7.9 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 orthodox inter­pretation, the square magnitudes of the coefficients should give the proba­bilities of getting the corre­sponding values of position and 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.

7.9.1 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 $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\underline x}$ 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 7.10.

Figure 7.10: 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, $f({\underline x})$:

$$\langle \delta(x-{\underline x})\vert f(x)\rangle = \int... ...erline x}) f({\underline x}) {\,\rm d}x = f({\underline x})$$ (7.49)

(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:

$$\fbox{$\displaystyle R_{{\underline x}{\underline y}{\un... ...v \delta^3({\skew0\vec r}-{\underline{\skew0\vec r}}) $} %$$ (7.50)

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

$$c_{{\underline x}{\underline y}{\underline z}}(t) = \Psi({\underline x},{\underline y},{\underline z};t)$$ (7.51)

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 an infinite number of digits accurate would be zero. So, the properly normalized eigen­functions would have been useless anyway.

Instead, according to Born's statistical inter­pretation of chapter 3.1, 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.

7.9.2 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. 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 non­quantum ones, are likely to make a different choice.)

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

$$\fbox{$\displaystyle P_{p_xp_yp_z}(x,y,z) = \frac{1}{\... ...i\hbar}^3} e^{{\rm i}(p_x x + p_y y + p_z z)/\hbar} $} %$$ (7.52)

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:

$$\fbox{$\displaystyle \Phi(p_x,p_y,p_z;t) = \frac{1}{\sqr... ...{{\rm i}(p_x x + p_y y + p_z z)/\hbar} \vert \Psi\rangle $}$$ (7.53)

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:

$$\fbox{$\displaystyle \Psi(x,y,z;t) = \frac{1}{\sqrt{2\pi... ...p_y y + p_z z)/\hbar} \vert \Phi\rangle_{{\skew0\vec p}} $}$$ (7.54)

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:

$$\fbox{$\displaystyle \Psi(x,y,z;t) = \frac{1}{\sqrt{2\pi... ... + p_z z)/\hbar} {\,\rm d}p_x {\rm d}p_y {\rm d}p_z $} %$$ (7.55)

Mathema­ticians prove this formula under the name “Fourier Inversion Theorem”, {A.26}. 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 $\Psi$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\sum_nc_n\psi_n$. In this case, the coefficients are given by $\Phi$ and the eigen­functions by the exponential (7.52). The only real difference is that the sum has become an integral since ${\skew0\vec p}$ 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.