Free Particle (Python Notebook)
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Describing the Free Particle
We start by describing a free particle: a particle that is not under the influence of a potential. As any other particle, the state of a Free Particle is described with a wavefunction ψ(x). In order to learn about this particle (i.e., measure its properties) we must construct Hermitian Operators. For example, what is the momentum operator ˆP?
The momentum operator most obey the following property (eigenfunction/eigenvalue equation):
ˆPψk(x)=λψk(x)
where λ is an eigenvalue of a Hermitian operator and therefore it is a real number.
In the x representation, using the momentum operator as ˆP=−iℏ∂∂x, we can solve Equation ??? by proposing a function to represent ψk(x) as ψk(x)=c eikx, where k is a real number.
Let's see if this functional form for the wavefunction works (i.e., solved Equation ???):
ˆPψk(x)=pψk(x)
−iℏ∂c eikx∂x=−iℏ c ik eikx
ℏk c eikx=ℏk ψk(x)
with p=ℏk. Although ψk(x) can not be normalized, since it is not a square-integral function (i.e., the integral of the square of the absolute value is not finite). However, the constant c is chosen in such a way (see note below) that the function becomes:
ψk(x)=1√2πeikx
The Normalization Factor and Dirac's Delta Function
The normalization of a bra-ket is given by
⟨ψk′(x)|ψk(x)⟩=⟨k′|k⟩=∫+∞−∞ψ∗k′(x)ψk(x)dx=∫+∞−∞c∗e−ik′xceikxdx=|c|2∫+∞−∞ei(k−k′)xdx=|c|22π δ(k−k′)
Where we used the definition of the Dirac's delta function:
δ(x)=12π∫+∞−∞eivxdv
which obeys the following properties:
δ(x−a)=0 ifx≠a
∫∞−∞f(x) δ(x−a) dx=f(a)
The normalization constant c is chosen as c=1√2π in such a way that:
⟨ψk′(x)|ψk(x)⟩=δ(k−k′)
Once we know the normalized eigenfunctions, we can ask some questions:
Questions
- Q1: What is the momentum of a free particle?
- Q2: What are the units of momentum?
- Q3: Is the momentum of a Free Particle quantized?
In SI units, k has units of 1m and ℏ has units of J.s=kg.m2s, therefore from Equation ??? we can see that p=ℏk and the units are J.sm=Kg.ms.
To visualize the wavefunction ψk(x) we use Euler relationship
eiθ=cos(θ)+isin(θ)
and graph separately the Real and Imaginary contributions to the wavefunction (even though the wavefunction is complex, the ˆP is a Hermitian operator, thus the eigenvalues are measurable and real numbers).
On the other hand, the Probability Density (ψ∗k(x)ψk(x)) is a constant, since:
ψ∗k(x)ψk(x)=1√2πe−ikx1√2πeikx=12π
You can use the cell below to enter different values for the momentum wavevector k and the length of space x
Interactive Element
What do we learn from this graphical representaton?: The first plot shows that that the wavefunction is completely delocalized in the x coordinate. However, wherever we look, the value of the probability density (|ψ∗k(x)ψk(x)|) is the same.
An important question to ask is: what is the probability to find a free particle within a Δx region in space?
This is crucial for measurements, since we would never measure ALL space at the same time, but instead, our instrument will measure a section of x space
Ψ∗k(x)Ψk(x)Δx=|Ψk(x)|2Δx=(1√2πeikx)∗1√2πeikxΔx=12πΔx
The probability of finding it within a Δx, at any place in space, is the same.
Thus when the particle has a well-defined momentum (ℏk), the probability of finding it anywhere in space is the same.
Localization of the particle can be obtained by considering not just a single value of momentum (ℏk), but a range of values.
Consider the superposition of waves with a range of momenta given by
p=p0±Δp
or
ℏk=ℏ(k0±Δk)
ΨΔk(x)=N∫ko+Δkko−Δk eikxdk
ΨΔk(x)=2Nsin(Δkx)xeikox
where N is a normalization constant and is equal to
N=12√πΔk
To understand this new wavefunction, we plot its Real and Imaginary components separately.
Interactive Element
Assuming ko>Δk, the cos(kox) and sin(kox) components oscillate with a larger frequency (ko) and they are modulated by a sinusoidal component with smaller frequency (Δk).
If Δk is very small, the particle is still delocalized. As Δk increases, the particle becomes more localized in a specific region around x=0. This leads us to conclude that as the uncertainty in the momentum increases (larger Δk), the uncertainty in the position decreases, and the particle becomes more localized.
The superposition of waves can be accomplished by giving different contributions to different values of momenta.
For example, we can construct the superposition with a Gaussian weighting function. In this case, we modulate the contribution of each k-wave with a value given by a Gaussian distribution (maximum contribution from ko, with smaller contributions from waves with other k values).
ΨΔk(x)=N∫+∞−∞e−(k−ko)22Δk2 eikxdk,
where Δk is related to the width of the Gaussian distribution.
After integration, we obtain:
ΨΔk(x)=√2πNΔke−12x2Δk2eikox
where N is a normalization constant and is equal to
N=1√2Δk√π3
Interactive Element
We can compare the effect of the Gaussian distribution with the equally-weigthed k values:
Questions
- Q1: What do you expect to see for a particle with a momentum (ko) with contributions from waves with very different momenta? (large value of Δk)
- Q2: What do you expect to see for a a particel with a momentum (ko) with contributions from waves with similar momenta? (small value of Δk)
Interactive Element
What can we conclude: From the graphical representation we can learn that for Δk very small compared to ko (try plotting for ko=4 and Δk=0.04), the particle is still delocalized. As Δk increases the particle becomes more localized in a specific region around x=0 (i.e ko=4 and Δk=0.4). When Δk is large in comparison to ko (i.e. ko=4 and Δk=16), we observe the probability density concentrated around x=0.
This leads us to conclude that as the uncertainty in the momentum increases (larger Δk), the uncertainty in the position decreases, and the particle becomes more localized. This is the exactly the behavior predicted by the Heisenberg's uncertainty principle.
What other properties can we measure?
A Free particle has no potential (no acting force), thus the Hamiltonian only has a Kinetic term. The Schrödinger equation is:
−ℏ22m∂2ψ(x)∂x2=Eψ(x)
Since we already know an eigenfunction of the momentum operator, let's work a bit with commutator properties, and how they can be applied to the Free Particle wavefunction.
One commutator property says that if two Hermitian operators (ˆA and ˆB) commute, then they share a common set of eigenfunctions.
Let's now check if the Hamiltonian and the momentum operators commute: is
[ˆP,ˆH]=0?.
with ˆH=ˆP22m. This required evaluating
[ˆP,ˆH]=ˆP⋅ˆH−ˆH⋅ˆP=ˆP⋅ˆP22m−ˆP22m⋅ˆP=12m(ˆP⋅ˆP2−ˆP⋅ˆP2)=0
The two operators commute, leding to the existance of a complete set of eigenfunctions common to both operators. Since we already know the eigenfunctions of $\hat P$, we can try to see if they are also eigenfunctions of ˆH. For |k⟩=1√2πeikx, let's evaluate what is the result of ˆH|k⟩:
(note that we switch to a notation where Ψk(x)=|k⟩) :
ˆH|k⟩=−ℏ22m∂2|k⟩∂x2=−ℏ22m∂2∂x2(1√2πeikx)=−ℏ22m√2π∂2eikx∂x2=−ℏ22m√2π(ik)2eikx=ℏ2k22m(1√2πeikx)=p22m|k⟩
The first and last terms in this derivation show that the momentum eigenfunctions are also eigenfunctions of the Hamiltonian, with eigenvalues equal to the classical form of the kinetic energy. However, one has to be beware that although the value of the energy might look classical, the behavior of the particles are very different. For the quantum mechanical systems, the particle can never be at rest (if Δk=0 the particle is deloclaized; if the particle is very localized, then its momentum has many different values).
We are now ready to tackle "A Particle in a box with infinite-potential walls"