Position and Momentum Representations

1.2 Position and Momentum Representations

Summary: The position-space representation allows us to make contact with quantum mechanics expressed in terms of wavefunctions

Working in one dimension for the moment, it is convenient to deﬁne the state $| x_{0} ⟩$ which is an eigenfunction of the position operator: $\hat{x} | x_{0} ⟩ = x_{0} | x_{0} ⟩$ . Here we have used $x_{0}$ to indicate a speciﬁc value of $x$ – say 3 m right of the origin - but we also write $| x ⟩$ when we want to leave the exact value open. It’s like the diﬀerence between $| ω_{1} ⟩$ and $| ω_{i} ⟩$ .
The set ${| x ⟩}$ form a complete set as $\hat{x}$ is Hermitian, but now when we expand another state in terms of these, we have to integrate rather than sum, and rather than a discrete set of coeﬃcients ${a_{i}}$ we have a continuous set or function, $| α ⟩ = \int α (x^{'}) | x^{'} ⟩ d x^{'}$ . (Integration limits are $- \infty \to \infty$ unless otherwise speciﬁed.)
We choose the normalisation to be $⟨ x | x^{'} ⟩ = δ (x - x^{'})$ . Then $α (x) = ⟨ x | α ⟩$ . This is called the wave function in position space of the state $| α ⟩$ . With this normalisation the identity operator is $\int_{- \infty}^{\infty} | x^{'} ⟩ ⟨ x^{'} | d x^{'} .$
These states have inﬁnite norm and so don’t really belong to our vector space, but we don’t need to worry about this! (See PHYS20602 notes for much more detail.)
Since $| α ⟩ = \int α (x^{'}) | x^{'} ⟩ d x^{'}$ , from the fourth postulate we might expect that $P (x) \equiv | α (x) |^{2}$ is the probability of ﬁnding the particle at position $x$ . But we also need $\int P (x^{'}) d x^{'} = 1$ – the probability of ﬁnding the particle at some $x$ is unity. The transition from sums to integrals corresponds to the transition from discrete probabilities to probability distributions, and so the correct interpretation is that $P (x) d x$ is the probability of ﬁnding the particle in an inﬁnitesimal interval between $x$ and $x + d x$ .
Since $⟨ x | \hat{x} | α ⟩ = x ⟨ x | α ⟩ = x α (x)$ , we can say that the position operator in the position space representation is simply $x$ , and operating with it is equivalent to multiplying the wave function by $x$ .
If we make the hypothesis that $⟨ x | \hat{p} | α ⟩ = - i ℏ d α (x) ∕ d x$ , we can see that the required commutation relation $[\hat{x}, \hat{p}] = i ℏ$ is satisﬁed.
We can equally imagine states of deﬁnite momentum, $| p ⟩$ , with $\hat{p} | p ⟩ = p | p ⟩$ . Then $\tilde{α} (p) = ⟨ p | α ⟩$ is the wave function in position space; in this representation $\hat{p}$ is given by $p$ and $\hat{x}$ by $(i ℏ) d ∕ d p$ .
Let’s write the position-space wave function of $| p ⟩$ , $⟨ x | p ⟩$ , as $ϕ_{p} (x)$ . Since $⟨ x | \hat{p} | p ⟩$ can be written as either $p ϕ_{p} (x)$ or $- i ℏ d ϕ_{p} (x) ∕ d x$ , we see that $ϕ_{p} (x) = (1 ∕ \sqrt{2 π ℏ}) e^{i p x ∕ ℏ}$ . This is just a plane wave, as expected. (The normalisation is determined by $⟨ p | p^{'} ⟩ = δ (p - p^{'})$ .)
This implies that $α (x) = (1 ∕ \sqrt{2 π ℏ}) \int e^{i p^{'} x ∕ ℏ} \tilde{α} (p^{'}) d p^{'}$ , that is the position- and momentum-space wave functions are Fourier transforms of each other. (The $1 ∕ \sqrt{ℏ}$ is present in the prefactor because we are using $p$ and not $k = p ∕ ℏ$ as the conjugate variable.)
Strictly speaking we should write the position-space representations of $\hat{x}$ and $\hat{p}$ as $⟨ x | \hat{x} | x^{'} ⟩ = x^{'} δ (x - x^{'})$ and $⟨ x | \hat{p} | x^{'} ⟩ = (i ℏ) d δ (x - x^{'}) ∕ d x^{'}$ . These only make sense in a larger expression in which integration over $x^{'}$ is about to be performed.
The extension to three dimensions is trivial, with the representation of the vector operator $\hat{p}$ in position space being $- i ℏ \nabla$ . Since diﬀerentiation with respect to one coordinate commutes with multiplication by another, $[{\hat{x}}_{i}, {\hat{p}}_{j}] = i ℏ δ_{i j}$ as required. (We have used $\hat{x}$ as the position operator. We will however use r as the position vector. The reason we don’t use $\hat{r}$ is that it is so commonly used to mean a unit vector. Thus $\hat{x} | r ⟩ = r | r ⟩$ .)
The generalisation of $ϕ_{p} (x)$ is $ϕ_{p} (r) = (1 ∕ \sqrt{2 π ℏ}) e^{i p \cdot r ∕ ℏ}$ , which is a plane wave travelling in the direction of p.
In the position representation, the Schrödinger equation reads $- \frac{ℏ^{2}}{2 m} \nabla^{2} ψ (r, t) + V (r) ψ (r, t) = i ℏ \frac{\partial}{\partial t} ψ (r, t)$
Note though that position and time are treated quite diﬀerently in quantum mechanics. There is no operator corresponding to time, and $t$ is just part of the label of the state: $ψ (r, t) = ⟨ r | ψ (t) ⟩$ .
Together with the probability density, $ρ (r) = {|ψ (r)|}^{2}$ , we also have a probability ﬂux $j (r) = (- i ℏ ∕ 2 m) (ψ {(r)}^{*} \nabla ψ (r) - ψ (r) \nabla ψ {(r)}^{*})$ . The continuity equation $\nabla \cdot j = - \partial ρ ∕ \partial t$ which ensures local conservation of probability density follows from the Schrödinger equation.
A two-particle state has a wave function which is a function of the two positions, $Φ (r_{1}, r_{2})$ , and the basis kets are direct product states $| r_{1} ⟩ \otimes | r_{2} ⟩$ . For states of non-interacting distinguishable particles where it is possible to say that the ﬁrst particle is in single-particle state $| ψ ⟩$ and the second in $| ϕ ⟩$ , the state of the system is $| ψ ⟩ \otimes | ϕ ⟩$ and the wave function is $Φ (r_{1}, r_{2}) = (⟨ r_{1} | \otimes ⟨ r_{2} |) (| ψ ⟩ \otimes | ϕ ⟩) = ⟨ r_{1} | ψ ⟩ ⟨ r_{2} | ϕ ⟩ = ψ (r_{1}) ϕ (r_{2})$ .
From $i ℏ \frac{d}{d t} | ψ (t) ⟩ = \hat{H} | ψ (t) ⟩$ we obtain $| ψ (t + d t) ⟩ = (1 - \frac{i}{ℏ} \hat{H} d t) | ψ (t) ⟩$ . Thus $| ψ (t) ⟩ = lim_{N \to \infty} {(1 - \frac{i}{ℏ} \frac{(t - t_{0})}{N} \hat{H})}^{N} | ψ (t_{0}) ⟩ = e^{- i \hat{H} (t - t_{0}) ∕ ℏ} | ψ (t_{0}) ⟩ \equiv U (t, t_{0}) | ψ (t_{0}) ⟩$
where the exponential of an operator is deﬁned as $e^{\hat{Ω}} = \sum_{n} \frac{1}{n!} {\hat{Ω}}^{n}$ . If the Hamiltonian depends explicitly on time, we have $U (t, t_{0}) = T exp (- i \int_{t_{0}}^{t} \hat{H} (t^{'}) d t^{'} ∕ ℏ)$ , where the time-ordered exponential denoted by $T exp$ means that in expanding the exponential, the operators are ordered so that $\hat{H} (t_{1})$ always sits to the right of $\hat{H} (t_{2})$ (so that it acts ﬁrst) if $t_{1} < t_{2}$ . (This will be derived 4.7, and is given here for completeness.)

References

Shankar ch 1.10
Mandl ch 12.4

[next] [prev] [up] [top]