therefore spherically symmetric, and the function $westward (ρ) = {westward}_{0}$ is but a constant. Hence $u (ρ) = ρ {eastward}^{- ρ} w_{0}$ and the actual radial wave function is this divided past $r,$ and of course suitably normalized.

To write the wave office in terms of $r,$ we need to find $κ$ (the energy-dependent scaling factor nosotros used in going to a dimensionless variable). Putting together $ρ = κ_{n} r$ , $κ_{due north} = \sqrt{- 2 m {East}_{northward} / ℏ^{2}}$ and $E_{due north} = - \frac{m e^{four}}{2 ℏ^{2}} \frac{1}{n^{2}}$ ,

$κ_{n} = \sqrt{2 m \frac{m e^{four}}{ii ℏ^{2}} \frac{1}{n^{2}}} / ℏ = \frac{m e^{2}}{ℏ^{2} n} = \frac{one}{a_{0} n},$

where the length

$a_{0} = \frac{ℏ^{2}}{m e^{ii}} = 0.529 \times 10^{- 10} k .$

is chosen the Bohr radius: it is in fact the radius of the lowest orbit in Bohr's model.

Exercise : check this last argument.

It is worth noting at this signal that the energy levels can be written in terms of the Bohr radius a ₀:

$E_{n} = - \frac{{east}^{2}}{2 a_{0}} \frac{one}{n^{ii}} .$

(This is really obvious: remember that the energies ${East}_{n}$ are identical to those in the Bohr model, in which the radius of the $n^{th}$ orbit is ${northward}^{2} a_{0},$ and so the electrostatic potential energy is $- e^{ii} / n a_{0},$ etc.)

Moving on to the excited states: for $n = 2,$ we have a selection: either the radial function $due west (ρ)$ can have one term, as earlier, but now the angular momentum $l = i$ (since $n = chiliad + fifty + 1$ );or $w (ρ)$ can accept two terms (so $thousand = i$ ), and $50 = 0.$ Both options give the same free energy, -0.25 Ry, since $n$ is the same, and the energy just depends on $due north .$

In fact, there are four states at this energy, since $l = 1$ has states with $grand = i, 1000 = 0$ and $m = - 1,$ and $fifty = 0$ has the 1 state $m = 0.$ (For the moment, nosotros are not counting the extra factor of ii from the two possible spin orientations of the electron.)

For $n = 3,$ there are 9 states altogether: gives one, $l = 1$ gives iii and $l = 2$ gives v dissimilar $m$ values. In fact, for principal quantum number $northward$ there are ${north}^{2}$ degenerate states. ( $n^{2}$ being the sum of the first $n$ odd integers.)

The states tin be mapped out, free energy vertically, angular momentum horizontally:

The energy $Eastward = - 1 / {north}^{2},$ the levels are labeled $n l, n$ being the principal quantum number and the traditional notation for athwart momentum $50$ is given at the bottom of the diagram. The 2 red vertical arrows are the first two transitions in the spectroscopic Balmer series, 4 lines of which gave Bohr the clue that led to his model. The respective series of transitions to the $1 s$ ground country are in the ultraviolet, they are called the Lyman serial.

Moving ridge Functions for some Depression-northward States

From at present on, we label the moving ridge functions with the quantum numbers, $ψ_{due north l chiliad} (r, θ, ϕ)$ , and then the ground state is the spherically symmetric $ψ_{100} (r)$ .

For this state $R (r) = u (r) / r$ , where $u (ρ) = e^{- ρ} ρ^{l + i} westward (ρ) = e^{- ρ} ρ w_{0}$ , with ${due west}_{0}$ a constant, and $ρ = κ_{1} r = r / a_{0}$ .

So, equally a function of $r,$ $ψ_{100} (r) = N e^{- r / a_{0}}$ with $N$ an hands evaluated normalization constant:

$ψ_{100} (r) = {(\frac{1}{π a_{0}^{3}})}^{1 / two} e^{- r / a_{0}} .$

For $n = 2, l = 1$ the role $due west (ρ)$ is all the same a unmarried term, a constant, only at present $u (ρ) = e^{- ρ} ρ^{50 + one} w (ρ) = e^{- ρ} ρ^{two} w_{0}$ , and, for $n = 2, ρ = κ r = r / ii a_{0}$ , remembering the free energy-dependence of $κ .$

Therefore $ψ_{210} (r, θ, ϕ) = Due north (\frac{r}{a_{0}}) {due east}^{- r / 2 a_{0}} \cos θ$ . Once more, evaluating the normalization constant is routine, yielding

$ψ_{210} (r, θ, ϕ) = {(\frac{one}{32 π a_{0}^{3}})}^{one / ii} (\frac{r}{a_{0}}) {east}^{- r / ii a_{0}} \cos θ$ .

The wave functions for the other $thousand$ -values, $ψ_{21 \pm ane} (r, θ, ϕ),$ have the $\cos θ$ in $ψ_{210}$ replaced by $\mp (i / \sqrt{ii}) \sin θ {east}^{\pm i ϕ}$ respectively (from the before discussion of the $Y_{fifty}^{grand}$ 's).

The other $n = ii$ state has $l = 0,$ so from $north = yard + l + one,$ nosotros have $k = 1$ and the series for $w$ has two terms, $g = 0$ and $k = ane,$ the ratio existence

$\frac{{west}_{k + 1}}{{due west}_{thou}} = \frac{two (k + 50 + 1 - northward)}{(thou + 1) (g + 2 (50 + 1))} = - i$

for the relevant values: $chiliad = 0, l = 0, n = 2.$ So $w_{1} = - w_{0}, west (ρ) = {westward}_{0} (1 - ρ)$ . For $n = 2,$ $ρ = r / 2 a_{0}$ , the normalized wave function is

$ψ_{200} (r) = {(\frac{1}{32 π a_{0}^{3}})}^{1 / 2} (two - \frac{r}{a_{0}}) e^{- r / ii a_{0}} .$

Notation that the zero athwart momentum wave functions are nonzero and have nonzero slope at the origin. This ways that the total three dimensional wave functions have a slope discontinuity at that place! Only this is fine $—$ the potential is infinite at the origin. (Really, the proton is not a point charge, so really the kink volition be smoothed out over a volume of the size of the proton $—$ a very tiny effect.)

General Solution of the Radial Equation

In exercise, the first few radial functions $westward (ρ)$ can be constructed fairly hands using the method presented higher up, but it should be noted that the differential equation for $w (ρ)$

$ρ \frac{d^{ii} west (ρ)}{d ρ^{two}} + 2 (l + 1 - ρ) \frac{d w (ρ)}{d ρ} + 2 (n - (l + 1)) w (ρ) = 0$

is in fact Laplace's equation, usually written

$(z \frac{d^{ii}}{d z^{2}} + (m - one - z) \frac{d}{d z} + p) 50_{p}^{g} (z) = 0$

where $yard, p$ are integers, and ${Fifty}_{p}^{thou} (z)$ is a Laguerre polynomial (Messiah, folio 482).

The 2 equations are the same if z = 2 ρ , and the solution to the radial equation is therefore

$w_{n l} (ρ) = {Fifty}_{due north - l - ane}^{2 l + one} (ii ρ) .$

Quoting Messiah, the Laguerre polynomials $L_{p}^{0} (z)$ , and the associated Laguerre polynomials $L_{p}^{chiliad} (z)$ are given by:

$\begin{array}{l} L_{p}^{0} (z) = e^{z} \frac{d^{p}}{d z^{p}} {due east}^{- z} z^{p} \\ L_{p}^{m} (z) = {(- i)}^{k} \frac{d^{k}}{d z^{one thousand}} L_{p + k}^{0} (z) . \end{array}$

(These representations can be institute neatly by solving Laplace'southward equation using $-$ surprise $-$ a Laplace transform. Come across Merzbacher for details.) The polynomials satisfy the orthonormality relations (with the mathematicians' normalization convention)

$\int_{0}^{\infty} {due east}^{- z} z^{k} 50_{p}^{grand} L_{q}^{k} d z = \frac{{[(p + k)!]}^{iii}}{p!} δ_{p q} .$

Only what exercise they look similar? The function $e^{- z} z^{p}$ is zero at the origin (apart from the niggling example $p = 0$ ) and cipher at infinity, always positive and having nonzero slope except at its maximum value, $z = p$ . The $p$ derivatives bring in $p$ separated zeroes, easily checked past sketching the curves generated by successive differentiation. Therefore, $L_{p}^{0} (z)$ , a polynomial of degree $p,$ has $p$ real positive zeroes, and value at the origin $50_{p}^{0} (0) = p!$ , since the simply nonzero term at $z = 0$ is that generated by all $p$ differential operators acting on $z^{p}$ .

The associated Laguerre polynomial $L_{p}^{1000} (z)$ is generated by differentiating $50_{p + k}^{0} (z) k$ times. At present $L_{p + thou}^{0} (z)$ has $p + k$ real positive zeroes, differentiating it gives a polynomial one degree lower, with zeroes which must be one in each interval between the zeroes of $L_{p + one thousand}^{0} (z)$ . This argument remains valid for successive differentiations, so $L_{p}^{m} (z)$ must have p existent separate zeroes.

Putting all this together, and translating back from $ρ$ to $r,$ the radial solutions are:

$R_{n l} (r) = Northward {due east}^{- r / northward a_{0}} {(\frac{r}{n a_{0}})}^{l} L_{north - l - 1}^{ii l + 1} (\frac{2 r}{n a_{0}})$

with $Due north$ the normalization abiding. Griffiths (page 141) gives more details, including the normalization constants worked out. We used those to plot the $n = 3$ states $—$ plotting here the functions $u (r) = r R (r)$ , since the normalization is $4 π \int_{0}^{\infty} {| u (r) |}^{2} d r = 1$ , $u (r)$ gives a better idea of at what distance from the proton the electron is nigh likely to be found.

Here are the three $northward = 3$ radial wave functions:

The number of nodes, the radial quantum number, is $iii - 50 - 1.$ (Note: The relative normalizations are correct hither, but not the overall normalization.)

For higher $n$ values, the wave functions become reminiscent of classical mechanics. for example, for $n = 10,$ the highest angular momentum state probability distribution peaks at $r = 100 a_{0}$ , the Bohr orbit radius:

whereas for $northward = 10, l = 0,$ we observe:

Notice this peaks just below twice the Bohr radius. This can be understood from classical mechanics: for an changed square force police, elliptical orbits with the same semimajor axis accept the aforementioned energy. The $l = n - 1$ orbit is a circle, the $l = 0$ orbit is a long thin ellipse (i finish close to the proton), so it extends well-nigh twice as far from the origin as the circle. Furthermore, the orbiting electron volition spend longer at the far altitude, since it will exist moving very slowly.

(Note: the normalizations in the above graphs are simply estimate.)

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