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Quantum Mechanics 05 - Translation in QM

August 14, 2020

In the 2nd and 3rd post, we discuss two forms of translations - rotations and time-evolutions. But wait, we haven’t even talked about linear translation!

The Position Space

Before even talking about the translation operator, we need to think about what space, or what basis states we should use. Obviously, translations are closely related to positions, and it would be convenient to have the position space. In this post, we will first deal with 1D space for simplicity. We define a position operator x^\hat x:

x^∣x⟩=x∣x⟩\hat x |x\rangle = x |x\rangle

where ∣x⟩|x\rangle is an eigenstate of x^\hat xin the position space, and xx is the 1D position. Unlike angular momentum space we’ve dealt with, the position space have infinite many eigenstates, and we need to be more careful with the mathematical notations.

For a general state ψ\psi, we can represent it as the superposition of position eigenstates:

∣ψ⟩=∫−∞∞dx∣x⟩⟨x∣ψ⟩|\psi\rangle = \int_{-\infty}^\infty dx |x\rangle\langle x|\psi\rangle

and its conjugate,

⟨ψ∣=∫−∞∞dx⟨ψ∣x⟩⟨x∣\langle \psi| = \int_{-\infty}^\infty dx \langle \psi|x\rangle\langle x|

If we apply this relationship to an eigenstate ∣x′⟩|x'\rangle, we will obtain an important relationship:

∣x′⟩=∫dx∣x⟩⟨x∣x′⟩⟹⟨x∣x′⟩=δ(x−x′)|x'\rangle = \int dx |x\rangle\langle x|x'\rangle \Longrightarrow \langle x|x'\rangle = \delta(x-x')

From here, we can also further explore the first postulate of QM we talked about in the first post, that a quantum mechanical state is represented by a probability wave function.

Let ψ\psi be a normalized state, i.e. ⟨ψ∣ψ⟩=1\langle \psi|\psi\rangle = 1, we have

1=⟨ψ∣ψ⟩=∫∫dx′dx⟨ψ∣x′⟩⟨x′∣x⟩⟨x∣ψ⟩=∫∫dx′dx⟨ψ∣x′⟩δ(x−x′)⟨x∣ψ⟩=∫dx⟨ψ∣x⟩⟨x∣ψ⟩=∫dx∣⟨x∣ψ⟩∣2\begin{aligned} 1 = \langle \psi|\psi\rangle &= \int \int dx' dx \langle \psi|x'\rangle\langle x'|x\rangle\langle x|\psi \rangle \\ &= \int \int dx' dx \langle \psi|x'\rangle \delta(x-x') \langle x|\psi \rangle \\ &= \int dx \langle \psi|x\rangle \langle x|\psi \rangle \\ &= \int dx |\langle x|\psi \rangle|^2 \end{aligned}

This is where we get the physical explanation that ⟨x∣ψ⟩2\langle x|\psi \rangle^2 is the probability that the particle is at position xx. Further, notice that ⟨x∣ψ⟩\langle x|\psi \rangle, the probability of amplitude, is a function of xx. Here we have the wave function:

ψ(x)=⟨x∣ψ⟩\psi(x) = \langle x|\psi \rangle

The Translation Operator

Now we can talk about the translation operator T^\hat T, which should act as follows:

T^(a)∣x⟩=∣x+a⟩\hat T(a) |x\rangle = |x+a\rangle

Literally, T^(a)\hat T(a) should “move” the state by aa.

We can act T^\hat T on ∣ψ⟩|\psi\rangle to obtain

∣ψ′⟩=T^∣ψ⟩=∫dx′∣x′+a⟩⟨x′∣ψ⟩|\psi '\rangle = \hat T |\psi\rangle = \int dx'|x'+a\rangle \langle x'|\psi\rangle

with this definition and see how ψ(x)\psi (x) changes to ψ′(x)\psi' (x):

ψ′(x)=⟨x∣ψ′⟩=∫dx′⟨x∣x+a⟩⟨x′∣ψ⟩=∫dx′δ[x−(x′+a)]⟨x′∣ψ⟩=⟨x−a∣ψ⟩=ψ(x−a)\begin{aligned} \psi '(x) = \langle x|\psi '\rangle &= \int dx'\langle x|x+a\rangle \langle x'|\psi\rangle \\ &= \int dx' \delta[x-(x'+a)] \langle x'|\psi\rangle \\ &= \langle x-a | \psi\rangle = \psi(x-a)\\ \end{aligned}

This may seem wrong, but think about what happens when you shift a function f(x)f(x) to the left by aa: you obtain f(x−a)f(x-a).

Now let’s investigate the representation of the translation operator. Think about the rotation operator Ri^\hat{R_i} and the time-evolution operator U^\hat U that we’ve discussed in the 2nd and 3rd posts, it shouldn’t be a surprise that we need a generator of translation, which happens to be the momentum operator px^\hat{p_x}.

T^(dx)=1−iℏpx^dx    T^(a)=e−ipx^a/ℏ\hat T(dx) = 1 - \frac{i}{\hbar} \hat{p_x} dx \; \; \hat T(a) = e^{-i\hat{p_x} a/\hbar}

The Commutator of the Position and Momentum Operator

The commutator of x^\hat x and px^\hat{p_x} very useful and fundamental relationship to obtain:

By applying this infinitesimal limit definition above, we have:

(x^T^(δx)−T^(δx)x^)∣ψ⟩=(−iδxℏ)[x^,px^]∣ψ⟩\Big(\hat x \hat T(\delta x) - \hat T(\delta x) \hat x\Big) |\psi\rangle = \Big(\frac{-i\delta x}{\hbar}\Big)\Big[\hat x, \hat{p_x}\Big] |\psi\rangle

By applying the translation operator and position operator directly, we obtain

(x^T^(δx)−T^(δx)x^)∣ψ⟩=(x^T^(δx)−T^(δx)x^)∫dx∣x⟩⟨x∣ψ⟩=x^∫dx∣x+δx⟩⟨x∣ψ⟩−T^(δx)∫dx  x∣x⟩⟨x∣ψ⟩=∫dx(x+δx)∣x+δx⟩⟨x∣ψ⟩−∫dx  x∣x+δx⟩⟨x∣ψ⟩=δx∫dx∣x+δx⟩⟨x∣ψ⟩=δx∫dx′∣x’⟩⟨x′−δx∣ψ⟩=δx∣ψ⟩\begin{aligned} \Big(\hat x \hat T(\delta x) - \hat T(\delta x) \hat x\Big) |\psi\rangle &= \Big(\hat x \hat T(\delta x) - \hat T(\delta x) \hat x\Big) \int dx|x\rangle \langle x|\psi\rangle \\ &= \hat x \int dx|x+\delta x\rangle \langle x|\psi\rangle - \hat T(\delta x) \int dx \; x |x\rangle \langle x|\psi\rangle \\ &= \int dx (x+\delta x) |x+\delta x\rangle \langle x|\psi\rangle - \int dx \; x |x+\delta x\rangle \langle x|\psi\rangle \\ &= \delta x \int dx |x+\delta x\rangle \langle x|\psi\rangle \\ &= \delta x \int dx' |x’\rangle \langle x'-\delta x|\psi\rangle = \delta x |\psi\rangle\\ \end{aligned}

where in the last step we keep only the leading-order term of the wave function. Equate the two results, we find that:

[x^,px^]=iℏ⟹ΔxΔpx≥ℏ2\Big[\hat x, \hat{p_x}\Big] = i\hbar \Longrightarrow \Delta x \Delta p_x \geq \frac{\hbar}{2}

The Ehrenfest Theorem and The Correspondence Principle

Though your intuition for QM now should be enough to convince you that p^x\hat p_x is the momentum operator, let’s validate this by calculating some momentum related quantities. Specifically, let’s start with the energy.

In classical mechanics, we have total energy is equal to mechanical energy and potential energy, or E=px22m+V(x)E = \frac{p^2_x}{2m} + V(x), which is translated to the Hamiltonian with the corresponding eigenvalues:

H^=px^22m+V(x^)\hat H = \frac{\hat{p_x}^2}{2m} + V(\hat x)

Recall the result about the time dependence of expectation from post 3,

ddt⟨A⟩=iℏ⟨ψ(t)∣[H^,A^]∣ψ(t)⟩+⟨ψ(t)∣∂A^∂t∣ψ(t)⟩\frac{d}{dt} \langle A\rangle = \frac{i}{\hbar} \langle\psi(t)| \Big[ \hat H, \hat A \Big] |\psi(t)\rangle + \langle\psi(t)| \frac{\partial \hat A}{\partial t}|\psi(t)\rangle

Using the fact that operators x^\hat x and px^\hat{p_x} are time-independent, we can obtain two relationships,

d⟨x⟩dt=iℏ⟨ψ∣[H^,x^]∣ψ⟩=i2mℏ⟨ψ∣[px2^,x^]∣ψ⟩=i2mℏ⟨ψ∣px^[px^,x^]−[x^,px^]px^∣ψ⟩=1m⟨ψ∣px^∣ψ⟩=⟨px⟩m\begin{aligned} \frac{d\langle x\rangle}{dt} &= \frac{i}{\hbar} \langle\psi| \Big[ \hat H, \hat x \Big] |\psi\rangle \\ &= \frac{i}{2m\hbar} \langle\psi| \Big[ \hat{p_x^2}, \hat x \Big] |\psi\rangle \\ &= \frac{i}{2m\hbar} \langle\psi| \hat{p_x} \Big[ \hat{p_x}, \hat x \Big] - \Big[ \hat x, \hat{p_x} \Big] \hat{p_x} |\psi\rangle \\ &= \frac{1}{m} \langle\psi| \hat{p_x} |\psi\rangle = \frac{\langle p_x\rangle}{m} \end{aligned}
d⟨px⟩dt=iℏ⟨ψ∣[H^,px^]∣ψ⟩=iℏ⟨ψ∣[V(x^),x^]∣ψ⟩=⟨−dVdx⟩\begin{aligned} \frac{d\langle p_x\rangle}{dt} &= \frac{i}{\hbar} \langle\psi| \Big[ \hat H, \hat{p_x} \Big] |\psi\rangle \\ &= \frac{i}{\hbar} \langle\psi| \Big[ V(\hat x), \hat x \Big] |\psi\rangle \\ &= \langle -\frac{dV}{dx}\rangle \end{aligned}

The next to last step of the second derivation requires writing out V(x^)V(\hat x) as Taylor series and find out [x^n,px^]\Big[ \hat x^n, \hat{p_x} \Big] by the mathematical induction.

These two relationships together,

d⟨x⟩dt=⟨px⟩m    d⟨px⟩dt=⟨−dVdx⟩\frac{d\langle x\rangle}{dt} = \frac{\langle p_x\rangle}{m} \; \; \frac{d\langle p_x\rangle}{dt} = \langle -\frac{dV}{dx}\rangle

are called the Ehrenfest theorem. It may seem to suggest the expectation values ⟨x⟩,⟨px⟩\langle x\rangle, \langle p_x\rangle satisfy the Newton’s second law, but that will require ⟨V(x)⟩=V(⟨x⟩)\langle V(x)\rangle = V(\langle x\rangle), which is not always true. Nevertheless, for the systems widely studied by the classical mechanics, this holds true. The Ehrenfest theorem manifests the correspondence principle, the principle that quantum mechanics reproduce the classical mechanics results in large limits.

The Momentum Operator in the Position Space

Last but not the least, let’s consider the momentum operator’s representation in the position space.

First, we directly apply the translation operator in the position space:

T^(δx)∣ψ⟩=∫dx∣x+δx⟩⟨x∣ψ⟩=∫dx′∣x⟩⟨x−δx∣ψ⟩=∫dx′∣x⟩⟨x∣ψ⟩−δx∂∂x⟨x′∣ψ⟩=⟨x∣ψ⟩−∫dx′δx∂∂x⟨x′∣ψ⟩\begin{aligned} \hat T(\delta x)|\psi\rangle &= \int dx |x+\delta x\rangle\langle x|\psi\rangle \\ &= \int dx' |x\rangle\langle x - \delta x|\psi\rangle \\ &= \int dx' |x\rangle\langle x|\psi\rangle - \delta x \frac{\partial}{\partial x} \langle x'|\psi\rangle \\ &= \langle x|\psi\rangle - \int dx' \delta x \frac{\partial}{\partial x} \langle x'|\psi\rangle \end{aligned}

Then, observe the definition of the translation operator:

T(δx)=1−iℏpx^δxT(\delta x) = 1 - \frac{i}{\hbar} \hat{p_x} \delta x

We conclude that, in the position space:

px^=ℏi∂∂x\hat{p_x} = \frac{\hbar}{i}\frac{\partial}{\partial x}

The Momentum Space

In the momentum space, we have basis states as the eigenstates of the momentum operator:

px^∣p⟩=p∣p⟩\hat{p_x}|p\rangle = p|p\rangle

Similarly to the position space, we have the following representations and relationships:

∣ψ⟩=∫dp∣p⟩⟨p∣ψ⟩|\psi\rangle = \int dp |p\rangle\langle p|\psi\rangle ⟨p∣p′⟩=δ(p−p′)\langle p|p'\rangle = \delta(p-p') ∫dp∣⟨p∣ψ⟩∣2=1\int dp |\langle p|\psi \rangle|^2 = 1

What’s interesting and useful is to find out ⟨x∣p⟩\langle x|p\rangle so that we can make transitions between the position space and the momentum space.

Luckily, we know the representation of the momentum operator in both spaces:

⟨x∣px^∣p⟩=p⟨x∣p⟩=ℏi∂∂x⟨x∣p⟩\langle x|\hat{p_x}|p\rangle = p\langle x|p\rangle = \frac{\hbar}{i}\frac{\partial}{\partial x}\langle x|p\rangle

Solving the differential equation, we get:

⟨x∣p⟩=Neipx/ℏ\langle x|p\rangle = N e^{ipx/\hbar}

where NN is a constant. From this expression, we can already derive the de Broglie wavelength. Notice that the expression is a oscillatory function of xx with the complex phase p/ℏp/\hbar. Hence the wavelength λ\lambda is 2πp/ℏ=2πℏp\frac{2\pi}{p/\hbar} = \frac{2\pi\hbar}{p}, or simply

λ=hp\lambda = \frac{h}{p}

To find the value of the constant, we represent a momentum eigenstate in the position space:

∣p⟩=∫dx∣x⟩⟨x∣p⟩|p\rangle = \int dx |x\rangle\langle x|p\rangle ⟨p′∣p⟩=∫dx⟨p′∣x⟩⟨x∣p⟩=N∗N∫dxei(p−p′)x/ℏ=N∗N(2πℏ)δ(p−p′)=δ(p−p′)\begin{aligned} \langle p'|p\rangle &= \int dx \langle p'|x\rangle\langle x|p\rangle \\ &= N^*N \int dx e^{i(p-p')x/\hbar} \\ &= N^*N (2\pi\hbar) \delta(p-p') = \delta(p-p') \end{aligned}

where we use the Dirac delta’s representation δ(x)=12π∫dk eikx\delta(x) = \frac{1}{2\pi} \int dk \ e^{ikx}.

Now we’ve found that N=12πℏN = \frac{1}{\sqrt{2\pi\hbar}} (I know 2πℏ=h2\pi\hbar = h, but we the quantum physicists just don’t like hh…), we can write:

⟨x∣p⟩=12πℏeipx/ℏ\langle x|p\rangle = \frac{1}{\sqrt{2\pi\hbar}} e^{ipx/\hbar}

The Solutions to the Schrödinger Equation

Now we have a representation of the momentum operator in the position space, we can first write out the Hamitonian in the position space:

H^=−ℏ22m∂2∂x2+V(x^)\hat H = - \frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(\hat x)

So the time-dependent Schrödinger Equation in the position space is:

[−ℏ22m∂2∂x2+V(x^)]ψ(x,t)=iℏ∂∂tψ(x,t)\Big[ - \frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(\hat x) \Big] \psi(x, t) = i\hbar \frac{\partial}{\partial t} \psi(x, t)

Suppose ψ(x)\psi(x) is an energy eigenfunction of the Hamiltonian and ψ(x,t)=ψ(x)e−iEt/ℏ\psi(x, t) = \psi(x) e^{-iEt/\hbar}, we have the time-independent Schrödinger Equation:

[−ℏ22m∂2∂x2+V(x^)]ψ(x)=Eψ(x)\Big[ - \frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(\hat x) \Big] \psi(x) = E \psi(x)

With this equation and a specific potential function, we can solve the wave equation.

There are many typical 1D problems to solve such as the finite and infinite potential well, scattering, tunneling, the delta function potential, the quantum oscillator. Check out resources online like this post from UVA for more references.

References

[1] Zajc, William. PHYS GU4021-4022 Quantum Mechanics I-II, Department of Physics, Columbia University. 2019-2020.

[2] Townsend, John. A Modern Approach to Quantum Mechanics. 2nd Edition.

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