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{{Quantum Mechanics A}}
{{Quantum Mechanics A}}
The motivation for exploring eigenvalue quantization comes form wanting to solve the energy eigenvalue problem.  It is not possible, in general, to specify and measure more than one component <math> \hat{\mathbf{n}}\cdot\mathbf{L} </math> of orbital angular momentum.  Yet it is possible to specify <math> \mathbf{L}^2 </math> simulataneously with any one component of <math> \mathbf{L} </math>. Normally <math>\ L_z </math> is choosen. A central force potential has a Hamiltonian that commutes with <math> \mathbf{L} </math>, and in that case one can require the energy eigenstates of the system to also be eigenvectors of <math>  \mathbf{L}^2 </math> and one component of <math> \mathbf{L} (L_z) </math>.
The motivation for exploring eigenvalue quantization comes form wanting to solve the energy eigenvalue problem for a particle in a central potential.  It is not possible, in general, to specify and measure more than one component <math> \hat{\mathbf{n}}\cdot\hat{\mathbf{L}}</math> of the orbital angular momentum.  It is, however, possible to specify <math> \hat{\mathbf{L}}^2 </math> simulataneously with any one component of <math>\hat{\mathbf{L}},</math> since <math>\hat{\mathbf{L}}^2</math> commutes with all of its Cartesian components, as we saw earlier. We typically choose <math>\hat{L}_z.</math>  A central potential yields a Hamiltonian that commutes with <math>\hat{\mathbf{L}},</math> and thus the energy eigenstates of the system may be chosen to be eigenvectors of <math>  \hat{\mathbf{L}}^2 </math> and one component of <math>\hat{\mathbf{L}},</math> usually <math>\hat{L}_z.</math>


The quantization of angular momentum follows simply from the above commutation relations. Define <math>\mathbf{L}^2\!</math> by:
The quantization of angular momentum follows simply from the [[Commutation Relations|commutation relations]] derived earlier. Recall that <math>\mathbf{L}^2\!</math> is given by
:<math>\mathbf{L}^2=L_x^2+L_y^2+L_z^2</math>


Since <math>\mathbf{L}^2\!</math> is a scalar, it commutes with each component of angular momentum.
<math>\hat{\mathbf{L}}^2=\hat{L}_x^2+\hat{L}_y^2+\hat{L}_z^2.</math>


Now Define a change of operators as follows:
Let us now define the operators, <math>\hat{L}_\pm=\hat{L}_x\pm i\hat{L}_y.\!</math> Note that <math>\hat{L}_+</math> and <math>\hat{L}_-</math> are Hermitian conjugates of each other.
:<math>L_+=L_x+iL_y\!</math>
:<math>L_-=L_x-iL_y\!</math>


The choice of these two operators are choosen because their commutation relations with <math> L_z \!</math> and one another only result in soultions with <math>  L_- \!</math>, <math>  L_+ \!</math>, and <math> L_z \!</math>.
We choose to work with these operators because, as we will see shortly, the operators <math>\hat{L}_\pm</math> function as raising and lowering operators for eigenvectors of <math>\hat{L}_z.</math>


From the commutation relations we get
We may use the commutation relations derived earlier to show that
:<math>L_+L_-=\mathbf{L}^2-L_z^2+\hbar L_z</math>


Similarly,
<math>\hat{L}_+\hat{L}_-=\hat{\mathbf{L}}^2-\hat{L}_z^2+\hbar\hat{L}_z</math>
:<math>L_-L_+=\mathbf{L}^2-L_z^2-\hbar L_z</math>
 
and
 
<math>\hat{L}_-\hat{L}_+=\hat{\mathbf{L}}^2-\hat{L}_z^2-\hbar\hat{L}_z.</math>
                                                              
                                                              
Thus,
Therefore,
:<math>[L_+,L_-]=L_+L_--L_-L_+=2\hbar L_z</math>
 
<math>[\hat{L}_+,\hat{L}_-]=\hat{L}_+\hat{L}_--\hat{L}_-\hat{L}_+=2\hbar\hat{L}_z.</math>
 
We may also show that
<math>[\hat{L}_z,\hat{L}_\pm]=\hat{L}_z\hat{L}_\pm-\hat{L}_\pm\hat{L}_z=\pm\hbar\hat{L}_\pm.</math>
 
We may also easily see that <math>[\hat{\mathbf{L}}^2,\hat{L}_\pm]=0.</math>
 
Let <math>|\beta,m\rangle\!</math> be a normalized eigenstate of <math>\hat{\mathbf{L}}^2</math> with eigenvalue <math>\beta\!</math> and of <math>\hat{L}_z\!</math> with eigenvalue <math>m\hbar.</math>  Let us first determine what the effects of the operators, <math>\hat{L}_\pm,</math> are.  By definition,
 
<math>\hat{L}_z|\beta,m\rangle=m\hbar|\beta,m\rangle.</math>
 
If we now act on this expression from the left with <math>\hat{L}_\pm</math> and use the above commutation relations, we find that
 
<math>\hat{L}_z\hat{L}_\pm|\beta,m\rangle=(m\pm 1)\hbar\hat{L}_\pm|\beta,m\rangle.</math>
 
We therefore see that <math>\hat{L}_\pm|\beta,m\rangle</math> is also an eigenvector of <math>\hat{L}_z,</math> but with an eigenvalue of <math>(m\pm 1)\hbar.</math>  In other words,
 
<math>\hat{L}_\pm|\beta,m\rangle=\alpha_\pm|\beta,m\pm 1\rangle,</math>
 
where <math>\alpha_\pm\!</math> is a normalization constant.  We see now that, as asserted earlier, the operators <math>\hat{L}_\pm</math> are raising and lowering operators for eigenstates of <math>\hat{L}_z.</math>
 
To find the normalization constant, we will evaluate the norms of these states.  The norm for <math>\hat{L}_-|\beta,m\rangle</math> is, using the expressions derived above,
 
<math>\langle\beta,m|\hat{L}_+\hat{L}_-|\beta,m\rangle=\beta-m(m-1)\hbar^2.</math>
 
The right-hand side of this expression is equal to <math>|\alpha_-|^2.\!</math>  If we now take <math>\alpha_-\!</math> to be real, then
 
<math>\alpha_-=\sqrt{\beta-m(m-1)\hbar^2}.</math>
 
Similarly, the norm for <math>\hat{L}_+|\beta,m\rangle</math> is
 
<math>\langle\beta,m|\hat{L}_-\hat{L}_+|\beta,m\rangle=\beta-m(m+1)\hbar^2,</math>
 
and thus, again taking <math>\alpha_+\!</math> to be real,
 
<math>\alpha_+=\sqrt{\beta-m(m+1)\hbar^2}.</math>
 
In summary,
 
<math>\hat{L}_\pm|\beta,m\rangle=\sqrt{\beta-m(m\pm 1)\hbar^2}|\beta,m\pm 1\rangle.</math>
 
We may now restrict the possible values of <math>\beta\!</math> as follows.  Obviously, it is impossible for <math>\langle\hat{L}_z^2\rangle</math> to be larger than <math>\langle\hat{\mathbf{L}}^2\rangle.</math>  This means that
 
<math>\beta\geq m^2\hbar^2.</math>
 
This implies that there must be both a lower bound and an upper bound on the allowed values of <math>m\!</math> for a given value of <math>\beta.\!</math>  Let <math>l\!</math> be the upper bound on <math>m.\!</math>  Then
 
<math>\hat{L}_+|\beta,l\rangle=0.</math>
 
This means that
 
<math>\sqrt{\beta-l(l+1)\hbar^2}=0,</math>
 
or
 
<math>\beta=l(l+1)\hbar^2>l^2\hbar^2.</math>
 
The value of <math>\beta\!</math> that we obtained is therefore a suitable value, since it is larger than the assumed upper bound on <math>m^2\hbar^2.</math>  If we now consider the lowering operator, then, using this value of <math>\beta,\!</math>


And
<math>\hat{L}_-|\beta,m\rangle=\hbar\sqrt{l(l+1)-m(m-1)}|\beta,m-1\rangle.</math>
:<math>[L_z,L_-]=L_zL_--L_-L_z=-\hbar L_-</math>


Also, It is easy to show that:
We see by direct substitution that, if there is a lower bound, then it must be equal to <math>-l.\!</math>  In order for us to be able to lower a state <math>|\beta,l\rangle</math> to <math>|\beta,-l\rangle</math> by repeated application of the lowering operator, then <math>l\!</math> and <math>-l\!</math> must differ by an integer:
:<math>[L^2,L_\pm]=0</math>


Let <math>L'_z\!</math> be an eigenvalue of <math>L_z\!</math>.
<math>l=-l+n\!</math>
:<math>\langle L_z'|L_+L_-|L_z'\rangle=(\beta-L_z'^2+\hbar L_z')\langle L_z'|L_z'\rangle</math>
                                                             
Since the left hand side of the above equation is the square of the length of a ket, it has to be non-negative. Therefore
:<math>\mathbf{L}^2-L_z'^2+\hbar L_z'\ge0</math>
:<math>\Rightarrow \beta+\frac{\hbar^2}{4}\ge (L_z'-\frac{\hbar}{2})^2</math>
:<math>\Rightarrow \beta+\frac{\hbar^2}{4}\ge 0</math>


Defining the number <math>k\!</math> by
This means that the possible values of <math>l\!</math> are quantized in half-integer steps; i.e.,
:<math>k+\frac{\hbar}{2}=\sqrt{\beta+\frac{\hbar^2}{4}}</math>
:<math>\Rightarrow k\ge -\frac{\hbar}{2}</math>


The inequality 5.1.13 becomes
<math>l=0,\tfrac{1}{2},1,\tfrac{3}{2},\ldots</math>
:<math>k+\frac{\hbar}{2}\ge |L_z'-\frac{\hbar}{2}|</math>
:<math>\Rightarrow k+\hbar\ge L_z'\ge -k</math>


Similarly, from equation 5.1.10, we get
If we chose any other values of <math>l,\!</math> then it would be possible to construct states with arbitrarily low values of <math>m,\!</math> which is impossible.  The allowed values of <math>m\!</math> are all integers such that <math>-l\leq m\leq l\!</math> for an integer value of <math>l,\!</math> or all half-integers satisfying the same constraint if <math>l\!</math> is a half-integer.
:<math>\langle L_z'|L_-L_+|L_z'\rangle=(\beta-L_z'^2-\hbar L_z')\langle L_z'|L_z'\rangle</math>
:<math>\Rightarrow \mathbf{L}^2-L_z'^2-\hbar L_z'\ge0</math>
:<math>\Rightarrow k\ge L_z'\ge -k-\hbar</math>


This result, combined with 5.1.15 shows that
From this point forward, we will use the notation, <math>|l,m\rangle,</math> for an eigenstate of <math>\hat{\mathbf{L}}^2</math> with eigenvalue <math>l(l+1)\hbar^2</math> and of <math>\hat{L}_z</math> with eigenvalue <math>m\hbar.</math> The quantum number <math>l\!</math> is often called the orbital quantum number and <math>m\!</math> the magnetic quantum number, the latter being so named because it characterizes the energy shift of, say, an atom in the presence of a magnetic field.
:<math>k\ge 0</math>
and                                
:<math>k\ge L_z'\ge -k</math>


From 5.1.12
==Problem==
:<math>L_z L_-|L_z'\rangle=(L_- L_z-\hbar L_-)|L_z'\rangle=(L_z'-\hbar)L_-|L_z'\rangle</math>


Now, if <math>L_z'\ne 0</math>, then <math>L_-|L_z'\rangle</math> is an eigenket of <math>L_z\!</math> belonging to the eigenvalue <math>L_z'-\hbar</math>.
Evaluate the following expressions:
Similarly, if <math>L_z'-\hbar\ne -k</math>, then <math>L_z'-2\hbar</math> is another eigenvalue of <math>L_z\!</math>, and so on. In this way, we get a series of eigenvalues which must terminate from 5.1.16, and can terminate only with the value <math>\!-k</math>.
Similarly, using the complex conjugate of 5.1.12, we get that <math>L_z',L_z'+\hbar,L_z'+2\hbar,...\!</math> are eigenvalues of L'z. Thus we may conclude that <math>2k\!</math> is an integral multiple of the Planck's constant, and that the eigenvalues are:
:<math>k, k-\hbar,k-2\hbar,...,-k+\hbar,-k</math>


If <math>|m\rangle\!</math> is an eigenstate of <math>L_z\!</math> with eigenvalue <math>m\hbar\!</math>, then
'''(a)''' <math>\left \langle {l,m\left |{\hat{L}_{x}^{2}} \right |l,m} \right \rangle </math>
:<math>
\begin{align}
L_z L_\pm |m\rangle &= ([L_z,L_\pm]+L_\pm L_z)|m\rangle=(\pm\hbar L_\pm+L_\pm m)|m\rangle \\
&=(m\pm 1)\hbar(L_\pm |m\rangle)
\end{align}
</math>


Which means that <math>L_+\!</math> or <math>L_-\!</math> raises or lowers the <math>z\!</math> component of the angular momentum by <math>\hbar\!</math>.
'''(b)''' <math>\left \langle {l,m\left |{\hat{L}_{x}\hat{L}_{y}} \right |l,m} \right \rangle </math>


[[Phy5645/angularmomcommutation/|Calculation of two angular momentum expressions]]
[[Phy5645/angularmomcommutation/|Solution]]

Latest revision as of 22:59, 31 August 2013

Quantum Mechanics A
SchrodEq.png
Schrödinger Equation
The most fundamental equation of quantum mechanics; given a Hamiltonian Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{H}} , it describes how a state Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\Psi\rangle} evolves in time.
Basic Concepts and Theory of Motion
UV Catastrophe (Black-Body Radiation)
Photoelectric Effect
Stability of Matter
Double Slit Experiment
Stern-Gerlach Experiment
The Principle of Complementarity
The Correspondence Principle
The Philosophy of Quantum Theory
Brief Derivation of Schrödinger Equation
Relation Between the Wave Function and Probability Density
Stationary States
Heisenberg Uncertainty Principle
Some Consequences of the Uncertainty Principle
Linear Vector Spaces and Operators
Commutation Relations and Simultaneous Eigenvalues
The Schrödinger Equation in Dirac Notation
Transformations of Operators and Symmetry
Time Evolution of Expectation Values and Ehrenfest's Theorem
One-Dimensional Bound States
Oscillation Theorem
The Dirac Delta Function Potential
Scattering States, Transmission and Reflection
Motion in a Periodic Potential
Summary of One-Dimensional Systems
Harmonic Oscillator Spectrum and Eigenstates
Analytical Method for Solving the Simple Harmonic Oscillator
Coherent States
Charged Particles in an Electromagnetic Field
WKB Approximation
The Heisenberg Picture: Equations of Motion for Operators
The Interaction Picture
The Virial Theorem
Commutation Relations
Angular Momentum as a Generator of Rotations in 3D
Spherical Coordinates
Eigenvalue Quantization
Orbital Angular Momentum Eigenfunctions
General Formalism
Free Particle in Spherical Coordinates
Spherical Well
Isotropic Harmonic Oscillator
Hydrogen Atom
WKB in Spherical Coordinates
Feynman Path Integrals
The Free-Particle Propagator
Propagator for the Harmonic Oscillator
Differential Cross Section and the Green's Function Formulation of Scattering
Central Potential Scattering and Phase Shifts
Coulomb Potential Scattering

The motivation for exploring eigenvalue quantization comes form wanting to solve the energy eigenvalue problem for a particle in a central potential. It is not possible, in general, to specify and measure more than one component Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{\mathbf{n}}\cdot\hat{\mathbf{L}}} of the orbital angular momentum. It is, however, possible to specify Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{\mathbf{L}}^2 } simulataneously with any one component of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{\mathbf{L}},} since Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{\mathbf{L}}^2} commutes with all of its Cartesian components, as we saw earlier. We typically choose Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{L}_z.} A central potential yields a Hamiltonian that commutes with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{\mathbf{L}},} and thus the energy eigenstates of the system may be chosen to be eigenvectors of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{\mathbf{L}}^2 } and one component of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{\mathbf{L}},} usually Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{L}_z.}

The quantization of angular momentum follows simply from the commutation relations derived earlier. Recall that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{L}^2\!} is given by

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{\mathbf{L}}^2=\hat{L}_x^2+\hat{L}_y^2+\hat{L}_z^2.}

Let us now define the operators, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{L}_\pm=\hat{L}_x\pm i\hat{L}_y.\!} Note that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{L}_+} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{L}_-} are Hermitian conjugates of each other.

We choose to work with these operators because, as we will see shortly, the operators Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{L}_\pm} function as raising and lowering operators for eigenvectors of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{L}_z.}

We may use the commutation relations derived earlier to show that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{L}_+\hat{L}_-=\hat{\mathbf{L}}^2-\hat{L}_z^2+\hbar\hat{L}_z}

and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{L}_-\hat{L}_+=\hat{\mathbf{L}}^2-\hat{L}_z^2-\hbar\hat{L}_z.}

Therefore,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [\hat{L}_+,\hat{L}_-]=\hat{L}_+\hat{L}_--\hat{L}_-\hat{L}_+=2\hbar\hat{L}_z.}

We may also show that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [\hat{L}_z,\hat{L}_\pm]=\hat{L}_z\hat{L}_\pm-\hat{L}_\pm\hat{L}_z=\pm\hbar\hat{L}_\pm.}

We may also easily see that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [\hat{\mathbf{L}}^2,\hat{L}_\pm]=0.}

Let Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\beta,m\rangle\!} be a normalized eigenstate of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{\mathbf{L}}^2} with eigenvalue Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \beta\!} and of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{L}_z\!} with eigenvalue Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m\hbar.} Let us first determine what the effects of the operators, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{L}_\pm,} are. By definition,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{L}_z|\beta,m\rangle=m\hbar|\beta,m\rangle.}

If we now act on this expression from the left with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{L}_\pm} and use the above commutation relations, we find that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{L}_z\hat{L}_\pm|\beta,m\rangle=(m\pm 1)\hbar\hat{L}_\pm|\beta,m\rangle.}

We therefore see that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{L}_\pm|\beta,m\rangle} is also an eigenvector of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{L}_z,} but with an eigenvalue of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (m\pm 1)\hbar.} In other words,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{L}_\pm|\beta,m\rangle=\alpha_\pm|\beta,m\pm 1\rangle,}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \alpha_\pm\!} is a normalization constant. We see now that, as asserted earlier, the operators Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{L}_\pm} are raising and lowering operators for eigenstates of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{L}_z.}

To find the normalization constant, we will evaluate the norms of these states. The norm for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{L}_-|\beta,m\rangle} is, using the expressions derived above,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle\beta,m|\hat{L}_+\hat{L}_-|\beta,m\rangle=\beta-m(m-1)\hbar^2.}

The right-hand side of this expression is equal to Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\alpha_-|^2.\!} If we now take Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \alpha_-\!} to be real, then

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \alpha_-=\sqrt{\beta-m(m-1)\hbar^2}.}

Similarly, the norm for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{L}_+|\beta,m\rangle} is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle\beta,m|\hat{L}_-\hat{L}_+|\beta,m\rangle=\beta-m(m+1)\hbar^2,}

and thus, again taking Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \alpha_+\!} to be real,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \alpha_+=\sqrt{\beta-m(m+1)\hbar^2}.}

In summary,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{L}_\pm|\beta,m\rangle=\sqrt{\beta-m(m\pm 1)\hbar^2}|\beta,m\pm 1\rangle.}

We may now restrict the possible values of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \beta\!} as follows. Obviously, it is impossible for to be larger than This means that

This implies that there must be both a lower bound and an upper bound on the allowed values of for a given value of Let be the upper bound on Then

This means that

or

The value of that we obtained is therefore a suitable value, since it is larger than the assumed upper bound on If we now consider the lowering operator, then, using this value of

We see by direct substitution that, if there is a lower bound, then it must be equal to In order for us to be able to lower a state to by repeated application of the lowering operator, then and must differ by an integer:

This means that the possible values of are quantized in half-integer steps; i.e.,

If we chose any other values of then it would be possible to construct states with arbitrarily low values of which is impossible. The allowed values of are all integers such that for an integer value of or all half-integers satisfying the same constraint if is a half-integer.

From this point forward, we will use the notation, for an eigenstate of with eigenvalue and of with eigenvalue The quantum number is often called the orbital quantum number and the magnetic quantum number, the latter being so named because it characterizes the energy shift of, say, an atom in the presence of a magnetic field.

Problem

Evaluate the following expressions:

(a)

(b)

Solution

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