Linear Operators, Eigenvalues and Eigenfunctions

Operators and Linear Operators

An operator is a mathematical construct which converts (maps) one function into another function. These can be entities such as matrices, or more commonly recognised operators such as integration and differentiation. In fact, calculus is really the study of these latter two operations, differentiation and integration. For example, the differentiation operator maps one function to its derivative.

`d/(dx) f(x) = g(x)`

We indicate an operator with the hat symbol, so we could recast the above differentiation operator as,

`hatD=d/(dx)`

A particularly important type of operator to quantum mechanics, is the linear operator. A linear operator is one for the distributive property applies - this means that the operator can be applied individually to a sum of functions and obtain the same result,

`hatO(alphaF(x)+betaG(x))=alphahatOF(x)+betahatOG(x)`

where `alpha` and `beta` are constants

Differentiation and integration are examples of linear operators, for example,

`hatD(3x^2+5sin(x))=3hatDx^2+5hatDsin(x)=6x+5cos(x)`

...but not all operators are linear. An example of an operator which is non-linear, consider,

`hatQF(x)=F(x)^2`

If we use, `F(x)=3x^2+5sin(x)`, then we can show the operator `hatQ` is non-linear, for example,

`hatQF(x)=(3x^2+5sin(x))^2=9x^4+30x^2sin(x)+25sin^2(x)`

Were we to distribute the operator inside the brackets we obtain,

`3hatQx^2+5hatQsin(x)=3x^4+5sin^2(x)`

Evidently, the two results are not equivalent, and so our operator `hatQ` is non-linear, thus,

`hatQ(3x^2+5sin(x))!=3hatQx^2+5hatQsin(x)`

In quantum mechanics, we do not have to concern ourselves with such non-linear operators, as the overwhelming majority of operators in quantum mechanics are linear.

Eigenvalues and Eigenfunctions

Another extremely important feature in quantum mechanics is the use of Eigenvalue-equations. An eigenvalue equation is an equation involving an operator and a function, which when evaluated returns the product of the original function and a new coefficient,

`hatOf(x)=lambdaf(x)`

According to this formulation, the function `f(x)` is called the eigenfunction of the operator `hatO`, and the constant `lambda` is an eigenvalue of the eigenfunction. Such equations are especially important in quantum mechanics since every observable (position, momentum, energy, spin etc.) is associated with an eigenvalue equation, involving the observables operator for which the measurable quantity is an eigenvalue. For example, the energy of a wave is the eigenvalue of the wave-function, when the Hamiltonian operator is applied, ie.,

`hatHPsi=EPsi`

Whilst it may be tempting to eliminate `Psi` from both sides through division, this cannot be done. An operator cannot be separated from the function on which it is operating - an operator has no solutions when applied to nothing. So what forms can such an operator take? Matrices are one such example, but also the differentiation operator. Consider the expression below,

`d/(dx)f(x)=lambdaf(x)`

We recognise this as a first-order differential equation. We also recognise that from its form, it is an eigenvalue-equation, since the action of the operator on the function `f(x)` returns the original function multiplied by a constant. We may wonder, what sort of function satisfies this equation? One which does, is said to be a solution, and one such function is the exponential function, `f(x)=e^(lambdax)`. We can demonstrate this quite simply by evaluating the derivative, through u-substitution, and we can show that,

`d/(dx)exp(lambdax)=lambdaexp(lambdax)`

Another such example, consider,

`d^2/(dx^2)f(x)=-lambdaf(x)`

Again, we recognise that this is an eigenvalue-equation, since the action of the operator on the function `f(x)` is to return the original function multiplied by the constant `lambda`. Whilst it may be slightly confusing that in this example we have a second-order differential operator, which as we know, is the differentiation operator applied twice, but there is no restriction on what the operator does, or how many times it does it. For example, we could re-cast the equation as expressed below, and the solutions remain the same,

`hatOf(x)=-lambdaf(x)` where `hatO=d^2/(dx^2)`

So what functions satisfy this equation? It is tempting to go straight for the exponential function again - it worked the first time! Using the eigenfunction `exp(kx)`, the first differential operation gives,

`d/(dx)exp(kx)=kexp(kx)`

Following this with the second differentiation affords,

`kd/(dx)exp(kx)=k^2exp(kx)`

Comparing this to the eigenvalue-equation, we can see that,

`-lambda=k^2` and therefore, `k=sqrt(-lambda)=isqrt(lambda)`

Thus our first guess worked, but this is not to say that this is only function which satisfies the eigenvalue-equation. In fact, there are an infinite number of solutions. Another wise guess would have been a trigonometric function. For example, consider what happens when we differentiate the sine or cosine functions. Over two differentiations, the function switches sine to cosine and vice versa, and back again, and the magnitude overall inverts from positive to negative and negative to positive. So either function `cos(x)` or `sin(x)` would also satisfy the eigenvalue-equation, with `lambda=1`.

Additionally, we can show that `cos(kx+phi)` satisfies the eigenvalue-equation. The first differentiation provides,

`d/(dx)cos(kx+phi)=-ksin(kx+phi)`

Then the second provides,

`-kd/(dx)sin(kx+phi)=-k^2cos(kx+phi)`

Therefore  `cos(kx+phi)` also satisfies,

`d^2/(dx^2)f(x)=-lambdaf(x)` with `lambda=k^2`

Where `-k^2` is the eigenvalue of the eigenfunction `cos(kx+phi)` associated with the operator `d^2/(dx^2)`

We've focused on the first and second-order differentiation operators as these are particularly pertinent in quantum mechanics. As we shall see, they appear in the Schrodinger wave-equation. Further, the two eigenfunctions considered (sinusoids and exponentials) are significant because they are wave-functions.