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# 2.3: Operations on Wave Functions

#### Position and Momentum Operators

In Equation 2.2.6 we expressed different forms that operators can take when they are expressed in terms of acting on the wave function. The position and momentum operators are the two special operators from which nearly all others are built, and we need to determine their precise forms in the position-space and momentum-space bases. We begin with the reminder that the state vectors are written in terms of the basis vectors for each space as follows:

$\begin{array}{l} position\; space: & \left|\;\Psi\;\right>=\left[\int \limits_{-\infty}^{+\infty} \left|\;x\;\right> \left<\;x\;\right| dx \right] \left|\;\Psi\;\right> = \int \limits_{-\infty}^{+\infty} dx\;\psi\left(x\right) \left|\;x\;\right> \\ momentum\; space: & \left|\;\Psi\;\right>=\left[\int \limits_{-\infty}^{+\infty} \left|\;k\;\right> \left<\;k\;\right| dk \right] \left|\;\Psi\;\right> = \int \limits_{-\infty}^{+\infty} dk\;\phi\left(k\right) \left|\;k\;\right> \end{array}$

The idea is now to replace an operator in Hilbert space acting on the vector $$\left|\;\Psi\;\right>$$ with an operator in the basis (constructed from functions and derivatives, as described in Equation 2.2.6). So it should look something like this:

$\begin{array}{l} position\; space: & \Omega\left|\;\Psi\;\right> = \int \limits_{-\infty}^{+\infty} dx\;\left[\widehat\Omega_x\;\psi\left(x\right)\right] \left|\;x\;\right> \\ momentum\; space: & \Omega\left|\;\Psi\;\right>= \int \limits_{-\infty}^{+\infty} dk\;\left[\widehat\Omega_k\;\phi\left(k\right)\right] \left|\;k\;\right> \end{array}$

We will use the notation that an operator in Hilbert space carries no special markings (the fact that it is an operator should be clear from context), but when that operator is expressed in a basis, it will come with a hat to remind us that it is not simply the variable that it resembles.

Although in both cases it is the same operator acting on the same state vector, giving the same result, the components in different bases are of course quite different, as are the operators that act on them.  Now let's look at the two special operators $$\widehat x$$ and $$\widehat p$$. We start with a simple, intuitive rule:

In their own basis, the operators $$\widehat x$$ and $$\widehat p$$ are simply the functions $$x$$ and $$p$$, respectively.

Where things get weird is when we want to express each of these operators in the other's basis. Keep in mind that the end result of the operation will be a new set of components for the state vector, and the components still must expressed as a function of $$x$$ in the position basis and $$k$$ in the momentum basis. It can be shown (we'll omit the details here, but the reader is encouraged to work it out) that the operator for the momentum in the position space basis that correctly satisfies Equation 2.3.2 and the link between position and momentum space expressed in Equation 2.1.10 is this:

$\widehat p_x = -i\hbar \dfrac{d}{dx}$

The operator for the position in the momentum space basis looks very similar:

$\widehat x_p = i\hbar \dfrac{d}{dp}$

We have strayed back to using the momentum variable here because these operators are so iconic, but in terms of the wave number we can simply replace the $$p$$'s with $$k$$'s, and replace the $$\hbar$$'s with 1's.

Let's recap to make sure we are clear on what we've done here. Suppose we operate on a quantum state with the momentum operator. This changes the state:

$\left|\;\widetilde\Psi\;\right> = P\left|\;\Psi\;\right>$

We can now view this new state in any basis we like. When we do so, the state will be expanded as a sum (integral) of components (wave functions) multiplied by unit vectors. So if we expand the case above into position space, we could write:

$\left|\;\Psi\;\right> = \int \limits_{-\infty}^{+\infty} \psi\left(x\right) \left|\;x\;\right> \;\;\; and \;\;\; \left|\;\widetilde\Psi\;\right> = \int \limits_{-\infty}^{+\infty} \widetilde\psi\left(x\right) \left|\;x\;\right>$

But the components (wave function) of the changed state vector can be written in terms of the components (wave function) of the unchanged vector. This can be done by performing an operation on the unchanged vector components, and this operation is specific to the basis the that wave functions are components in:

$\widetilde\psi\left(x\right) = \widehat p_x\;\psi\left(x\right) = -i\hbar\dfrac{\partial}{\partial x}\psi\left(x\right)$

Note that if we decide to work in the momentum space basis instead, Equation 2.3.5 is unchanged (because it is not working in any basis). The operation and the effect on the Hilbert space vector are exactly the same. But it is viewed differently in momentum space than in position space, and the next two equations change to:

$\left|\;\Psi\;\right> = \int \limits_{-\infty}^{+\infty} \phi\left(k\right) \left|\;k\;\right> \;\;\; and \;\;\; \left|\;\widetilde\Psi\;\right> = \int \limits_{-\infty}^{+\infty} \widetilde\phi\left(k\right) \left|\;k\;\right>$

$\widetilde\phi\left(k\right) = \widehat p_p\;\phi\left(k\right) = \hbar k\;\phi\left(k\right)$

#### Building More Operators

Our current goal is to get back to the Schrödinger equation, this time expressed in a basis. To do this, we need to build the kinetic and potential energy operators from our position and momentum operators. We use our knowledge of classical mechanics as a guide to do this building, by following the same functional form. The kinetic energy can be expressed in terms of the momentum and mass, so we have:

$\widehat {KE} = \dfrac{\widehat p^2}{2m}$

[Note: The mass is a c-number, so there is no need to go out of our way to express it as an operator.]

What does "squaring" an operator mean? Well, if a square operator acts on a wave function, then it essentially acts twice. That is, it acts once, creating a new wave function, then acts again, creating another. In momentum space, the kinetic energy operator is simply just the square of the momentum divided by twice the mass. But in position space, where the momentum vector is proportional to a derivative, it is proportionalk to the second derivative:

$\widehat {KE}_x = \dfrac{1}{2m} \left[ -i\hbar \dfrac{d}{dx}\right] \left[ -i\hbar \dfrac{d}{dx}\right] = -\dfrac{\hbar^2}{2m}\dfrac{d^2}{dx^2}$

What about the potential energy? Well, we can't build that unless we know the specific conditions to which the particle is subjected. One thing we will do, however, is assume that the potential energy function does not vary with momentum (or time). This means that it can be built strictly out of combinations of $$x$$'s:

$\widehat {PE}_x = V\left(x\right)$

#### Schrödinger's Equation

At last we can write down Schrödinger's equation in terms of the wave function. If we use position space, then we have the kinetic energy giving us a second derivative, while the potential energy operator becomes just a function of $$x$$. Including Equation 2.2.1 and Equation 2.28, we have:

$\widehat H \psi\left(x,t\right) = i\hbar \dfrac{\partial}{\partial t} \psi\left(x,t\right) \;\;\; \Rightarrow \;\;\; -\dfrac{\hbar^2}{2m}\dfrac{\partial^2}{\partial x^2}\psi\left(x,t\right) + V\left(x\right) \psi\left(x,t\right) = i\hbar \dfrac{\partial}{\partial t} \psi\left(x,t\right)$

You'll notice that the derivatives that came from the momentum in the kinetic energy are written as partial derivatives here. This is because the wave function is a function of two variables.

Schrödinger's equation can also be expressed in momentum space, though the potential operator can become troublesome, with the $$x$$ operators contained in the potential energy function having to be expressed as derivatives with respect to momentum. Technically it can be done, since whatever the function is, it can be expressed as a power series in $$x$$, and then the $$n^{th}$$ power of $$x$$ simply converts into $$n$$ consecutive derivatives. But this is likely to be a significantly tougher way to get the momentum space solution to Schrödinger's equation than solving the differential equation for the position space wave function, followed by taking a fourier transform to get the momentum space wave function.