5.4 Position and Linear Momentum

The subsequent sections will be looking at the time evolution of various quantum systems, as predicted by the Schrödinger equation. However, before that can be done, first the eigenfunctions of position and linear momentum must be found. That is something that the book has been studiously avoiding so far. The problem is that the position and linear momentum eigenfunctions have awkward issues with normalizing them.

These normalization problems have consequences for the coefficients of the eigenfunctions. In the normal orthodox interpretation, the square magnitudes of the coefficients should give the probabilities of getting the corresponding values of position, respectively linear momentum. But this statement will have to be modified a bit.

One good thing is that unlike the Hamiltonian, which is specific to a given system, the position operator

$${\skew 2\widehat{\skew{-1}\vec r}}= ({\widehat x}, {\widehat y}, {\widehat z})$$

and the linear momentum operator

$${\skew 4\widehat{\skew{-.5}\vec p}}= ({\widehat p}_x,{\wid... ...partial}{\partial y}, \frac{\partial}{\partial z} \right)$$

are the same for all systems. So, you only need to find their eigenfunctions once.

5.4.1 The position eigenfunction

The eigenfunction that corresponds to the particle being at a precise $x$-position ${\underline x}$, $y$-position ${\underline y}$, and $z$-position ${\underline z}$ will be denoted by $R_{{\underline x}{\underline y}{\underline z}}(x,y,z)$. The eigenvalue problem is:

$$\begin{eqnarray*} && {\widehat x}R_{{\underline x}{\underline y}{\underline z}... ...derline z}R_{{\underline x}{\underline y}{\underline z}}(x,y,z) \end{eqnarray*}$$

(Note the need in this analysis to use $({\underline x},{\underline y},{\underline z})$ for the measurable particle position, since $(x,y,z)$ are already used for the eigenfunction arguments.)

To solve this eigenvalue problem, try again separation of variables, where it is assumed that $R_{{\underline x}{\underline y}{\underline z}}(x,y,z)$ is of the form $X(x) Y(y) Z(z)$. Substitution gives the partial problem for $X$ as

$$x X(x) = {\underline x}X(x)$$

This equation implies that at all points $x$ not equal to ${\underline x}$, $X(x)$ will have to be zero, otherwise there is no way that the two sides can be equal. So, function $X(x)$ can only be nonzero at the single point ${\underline x}$. At that one point, it can be anything, though.

To resolve the ambiguity, the function $X(x)$ is taken to be the “Dirac delta function,”

$$X(x)=\delta(x-{\underline x})$$

The delta function is, loosely speaking, sufficiently strongly infinite at the single point $x={\underline x}$ that its integral over that single point is one. More precisely, the delta function is defined as the limiting case of the function shown in the left hand side of figure 5.6.

Figure 5.6: Approximate Dirac delta function $\delta_\varepsilon(x-\protect{\underline x})$ is shown left. The true delta function $\delta(x-\protect{\underline x})$ is the limit when $\varepsilon$ becomes zero, and is an infinitely high, infinitely thin spike, shown right. It is the eigenfunction corresponding to a position $\protect{\underline x}$.

The fact that the integral is one leads to a very useful mathematical property of delta functions: they are able to pick out one specific value of any arbitrary given function $f(x)$. Just take an inner product of the delta function $\delta(x-{\underline x})$ with $f(x)$. It will produce the value of $f(x)$ at the point ${\underline x}$, in other words, $f({\underline x})$:

$$\langle \delta(x-{\underline x})\vert f(x)\rangle = \int... ...erline x}) f({\underline x}) {\,\rm d}x = f({\underline x})$$ (5.41)

(Since the delta function is zero at all points except ${\underline x}$, it does not make a difference whether $f(x)$ or $f({\underline x})$ is used in the integral.) This is sometimes called the “filtering property” of the delta function.

The problems for the position eigenfunctions $Y$ and $Z$ are the same as the one for $X$, and have a similar solution. The complete eigenfunction corresponding to a measured position $({\underline x},{\underline y},{\underline z})$ is therefore:

$$\fbox{$\displaystyle R_{{\underline x}{\underline y}{\un... ...v \delta^3({\skew0\vec r}-{\underline{\skew0\vec r}}) $} %$$ (5.42)

Here $\delta^3({\skew0\vec r}-{\underline{\skew0\vec r}})$ is the three-dimensional delta function, a spike at position ${\underline{\skew0\vec r}}$ whose volume integral equals one.

According to the orthodox interpretation, the probability of finding the particle at $({\underline x},{\underline y},{\underline z})$ for a given wave function $\Psi$ should be the square magnitude of the coefficient $c_{{\underline x}{\underline y}{\underline z}}$ of the eigenfunction. This coefficient can be found as an inner product:

$$c_{{\underline x}{\underline y}{\underline z}}(t) = \lan... ...lta(y-{\underline y})\delta(z-{\underline z})\vert\Psi\rangle$$

It can be simplified to

$$c_{{\underline x}{\underline y}{\underline z}}(t) = \Psi({\underline x},{\underline y},{\underline z};t)$$ (5.43)

because of the property of the delta functions to pick out the corresponding function value.

However, the apparent conclusion that $\vert\Psi({\underline x},{\underline y},{\underline z};t)\vert^2$ gives the probability of finding the particle at $({\underline x},{\underline y},{\underline z})$ is wrong. The reason it fails is that eigenfunctions should be normalized; the integral of their square should be one. The integral of the square of a delta function is infinite, not one. That is OK, however; ${\skew0\vec r}$ is a continuously varying variable, and the chances of finding the particle at $({\underline x},{\underline y},{\underline z})$ to infinite number of digits accuracy would be zero. So, the properly normalized eigenfunctions would have been useless anyway.

Instead, according to Born's statistical interpretation of chapter 2.1, the expression

$$\vert\Psi(x,y,z;t)\vert^2 {\,\rm d}x {\rm d}y {\rm d}z$$

gives the probability of finding the particle in an infinitesimal volume ${\rm d}{x}{\rm d}{y}{\rm d}{z}$ around $(x,y,z)$. In other words, $\vert\Psi(x,y,z;t)\vert^2$ gives the probability of finding the particle near location $(x,y,z)$ per unit volume. (The underlines below the position coordinates are no longer needed to avoid ambiguity and have been dropped.)

Besides the normalization issue, another idea that needs to be somewhat modified is a strict collapse of the wave function. Any position measurement that can be done will leave some uncertainty about the precise location of the particle: it will leave $\Psi(x,y,z;t)$ nonzero over a small range of positions, rather than just one position. Moreover, unlike energy eigenstates, position eigenstates are not stationary: after a position measurement, $\Psi$ will again spread out as time increases.

Key Points

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Position eigenfunctions are delta functions.

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They are not properly normalized.

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The coefficient of the position eigenfunction for a position $(x,y,z)$ is the good old wave function $\Psi(x,y,z;t)$.

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Because of the fact that the delta functions are not normalized, the square magnitude of $\Psi(x,y,z;t)$ does not give the probability that the particle is at position $(x,y,z)$.

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Instead the square magnitude of $\Psi(x,y,z;t)$ gives the probability that the particle is near position $(x,y,z)$ per unit volume.

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Position eigenfunctions are not stationary, so localized particle wave functions will spread out over time.

5.4.2 The linear momentum eigenfunction

Turning now to linear momentum, the eigenfunction that corresponds to a precise linear momentum $(p_x,p_y,p_z)$ will be indicated as $P_{p_xp_yp_z}(x,y,z)$. If you again assume that this eigenfunction is of the form $X(x)Y(y)Z(z)$, the partial problem for $X$ is found to be:

$$\frac{\hbar}{{\rm i}} \frac{\partial X(x)}{\partial x} = p_x X(x)$$

The solution is a complex exponential:

$$X(x)= Ae^{{\rm i}p_x x/\hbar}$$

where $A$ is a constant.

Just like the position eigenfunction earlier, the linear momentum eigenfunction has a normalization problem too. In particular, since it does not become small at large $\vert x\vert$, the integral of its square is infinite, not one. The solution is to ignore the problem and to just take a nonzero value for $A$; the choice that works out best is to take:

$$A=\frac{1}{\sqrt{2\pi\hbar}}$$

(However, other books, in particular non-quantum ones, are likely to make a different choice.)

The problems for the $y$ and $z$-linear momentum have similar solutions, so the full eigenfunction for linear momentum takes the form:

$$\fbox{$\displaystyle P_{p_xp_yp_z}(x,y,z) = \frac{1}{\... ...i\hbar}^3} e^{{\rm i}(p_x x + p_y y + p_z z)/\hbar} $} %$$ (5.44)

The coefficient $c_{p_xp_yp_z}(t)$ of the momentum eigenfunction is very important in quantum analysis. It is indicated by the special symbol $\Phi(p_x,p_y,p_z;t)$ and called the “momentum space wave function.” Like all coefficients, it can be found by taking an inner product of the eigenfunction with the wave function:

$$\fbox{$\displaystyle \Phi(p_x,p_y,p_z;t) = \frac{1}{\sqr... ...{{\rm i}(p_x x + p_y y + p_z z)/\hbar} \vert \Psi\rangle $}$$ (5.45)

The momentum space wave function does not quite give the probability for the momentum to be $(p_x,p_y,p_z)$. Instead it turns out that

$$\vert\Phi(p_x,p_y,p_z;t)\vert^2 {\rm d}p_x {\rm d}p_y {\rm d}p_z$$

gives the probability of finding the linear momentum within a small momentum range ${\rm d}{p}_x{\rm d}{p}_y{\rm d}{p}_z$ around $(p_x,p_y,p_z)$. In other words, $\vert\Phi(p_x,p_y,p_z;t)\vert^2$ gives the probability of finding the particle with a momentum near $(p_x,p_y,p_z)$ per unit “momentum space volume.” That is much like the square magnitude $\vert\Psi(x,y,z;t)\vert^2$ of the normal wave function gives the probability of finding the particle near location $(x,y,z)$ per unit physical volume. The momentum space wave function $\Phi$ is in the momentum space $(p_x,p_y,p_z)$ what the normal wave function $\Psi$ is in the physical space $(x,y,z)$.

There is even an inverse relationship to recover $\Psi$ from $\Phi$, and it is easy to remember:

$$\fbox{$\displaystyle \Psi(x,y,z;t) = \frac{1}{\sqrt{2\pi... ...p_x x + p_y y + p_z z)/\hbar} \vert \Phi\rangle_{\vec p} $}$$ (5.46)

where the subscript on the inner product indicates that the integration is over momentum space rather than physical space.

If this inner product is written out, it reads:

$$\fbox{$\displaystyle \Psi(x,y,z;t) = \frac{1}{\sqrt{2\pi... ... + p_z z)/\hbar} {\,\rm d}p_x {\rm d}p_y {\rm d}p_z $} %$$ (5.47)

Mathematicians prove this formula under the name “Fourier Inversion Theorem”, {A.43}. But it really is just the same sort of idea as writing $\Psi$ as a sum of eigenfunctions $\psi_n$ times their coefficients $c_n$, as in $\Psi = \sum_n c_n\psi_n$. In this case, the coefficients are given by $\Phi$ and the eigenfunctions by the exponential (5.44). The only real difference is that the sum has become an integral since $\vec{p}$ has continuous values, not discrete ones.

Key Points

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The linear momentum eigenfunctions are complex exponentials of the form:

$$\frac{1}{\sqrt{2\pi\hbar}^3} e^{{\rm i}(p_x x + p_y y + p_z z)/\hbar}$$

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They are not properly normalized.

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The coefficient of the linear momentum eigenfunction for a momentum $(p_x,p_y,p_z)$ is indicated by $\Phi(p_x,p_y,p_z;t)$. It is called the momentum space wave function.

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Because of the fact that the momentum eigenfunctions are not normalized, the square magnitude of $\Phi(p_x,p_y,p_z;t)$ does not give the probability that the particle has momentum $(p_x,p_y,p_z)$.

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Instead the square magnitude of $\Phi(p_x,p_y,p_z;t)$ gives the probability that the particle has a momentum close to $(p_x,p_y,p_z)$ per unit momentum space volume.

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In writing the complete wave function in terms of the momentum eigenfunctions, you must integrate over the momentum instead of sum.

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The transformation between the physical space wave function $\Psi$ and the momentum space wave function $\Phi$ is called the Fourier transform. It is invertible.