Exercise 1 (Quantum mechanics, revisited)
When, in the generalized statistical interpretation, we revisited the postulates of quantum mechanics using the new language of linear algebra, we assumed the state of a quantum mechanical system is a vector living in the Hilbert space of the square-integrable functions ∣ψ⟩∈L2(−∞,+∞).
We then saw that there are two different sets of “orthonormal” basis states for this space. The position basis ∣ex⟩ and the momentum basis ∣ep⟩. The function we previously called a wavefunction ψ(x) is nothing but the inner product of the state ∣ψ⟩ with the ∣ex⟩ basis
ψ(x)=⟨ex∣ψ⟩. Then we interpret ∣ψ(x)∣2dx as the probability of finding the particle at x.
But there is nothing special about the ∣ex⟩ basis. If we take the inner product of the momentum basis ∣ep⟩ with ∣ψ⟩ we get another function
ϕ(p)=⟨ep∣ψ⟩. This new function is called the momentum-space wavefunction or momentum wavefunction for short. The quantity ∣ϕ(p)∣2dp will then be the probability of measuring the momentum of the particle to be p.
It is important to pay attention to the difference between the state ∣ψ⟩ and the wavefunction ψ(x). The state ∣ψ⟩ is a vector in the Hilbert space and is not a function of position or momentum. The functional dependency comes from representing the state in a specific basis such as the position ∣ex⟩ or momentum ∣ep⟩ basis
∣ψ⟩=−∞∫+∞ψ(x)∣ex⟩dx or
∣ψ⟩=−∞∫+∞ϕ(p)∣ep⟩dp. The different basis are related to each other by the relation
⟨ex∣ep⟩=ep(x)=2πℏ1eipx/ℏ and the momentum wavefunction is the Fourier transform of the position wavefunction
ϕ(p)=2πℏ1−∞∫+∞ψ(x)e−ipx/ℏdx. Using this interpretation, solve the following.
We always chose the position wavefunction of the system to be normalized therefore we had
⟨ψ∣ψ⟩=−∞∫+∞∣ψ(x)∣2dx=1 which was necessary for ∣ψ(x)∣2 to be a probability density. It is not immediately clear though whether ∣ϕ(p)∣2 is also a probability density in that sense. For it to be a probability density we must have
−∞∫+∞∣ϕ(p)∣2dp=−∞∫+∞ϕ∗(p)ϕ(p)dp=1. Put ϕ(p) from Eq. 6 and its conjugate ϕ∗(p) into the integral
−∞∫+∞∣ϕ(p)∣2dp to show that
−∞∫+∞∣ϕ(p)∣2dp=−∞∫+∞∣ψ(x)∣2dx which then means for any normalized wavefunction ψ(x), the corresponding momentum wavefunction ϕ(p) will also be normalized. Note that the relation in Eq. 10 is not only valid in quantum mechanics but is a mathematical feature of Fourier analysis and it is called the Parseval–Plancherel identity.
Hint: Use the relation proved from the Problem Set 5 that
2π1−∞∫+∞eiyzdy=δ(z). Also, be careful when multiplying ϕ(p) and ϕ∗(p) to use different letters for the variable under integration from Eq. 6.
Solution
We have
ϕ(p)=2πℏ1−∞∫+∞ψ(x)e−ipx/ℏdx, which means if we take the complex conjugate of the equation, we will have
ϕ∗(p)=2πℏ1−∞∫+∞ψ∗(x)e+ipx/ℏdx. Therefore, the integration in Eq. 8 will become
−∞∫+∞∣ϕ(p)∣2dp=−∞∫+∞ϕ∗(p)ϕ(p)dp=−∞∫+∞[2πℏ1−∞∫+∞ψ∗(x′)e+ipx′/ℏdx′][2πℏ1−∞∫+∞ψ(x)e−ipx/ℏdx]dp=−∞∫+∞[2πℏ1−∞∫+∞−∞∫+∞ψ∗(x′)ψ(x)e+ip(x′−x)/ℏdx′dx]dpAt this point, we change the order of integrations by pulling out the terms only involving x and x′, and integrating over p first. This means we have=−∞∫+∞−∞∫+∞ψ∗(x′)ψ(x)[−∞∫+∞2πℏ1e+ip(x′−x)/ℏdp]dx′dxNow we change variables in the integration over p to get to the representation of the delta function we are used to. If we change variables to q=p/ℏ, we will have=−∞∫+∞−∞∫+∞ψ∗(x′)ψ(x)[−∞∫+∞2π1e+iq(x′−x)dq]dx′dxAnd we can immediately see that the integration over q will just result in a delta function=−∞∫+∞−∞∫+∞ψ∗(x′)ψ(x)[δ(x′−x)]dx′dxThis allows us to use the delta function to get rid of one of the integrations by integrating over x′ and get=−∞∫+∞ψ∗(x)ψ(x)dx=−∞∫+∞∣ψ(x)∣2dx QED.