Chapter 10: Quantum mechanics in three dimensions

In this chapter we are now ready to move onto the study of quantum mechancis in three dimensions! Much of what we have learnt up until this stage will carry across in a rather straightforward manner, which is good news. Obviously the world around us is three dimensional, and so it is necessary to move broaden our application of quantum mechanics to three dimensions, so that we can go on to study real-world situations, for example, laying the foundations for the study of the Hydrogen atom — arguably the most famous quantum system of all, and one of its biggest early successes. While we won’t get all the way there in this unit, we will nevertheless introduce the basics of quantum mechancis in 3D, so that you will be able to go straight there next year.

10.1The position of a particle in 3D¶

We will largely follow the same path in 3D that we followed in 1D, and start off with the most basic of all properties of a particle: its position. One of the biggest changes in 3D compared to 1D is that many physical quantities now become vectors, and the position is the first of these. In order to specify a point in space, we now need to use the position vector $\mathbf{r}$ ,

\rvec = x \ivec + y \jvec + z \kvec = (x,y,z).

(10.1)

Just as before we associated a quantum state to each position $x$ , we will now introduce a 3D position states as $\ket{\rvec}$ , and an associated 3D spatial wavefunction $\psi(\rvec)$ , such that a general quantum state of a particle in three dimensions can be written as

\ket{\psi} = \infint \infint \infint \psi(\rvec) \ket{\rvec} d^3\rvec,

(10.2)

which should be compared to (1.4).

Notation for vectors and 3D integrals

We will write mostly use the notation $\rvec = (x,y,z)$ as convenient shorthand for 3D vectors, instead of $\rvec = x \ivec + y \jvec + z \kvec$ , which proves more cumbersome in general. In this way, we can write $\psi(\rvec)$ and $\psi(x,y,z)$ interchangably, as both should make sense. We will exclusively use boldface in the lecture notes for vectors, instead of $\vec{r}$ .

We will also use the notation $d^3\rvec$ for the volume element $dx dy dz$ . Previously you have seen this as $d\tau$ . Both are common (Griffiths for example uses the former, and we will follow their notation here).

Finally, we will often simply write $\int_V f(\rvec) d^3\rvec$ in place of $\infint \infint \infint f(\rvec) d^3\rvec$ , to indicate an integration over all space when this will not cause any confusion.

This state corresponds to a superposition of the particle being at position $\rvec$ . The 3D position states $\ket{\rvec}$ are unphysical states, which satisfy the orthogonality and normalisation condition

\inner{\rvec'}{\rvec} = \delta^{(3)}(\rvec-\rvec'),

(10.3)

where $\delta^{(3)}(\rvec-\rvec')$ is the 3D Dirac delta function, given by

\delta^{(3)}(\rvec-\rvec') = \delta(x-x')\delta(y-y')\delta(z-z'),

(10.4)

i.e. it is the product of three delta functions, one for each coordinate. Using the 3D delta function, as an exercise you will show that the normalisation condition $\| \ket{\psi} \|^2 = 1$ , in terms of the wavefunction $\psi(\rvec)$ becomes

\int_V|\psi(\rvec)|^2 d^3\rvec = 1,

(10.5)

which should be compared to (1.7). The interpretation of this equation is in fact identical to the interpretation we gave to (1.7). However, to arrive at this, we first need to introduce operators corresponding to the coordinates of the particle.

Finally, using the 3D Dirac delta function, as an exercise you will show that we have the direct analogue of (2.17) in 3D, namely

\inner{\rvec}{\psi} = \psi(\rvec),

(10.6)

which shows that the wavefunction is nothing but the scalar product between a 3D position state and the quantum state itself.

10.1.1Position and coordinate operators¶

One new question that arises in 3D with the introduction of vector quantities is how to associate operators to them? It turns out that the correct way to do this is to associate a separate operator to each component of the vector. In the context of the position, this means we will associate an operator to each coordinate of the particle. We thus introduce 3 operators, $\hat{X}$ , $\hat{Y}$ and $\hat{Z}$ . The position states $\ket{\rvec}$ are joint eigenstates of all 3 operators, satisfying the eigenvalue equations

\begin{align*} \hat{X}\ket{\rvec} &= x \ket{\rvec},& \hat{Y}\ket{\rvec} &= y \ket{\rvec},& \hat{Z}\ket{\rvec} &= z \ket{\rvec} \end{align*}

(10.7)

As you will confirm as an exercise, we now have a vast amount of degeneracy in these operators. Recall that degeneracy refers to the fact that there can be multiple eigenvectors with the same eigenvalue. We see that any two position states $\ket{\rvec}$ and $\ket{\rvec'}$ that have the same $x$ coordinate, but arbitrary $y$ and $z$ coordinates, will be degenerate as far as $\hat{X}$ is concerned (and similarly for $\hat{Y}$ and $\hat{Z}$ ). There are infinitely many such states.

Just as it is convenient to combine coordinates together into a vector in classical mechanics, it is also convenient and useful to combine operators together into a vector of operators. In particular, we can combine the coordinate operators to form a 3D position operator $\hat{\mathbf{R}}$ , given by

\hat{\mathbf{R}} = (\hat{X},\hat{Y},\hat{Z}).

(10.8)

With this notation, we therefore see that

\begin{align*} \hat{\mathbf{R}}\ket{\rvec} &= (\hat{X},\hat{Y},\hat{Z})\ket{\rvec},\\ &= (x,y,z)\ket{\rvec},\\ &= \rvec \ket{\rvec}, \end{align*}

(10.9)

where to obtain the second line we have used (10.7). Apart from the fact that $\hat{\mathbf{R}}$ and $\rvec$ are vectors, this is exactly in the form of an eigenvalue equation, and this is one reason why it is useful: we see that $\ket{\rvec}$ can be viewed as the eigenvector of the 3D position operator $\hat{\mathbf{R}}$ .

We can also introduce the coordinate operators acting on wavefunctions using the same prescription as before, (1.21). Using (10.7), we see for example that

\hat{X}\ket{\psi} = \int_V \psi(\rvec) x \ket{\rvec} d^3\rvec,

(10.10)

i.e. the action of the operator is to transform $\psi(\rvec)$ into $\psi'(\rvec) = x \psi(\rvec)$ , and so we have $\op{X} = x$ , just as before. That is, the $X$ coordinate operator multiplies a wavefunction by the coordinate $x$ , as we should expect. We similarly have $\op{Y} = y$ and $\op{Z} = z$ . Interesting, we can also combine these into a single vector operator. In particular, we see that

\begin{align*} \hat{\mathbf{R}}\ket{\psi} &= \hat{\mathbf{R}} \int_V \psi(\rvec)\ket{\rvec} d^3\rvec, \\ &= \int_V \psi(\rvec)\hat{\mathbf{R}} \ket{\rvec}d^3\rvec,\\ &= \int_V \psi(\rvec)\rvec \ket{\rvec}d^3\rvec, \end{align*}

(10.11)

Thus the action of the operator is to turn the wavefunction into the vector

\psi(\rvec)\rvec = (x\psi(\rvec), y\psi(\rvec), z\psi(\rvec)),

(10.12)

i.e. each component of $\rvec$ is multiplied by the the wavefunction $\psi(\rvec)$ (which is itself just a scalar quantity, being a single complex number for each position $\rvec$ ). Given (10.12), we therefore write

\op{\mathbf{R}} = \rvec

(10.13)

Finally, using the definitions of $\op{X}$ , $\op{Y}$ and $\op{Z}$ , it is also straightforward to see that the three coordinates are compatible observables that can therefore be simultaneously measured (in principle at least). Recall that the mathematical property behind compatibility is commutativity. We can easily check that the commutator of any two coordinate wavefunction operators vanishes, for example,

\begin{align*} [\op{X},\op{Y}]\psi(\rvec) &= (\op{X}\op{Y} - \op{Y}\op{X})\psi(\rvec),\\ &= \op{X}(y\psi(\rvec)) - \op{Y}(x\psi(\rvec)),\\ &= (xy - yx)\psi(\rvec),\\ &= 0, \end{align*}

(10.14)

from which it follows that $[\hat{X}, \hat{Y}] = 0$ , according to (3.11).

10.1.2Probability to find a particle somewhere¶

We can now think about measuring the position of the particle in 3D, and ask *what is the probability to find it in an (arbitrarily small) region around $\rvec$ ? That is, if we performed an arbitrarily good simultaneous measurement of the 3 coordinates of the particle? The answer is the direct generalisation of what we saw for a particle in one dimension: the probability density at position $\rvec$ $ is precisely

\pd(\rvec) = |\psi(\rvec)|^2,

(10.15)

so that the probability to find the particle in a small region of volume $dV$ around the point $\rvec$ is $|\psi(\rvec)|^2 dV$ . We can then ask similar questions about the probability of a measurement finding the particle in a finite region, and the total probability in any region is obtained by integrating the probability density over the region.

This then allows us to interpret the normalisation condition (10.5) in a natural way: it says that the total probability to find the particle somewhere (in 3D now!) must be unity.

10.2The momentum of a particle in 3D¶

Let us now turn our attention to the momentum of a particle in 3D. As with the position, the momentum is now a vector quantity, $\pvec = (p_x, p_y, p_z)$ . In Chapter 2: The momentum of a quantum particle we saw that if a particle had a definite momentum $p_x$ in the $x$ -direction, then the spatial wavefunction was the momentum eigenfunction $v_{p_x}(x) = \frac{1}{\sqrt{2\pi\hbar}} e^{ip_x x/\hbar}$ .

Even though we are now in 3D, we still want this to be the behaviour of the wavefunction as $x$ varies if the particle has a definite $x$ component of momentum. We can apply the same logic along the $y$ and $z$ directions, and this suggests that momentum eigenfunctions of momentum $\pvec$ should be given by

\begin{align*} v_{\pvec}(\rvec) &= \frac{1}{(\sqrt{2\pi\hbar})^3}e^{ip_x x/\hbar}e^{ip_y y/\hbar}e^{ip_z z/\hbar},\\ &= \frac{1}{(2\pi\hbar)^{3/2}} e^{i\pvec \cdot \rvec / \hbar} \end{align*}

(10.16)

where in the second line we have realised that $p_x x + p_y y + p_z z$ (that arises when combining the exponents of the three exponential functions) can be written as the scalar product between $\rvec$ and $\pvec$ , i.e. as $\pvec \cdot \rvec$ .

In what follows, we will see that this is the correct definition of the 3D momentum eigenfunctions. They correspond precisely to 3D complex plane waves with wave vector $\mathbf{k} = \pvec/\hbar$ , and definite wavelength $\lambda = 2\pi/|\mathbf{k}| = h/|\pvec|$ . This is precisely the de Broglie relation, for a particle with momentum of magnitude $|\pvec|$ .

Having defined the momentum eigenfunctions, we can now write down — using the standard prescription — the corresponding 3D momentum eigenstates,

\begin{align*} \ket{\pvec} &= \int_V v_{\pvec}(\rvec)\ket{\rvec}d^3\rvec,\\ &= \frac{1}{(2\pi\hbar)^{3/2}} \int_V e^{i\pvec \cdot \rvec / \hbar} \ket{\rvec} d^3\rvec. \end{align*}

(10.17)

These states are also unphysical and satisfy the same orthogonality-normalisation condition as the 3D position states $\ket{\rvec}$ , namely

\inner{\pvec'}{\pvec} = \delta^{(3)}(\pvec - \pvec').

(10.18)

We can use the momentum eigenstates in order to introduce the 3D momentum wavefunction, in exactly analogy to (2.12)

\ket{\psi} = \int_V \tilde{\psi}(\pvec) \ket{\pvec} d^3 \pvec,

(10.19)

i.e. it is the way that we can express an arbitrary state as a superposition of momentum states. Just as we saw above for the spatial wavefunction, we can similarly directly show that the momentum wavefunction is the scalar product of the quantum state with a 3D momentum state, namely

\inner{\pvec}{\psi} = \tilde{\psi}(\pvec).

(10.20)

10.2.1Momentum operator¶

Just as for the position, we will need to introduce one momentum operator for each component of momentum, and we can collect these together into a single vector momentum operator,

\hat{\mathbf{P}} = (\hat{P}_x, \hat{P}_y, \hat{P}_z).

(10.21)

Momentum eigenstates will then be — by definition — joint eigenstates of each component of the momentum operator, i.e.

\begin{align*} \hat{P}_x \ket{\pvec} &= p_x \ket{\pvec},& \hat{P}_y \ket{\pvec} &= p_y \ket{\pvec},& \hat{P}_z \ket{\pvec} &= p_z \ket{\pvec}. \end{align*}

(10.22)

We can again write this in a succint way as

\begin{align*} \hat{\mathbf{P}}\ket{\pvec} &= \pvec \ket{\pvec}, \end{align*}

(10.23)

just as we did in (10.9) for the 3D position operator $\hat{\mathbf{R}}$ .

What about the wavefunction operator $\op{\mathbf{P}}$ ? Well, for the position operator, we saw in (10.13) that we in fact kept the same wavefunction operator as we had in 1D. Namely, we just multiplied by the coordinate. This suggests that for momentum we might have the same: that each component of the momentum operator should be specified the same way as in 1D, as in (2.27). We will thus take

\begin{align*} \op{\mathbf{P}} &= \left(-i\hbar \frac{\partial}{\partial x}, -i\hbar \frac{\partial}{\partial y}, -i\hbar \frac{\partial}{\partial z}\right),\\ &= -i\hbar \boldsymbol{\nabla}. \end{align*}

(10.24)

As an exercise, you will show that the 3D momentum eigenfunctions $v_{\pvec}(\rvec)$ from (10.16) are indeed eigenfunctions of the 3D wavefunction momentum operator, satisfying

\op{\mathbf{P}}v_{\pvec}(\rvec) = \pvec v_{\pvec}(\rvec),

(10.25)

which is the same eigenvalue equation for the momentum operator as (10.23), just in terms of the wavefunction operator $\op{\mathbf{P}}$ instead of $\hat{\mathbf{P}}$ .

Just as the three coordinate operators were compatible observables, so too are the three components of momentum. We can see that this is the case by using the wavefunction operators. For example, for the $x$ -component and $y$ -component of momentum, we have

\begin{align*} [\hat{P}_{x,\mathrm{w}},\hat{P}_{y,\mathrm{w}}]\psi(\rvec) &= (\hat{P}_{x,\mathrm{w}}\hat{P}_{y,\mathrm{w}} - \hat{P}_{y,\mathrm{w}}\hat{P}_{x,\mathrm{w}})\psi(\rvec),\\ &= \left(-i\hbar \frac{\partial}{\partial x}\right)\left(-i\hbar \frac{\partial \psi(\rvec)}{\partial y}\right) - \left(-i\hbar \frac{\partial}{\partial y}\right)\left(-i\hbar \frac{\partial \psi(\rvec)}{\partial x}\right),\\ &= -\hbar^2 \frac{\partial^2\psi(\rvec)}{\partial x \partial y} + \hbar^2 \frac{\partial^2\psi(\rvec)}{\partial x \partial y} \\ &= 0, \end{align*}

(10.26)

where we used the fact in the third line that we can take a partial derivative in either order (i.e. partial differentiation commutes!). Similar calculations show that all components of momentum pairwise commute, and hence all 3 components are jointly compatible. This means it is (in principle) possible to measure the momentum $\pvec$ of a particle. Note that it isn’t a given that the components of a vector quantity are necessarily compatible — and is fact not true for the components of angular momentum, which is why the quantum mechanical treatment of angular momentum is in fact so rich an interesting (analogous somewhat to the incompatibility of kinetic and potential energy, and how this leads to interesting dynamics).

10.3Canonical commutation relations¶

In Chapter 3 we saw that the position and momentum operators are incompatible observables and satisfy the so-called canonical commutation relation (3.10). In 3D we have already seen above in that the coordinates of a particle are compatible, and that the components of momentum are also compatible. What about the coordinates and components of momentum? It turns out that is only the pairs of coordinate and momentum component in the same direction that are incompatible, all of which obey the canonical commutation relation:

\begin{align*} [\hat{X},\hat{P}_x] &= i\hbar,& [\hat{Y},\hat{P}_y] &= i\hbar,& [\hat{Z},\hat{P}_z] &= i\hbar, \end{align*}

(10.27)

while all other pairs are in fact compatible,

\begin{align*} [\hat{Y},\hat{P}_x] &= 0,& [\hat{X},\hat{P}_y] &= 0,& [\hat{X},\hat{P}_z] &= 0,\\ [\hat{Z},\hat{P}_x] &= 0,& [\hat{Z},\hat{P}_y] &= 0,& [\hat{Y},\hat{P}_z] &= 0, \end{align*}

(10.28)

You will prove one example of the above as an exercise, from which all of the others follow by direct analogy.

The physical significance of the above is that we have analogues of the HUP (3.5) for all of the coordinates, namely

\begin{align*} \Delta x \Delta p_x &\geq \frac{\hbar}{2},& \Delta y \Delta p_y &\geq \frac{\hbar}{2},& \Delta z \Delta p_z &\geq \frac{\hbar}{2}, \end{align*}

(10.29)

and as such, a particle cannot have a very well-defined coordinate and momentum along the same axis at the same time. However, since for example the $x$ -coordinate and the $y$ -component of momentum commute, there is no uncertainty principle for them, and a particle can have a very well defined $x$ -coordinate and a very well defined $y$ -component of momentum at the same time!

10.4Relationship between 3D spatial & momentum wavefunctions¶

In one dimension we saw in (2.19) and (2.20) that the relationship between the position and momentum wavefunctions was that they are a Fourier transform pair. This is very useful, as it provides a useful way of obtaining the momentum wavefunction from the spatial wavefunction and vice versa, without having to go via the quantum state $\ket{\psi}$ .

In 3D (hopefully unexpectedly at this stage of the chapter!) the same relationship holds: The spatial and momentum wavefunctions are 3D Fourier transform pairs. On the one hand, starting from (10.19), taking the scalar product with a 3D position state $\bra{\rvec}$ , and using the fact that $\inner{\rvec}{\pvec} = v_{\pvec}(\rvec)$ (as given in (10.16)), we readily obtain

\psi(\rvec) = \frac{1}{(2\pi\hbar)^{3/2}} \int_V \tilde{\psi}(\pvec) e^{i\pvec \cdot \rvec / \hbar} d^3\pvec.

(10.30)

Alternatively, starting from (10.2), taking the scalar product with a 3D momentum state $\bra{\pvec}$ , and using the fact that $\inner{\pvec}{\rvec} = (\inner{\rvec}{\pvec})^* = v_{\pvec}^*(\rvec)$ , we also readily obtain

\tilde{\psi}(\pvec) = \frac{1}{(2\pi\hbar)^{3/2}} \int_V \psi(\rvec) e^{-i\pvec \cdot \rvec / \hbar} d^3\rvec.

(10.31)

10.5The energy of a particle in 3D¶

Having covered the position and momentum of a particle above, we are now ready to consider the energy of a particle in 3D. The Hamiltonian still contains two contributions to the energy — kinetic and potential. The latter is rather straightforward, as we simply now have $V(\rvec)$ , i.e. a potential which depends upon the (3D) position of the particle. We will study explicit examples later (e.g. a 3D ‘box’ and the 3D harmonic oscillator potential).

For the kinetic energy, the situation is slightly different, given that momentum is now a vector quantity in 3D. The kinetic energy depends only upon the squared magnitude of the momentum, that is (classically for the moment),

K = \vph\frac{p_x^2 + p_y^2 + p_z^2}{2M}

(10.32)

We will thus form a kinetic energy operator by substituting each component of squared momentum by the corresponding squared operator, and thus arrive at the general Hamiltonian operator

\hat{H} = \frac{\hat{P}_x^2 + \hat{P}_y^2 + \hat{P}_z^2}{2M} + \hat{V}.

(10.33)

As always, a large part of our interest in 3D will be to find the eigenstates and eigenvalues of the Hamiltonian operator, in different physical situations, with different choices of $\hat{V}$ (specifying the forces acting on the particle).

The associated operator acting on wavefunctions is

\begin{align*} \op{H} &= \vph -\frac{\hbar^2}{2M}\left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} \right) + V(\rvec), \\ &= -\frac{\hbar^2}{2M}\nabla^2 + V(\rvec). \end{align*}

(10.34)

From this we arrive at the 3D TISE,

\op{H} u_E(\rvec) = Eu_E(\rvec),

(10.35)

the eigenvalue equation for the Hamiltonian operator in terms of energy eigenfunctions. This is our central tool for finding the energy eigenstates in 3D, just as the 1D TISE was our central tool for finding energy eigenstates (via energy eigenfunctions) in 1D.

As in 1D, there are two distinct physical situations which we encounter, when a particle is either bound (remaining in a finite region of space), or non-bound (in which it can escape off to infinity). In 3D, there is a potential subtlety: what if the particle was confined in one direction but not the others? what happens then? In this case, the region that the particle can explore will still be infinite in volume, and as such we find that the energy of the particle varies continuously. Only if the particle is confined in all three spatial dimensions will we find discrete energy levels.

One update is that previously we only ever encountered a single quantum number $n$ per energy level. That is, in both the infinite square well and the harmonic oscillator we were able to enumerate the energy levels using a single integer $n$ . In 3D, in contrast, we will typically see that we have multiple quantum numbers per energy level.

A common case is to have three quantum numbers, one per degree of freedom (i.e. each independent direction of travel). We will label these by $n_x$ , $n_y$ and $n_z$ respectively, and sometimes collect them into the vector $\mathbf{n} = (n_x,n_y,n_z)$ .

10.6Evolution in 3D¶

The equation of motion of a particle in 3D is no different from the equation of motion in 1D. It is given once again by the Schrödinger Equation, which takes an identical form to (5.1) when written out abstractly,

i\hbar \frac{d}{dt}\ket{\psi(t)} = \hat{H}\ket{\psi(t)}.

(5.1)

In 1D we saw that we never actually solve the SE directly — we instead used the superposition principle and the fact that energy eigenstates evolve simply to solve it indirectly. In 3D, nothing will change on this front either!

For example, assuming that the particle is bound and therefore that it has a discrete set of energy levels $\ket{E_\mathbf{n}}$ (specified by the vector of quantum numbers $\mathbf{n}$ ), this if this is the initial state of a particle at time $t = 0$ , i.e. $\ket{\psi(t=0)} = \ket{E_\mathbf{n}}$ , then the state at time $t$ will simply be

\ket{\psi(t)} = e^{-iE_{\mathbf{n}}t/\hbar}\ket{E_\mathbf{n}},

(10.36)

as can be directly verified by substituting into the SE.

For an initial state $\ket{\psi_\init}$ , it is again crucial to write it as a superposition of energy eigenstates as

\ket{\psi_\init} = \sum_{\mathbf{n}} \alpha_{\mathbf{n}} \ket{E_\mathbf{n}},

(10.37)

where the summation runs over all the possible values of $\mathbf{n}$ , e.g. this is really 3 summations, overs $n_x$ , $n_y$ and $n_z$ , all of which are potentially infinite.

In this form, we can directly apply the superposition principle to write down the evolution of the particle, which is given by

\ket{\psi(t)} = \sum_{\mathbf{n}} \alpha_\mathbf{n} e^{-iE_\mathbf{n} t/\hbar} \ket{E_\mathbf{n}}.

(10.38)

Taking the scalar product of both sides with $\bra{\rvec}$ , we arrive at the evolution of the 3D wavefunction in time,

\psi(\rvec,t) = \sum_{\mathbf{n}} \alpha_\mathbf{n} e^{-iE_\mathbf{n} t/\hbar} u_\mathbf{n}(\rvec),

(10.39)

where we have written the energy eigenfunction $u_\mathbf{n}(\rvec) = \inner{\rvec}{E_\mathbf{n}}$ of the energy level $E_\mathbf{n}$ in the same shorthand notation as previously (i.e. with a subscript $\mathbf{n}$ in place of $E_\mathbf{n}$ , just as we did for the energy levels of the infinite square well and harmonic oscillator).

How do we write an arbitrary initial quantum state $\ket{\psi_\init}$ in the form (10.37)? That is, how do we determine $\alpha_\mathbf{n}$ ? We use the direct analogy of (7.21):

\alpha_\mathbf{n} = \int_V \psi_\init(\rvec) u_\mathbf{n}^*(\rvec) d^3\rvec,

(10.40)

where $\psi_\init(\rvec) = \inner{\rvec}{\psi_\init}$ is the 3D wavefunction of the initial state $\ket{\psi}$ . We use this form, as typically it is the energy eigenfunctions that we know explicitly, and so this is the most useful form to have.

If we wanted to, we could combine (10.39) and (10.40) to write the 3D wavefunction at time $t$ directly in terms of the intial wavefunction as

\psi(\rvec,t) = \sum_{\mathbf{n}} \int_V \psi_\init(\rvec') u_\mathbf{n}^*(\rvec') u_\mathbf{n}(\rvec) e^{-iE_\mathbf{n} t/\hbar} d^3\rvec',

(10.41)

where we have been careful to change our integration variable to $\rvec'$ (as always!), and where this should be compared to (7.22). This can be used to numerically solve for the quantum wavefunction at any time $t$ , given any initial wavefunction.

10.7Exercises¶

Chapters

Ch. 9: Scattering

Chapters

Ch. 11: 3D infinite box