Raising and Lowering Operators for the Harmonic Oscillator

It is fairly easy to verify that the ground-state eigenfunction of a harmonic oscillator is a gaussian, $\exp(-\frac{\sqrt{m\,k}}{2\hbar}\,x^2)$, where $m$ is the mass of the particle and $k$ is the force constant of the harmonic oscillator.

A really slick scheme has been devised for generating all the other eigenvectors of the harmonic oscillator from the ground state, using `raising and lowering operators'.

This scheme turns out to be a useful tool in many areas of quantum chemistry. Similar tricks simplify the treatment of:

Rotational motion of a rigid molecule (`Rigid Rotor').

Spectroscopy, especially light-matter interactions and lasers.

The Pauli Exclusion Principle.

Solids, polymers and other `infinite' systems.

This section is highly mathematical and introduces some completely new mathematical techniques, so getting comfortable with it requires several readings. But once you get it, it will help a lot with later sections. It will probably be a bit overwhelming at first. The key thing to focus on is how to use the raising and lowering operators to produce new eigenvalues and eigenvectors.

The (fairly long) derivation of the expression for the Hamiltonian in terms of raising and lowering operators is not examinable. In order to help you focus on the important bits I've included hints on which bits are highly examinable. Exam questions will be straightforward applications of the raising and lowering operators, and will be very similar to the `examinable' exercises in these notes. You will be given the Hamiltonian in terms of the raising and lowering operators, and any necessary commutation relations. You will not be asked to derive the raising or lowering operators, or the expression for the Hamiltonian in terms of raising and lowering operators.

Commutators

Before we get going on the harmonic oscillator we need to define a commutator. The commutator $\mathbf{[A,B]}$ of two operators $A$ and $B$ is,

$$[A,B]= A\,B - B\,A$$

It follows that

$$[A,B]= -[B,A]$$

Example: $A = p,\;\;\;B = x$.
For any function $f(x)$,

$$\begin{eqnarray*}[p,x]f(x) & = & \left( p\,x - x\,p\right) f(x) \\ & = & \frac... ...d x} - x\frac{d f(x)}{d x}\right) \\ & = & \frac{\hbar}{i} f(x) \end{eqnarray*}$$

The commutator of $p$ and $x$ is,

Application: The generalized version of the Heisenberg uncertainty principle involves a commutator,

$$\Delta A\,\Delta B \ge \frac{1}{2}\left\vert\int \psi^\ast [A,B] \psi d\tau \right\vert$$

For operators $p$ and $x$ this becomes

$$\Delta p\,\Delta x \ge \frac{1}{2}\left\vert\int\psi^\ast \frac{\hbar}{i} \psi dx\right\vert = \frac{1}{2}\frac{\hbar}{i}$$

Simplified Hamiltonian

The Schrodinger equation for a one-dimensional harmonic oscillator is,

$$\left(\frac{1}{2m} p_z^2 + \frac{1}{2} k z^2\right)\psi = {\cal E} \psi$$

I'm using slightly weird variables ($z$ and ${\cal E}$) here, so as to save the `normal' variables for the simplified Hamiltonian. The mass $m$ and force-constant $k$ can be eliminated, or at least hidden, by rescaling the position and energy variables,

$$x = \sqrt{m\omega}z,\;\;\;\;\;\; E = \frac{{\cal E}}{\omega},\;\;\;\;\;\; \omega = \sqrt{\frac{k}{m}}$$

to give the simplified Hamiltonian and Schrodinger equation,

$$\begin{array}{\vert rcl\vert} \hline & & \\ [-2.0ex] H & = ... ...x^2 + x^2\right) \psi & = & E\psi \\ [1.0ex] \hline \end{array}$$

Class exercise: Check this.

From now on we'll use this simplified equation.

Factorizing the Hamiltonian

Classical Mechanics

Because the Hamiltonian

$$H = \frac{1}{2}\left(p^2 + x^2\right)$$

is a quadratic function of $x$ and $p$, it is easy to factorize. Recall the procedure for factorizing a quadratic;

$$a^2 + b^2 = a^2 - (i\,b)^2 = (a - i \,b)(a + i \,b)\;\;\;\;\;\;\;\; \mbox{for any two numbers $a$\ and $b$}$$

Applying this to $H$ gives,

$$H_{\mbox{classical}} = \frac{1}{2}(x - i\,p)(x + i\,p)$$

This factorization of the Hamiltonian greatly simplifies the solution of the harmonic oscillator.

Special coordinates for the harmonic oscillator

Sometimes it is easier to use special coordinates like polar coordinates $(r,\theta)$ rather than cartesian coordinates $(x,y)$. The Hamiltonian for the harmonic oscillator is most simply expressed in a special coordinate system, but to do this we have to know a bit more about classical mechanics.

In the Hamiltonian formulation of classical mechanics $x$ and $p$ are both regarded as coordinates in phase space, $(x,p)$. For every coordinate $x$ in real space there are a pair of coordinates $(x,p)$ in phase space. For example, in real life there are three dimensions in real space $(x,y,z)$ and six dimensions in phase space $(x,y,z,p_x,p_y,p_z)$.

Returning to the one-dimensional harmonic oscillator, the usual phase-space coordinates are $(x,p)$, and a convenient set of special phase-space coodinates for the factorized Hamiltonian is $(x - i\,p,\,x + i\,p)$ .

Quantum Mechanics

In quantum mechanics a small complication arises when factorizing the Hamiltonian, because $p$ and $x$ don't commute;

$$\begin{eqnarray*} \frac{1}{2}(x - i\,p)(x + i\,p) & = & \frac{1}{2}\left( p^2 +... ...\right) - \frac{1}{2}i\,[p,x] \\ & = & H - \frac{1}{2}i\,[p,x] \end{eqnarray*}$$

Classically, $[p,x] = 0$.
But in quantum mechanics ${\displaystyle [p,x] = \frac{\hbar}{i}}$.
So in quantum mechanics,

$$\frac{1}{2}(x - i\,p)(x + i\,p) = H - \frac{1}{2}\hbar$$

So the quantum Hamiltonian is the sum of the factorized term and an additional `quantum' term;

$$H = \frac{1}{2}(x - i\,p)(x + i\,p) + \frac{1}{2}\hbar$$ (1)

Raising and Lowering Operators

The harmonic oscillator Hamiltonian is most simply expressed in terms of the special phase-space coordinates $(x - i\,p,\,x + i\,p)$. These special coordinates are called raising and lowering operators,

$$\begin{eqnarray*} \mathbf{ a} & = & \frac{1}{\sqrt{2\hbar}}\left( x + i\,p\right... ...ight)\;\;\;\;\;\;\mbox{\textcolor {blue}{\bf raising operator}} \end{eqnarray*}$$

For now the dagger symbol $^\dag \;$ just means complex conjugate, $i\rightarrow -i$. Later on we'll give the true definition, Hermitean conjugate.

The factor of $1/\sqrt{2\hbar}$ is included so that $a$ and $a^\dag$ have a very simple commutator,

Class exercise: Prove this, by expressing $a$ and $a^\dag$ in terms of $x$ and $p$, and then using $[p,x]$.

Later, we'll repeatedly make use of the commutator of $a$ and $a^\dag$ to reorder products like $a^\dag a$ as $a\,a^\dag -1$.

Hamiltonian expressed in terms of raising and lowering operators

The Hamiltonian (Eq. (1)) expressed in terms of the raising and lowering operators is,

Exercise: Make sure you understand this.

The operator

is called the number operator, because its eigenvalues turn out to be $0,1,2,3\ldots$ and label the eigenvectors in order of increasing energy. I put a hat over the number operator to make it clear that it's an operator. The Hamiltonian is often written as,

Changing back and forth between the $(x,p)$ and $(a,a^\dag )$ coordinates

Occasionally it may be necessary to revert back to the original $(x,p)$ coordinates. For this we need the inverse coordinate transformation,

$$\begin{eqnarray*} x & = & \sqrt{\frac{\hbar}{2}} (a + a^\dag ) \\ p & = & -i \sqrt{\frac{\hbar}{2}} (a - a^\dag ) \end{eqnarray*}$$

Class exercise: Derive this, starting from the expression for $a$ and $a^\dag$ in terms of $x$ and $p$.

Eigenvalues and Eigenvectors

Consider an arbitrary eigenfunction of $\hat{n}$ with eigenvalue $\lambda$. For simplicity denote the eigenfunction by $\vert\lambda\rangle$,

$$\hat{n} \vert\lambda\rangle = \lambda \vert\lambda\rangle$$

Class exercise: Prove that $\vert\lambda\rangle$ is also an eigenfunction of $H$,

$$H \vert\lambda\rangle = \hbar(\lambda + \frac{1}{2}) \vert\lambda\rangle$$ (2)

hint: Take the left hand side of the equation, and express $H$ in terms of $\hat{n}$. [highly examinable].

Comment: The eigenvalues of $H$ and $\hat{n}$ are proportional to each other. As $\lambda$ increases, the energy also increases.

The raising and lowering operators raise/lower the eigenvalue of $\hat{n}$ by $\pm 1$, ie

$$\begin{array}{\vert rcrr\vert} \hline \hat{n} \vert\lambda\r... ...& \mbox{\textcolor {blue}{\bf lowering}} \\ \hline \end{array}$$

Given an eigenvector $\vert\lambda\rangle$ of $\hat{n}$ with eigenvalue $\lambda$, we obtain a new eigenvector $a^\dag \vert\lambda\rangle$ with eigenvalue $\lambda +1$, and a new eigenvector $a \vert\lambda\rangle$ with eigenvalue $\lambda - 1$.

Class exercise: Prove that $\hat{n} a^\dag \vert\lambda\rangle = (\lambda + 1) a^\dag \vert\lambda\rangle$ by expressing $\hat{n}$ in terms of $a^\dag$ and $a$, and then using $[a,a^\dag ] = 1$ to reorder the left hand side of the equation. [very highly examinable].

By repeatedly applying the raising and lowering operators we can can obtain all the eigenvalues and eigenvectors.

The eigenvalues form a series $\ldots,\lambda-2,\lambda-1,\lambda,\lambda+1,\lambda+2,\ldots$.

Exercise: Make sure you understand this. [highly examinable].

The minimum eigenvalue of the number operator is zero

Next we need to show that $\lambda$ is an integer,

$$\lambda = 0,1,2,\ldots$$

The argument in this section is a bit flaky. The right way to do it is to show that $\int \psi^\ast\, a^\dag a\, \psi dx \ge 0$, but we can't show that until we've learned more about the Hermitean conjugate $^\dag \;$ .

The series $\ldots,\lambda-2,\lambda-1,\lambda,\lambda+1,\lambda+2,\ldots$ seems to go on forever in both directions, but if this were true it would imply that a particle in a harmonic potential could have infinitely negative energy, $\lambda-\infty$, which would be weird! In order to avoid infinitely negative energy, the series must terminate at some minimum value. The only way to break the infinite chain of eigenvalues is if at some stage the lowering operator gives zero,

$$a \vert\lambda_{\mbox{min}}\rangle = 0$$

We can rewrite this as

$$a \vert\lambda_{\mbox{min}}\rangle = 0\vert\lambda_{\mbox{min}}\rangle$$

since zero times anything is zero, so the right hand side is still zero. Multiplying both sides by $a^\dag$, and using $a^\dag 0 = 0$, gives,

$$\hat{n} \vert\lambda_{\mbox{min}}\rangle = 0 \vert\lambda_{\mbox{min}}\rangle$$

$\vert\lambda_{\mbox{min}}\rangle$ is an eigenvector of $\hat{n}$ with eigenvalue 0, so

$$\lambda_{\mbox{min}}=0$$

By repeatedly applying the raising operator to $\vert\lambda_{\mbox{min}}\rangle$ we obtain the complete set of eigenvectors and eigenvalues. In particular $\hat{n}$ has eigenvalues

$$n = 0,1,2,3,\ldots$$

and eigenvectors $\vert\rangle,\vert 1\rangle,\vert 2\rangle,\vert 3\rangle,\ldots$.

Exercise: Make sure you understand how to generate the series of eigenvalues. [highly examinable. Proving that $\lambda_{\mbox{min}}=0$ is not examinable.].

The eigenvalues of the Hamiltonian

The simplified Hamiltonian $H$ has eigenvalues

$$E = (n + \frac{1}{2})\hbar\;\;\;\;\;\;,n = 0,1,2,\ldots$$

, (this follows from Eq. (2).)

The original eigenvalues are ${\cal E} = E\omega$,

$${\cal E} = (n + \frac{1}{2})\hbar\omega\;\;\;\;\;\;,n = 0,1,2,\ldots$$

,

The eigenvalues are equally spaced, and the ground-state eigenvalue, also known as the zero point energy, is equal to half the spacing. [Qualitative aspects are highly examinable, though you don't have to memorize the formula.]

Compared with a `brute force' approach, a significant advantage of raising and lowering operators is that they enable us to determine the eigenvalues without ever having to find the eigenvectors!

The eigenvectors

It is fairly easy to verify that

$$\vert\rangle = e^{-\frac{x^2}{2\hbar}}$$

is an eigenfunction of $H$. This eigenfunction has no nodes, so we (correctly) suspect that it is the ground state. Alternatively, the ground state eigenfunction can be derived from the equation

$$a \vert\rangle = 0$$

Substituting in $a = \frac{1}{\sqrt{2\hbar}}\left( x + i\,p\right)$ and $p = \frac{\hbar}{i}\frac{d}{dx}$ gives a first order differential equation,

$$\left( x + \hbar\,\frac{d}{dx}\right) \vert\rangle = 0$$

with solution

$$\vert\rangle = e^{-\frac{x^2}{2\hbar}}$$

Exercise: What does the eigenvector $\vert\rangle$ look like in the classical limit where $\hbar\rightarrow 0$? What does the probability distribution (the square of the wavefunction) look like? Relate your answer to the known behaviour of a classical harmonic oscillator (eg a classical particle in a harmonic potential) in its ground state.

The first excited state can be obtained by applying the raising operator,

$$\vert 1\rangle = a^\dag \vert\rangle = \frac{1}{\sqrt{2\hbar... ...hbar}} + \frac{1}{\sqrt{2\hbar}} x\,e^{-\frac{x^2}{2\hbar}}$$

So the unnormalized state $\vert 1\rangle$ is just $x\,e^{-x^2/2}$. The second excited state, $\vert 2\rangle$, can be obtained similarly,

$$\vert 2\rangle = a^\dag \vert 1\rangle$$

and so on. .

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