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Ch. 7 Solutions

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Table of Contents

Basic

Medium

Advanced

Basic

B-1. Commutation Relations of Ladder Operators

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Preparation

Setting \(\alpha \equiv \sqrt{m\omega/(2\hbar)}\), we have \(\hat{a} = \alpha(\hat{x} + i\hat{p}/(m\omega))\) and \(\hat{a}^\dagger = \alpha(\hat{x} - i\hat{p}/(m\omega))\).

Calculation of \([\hat{a}, \hat{a}^\dagger]\)

\(\hat{a}\hat{a}^\dagger = \alpha^2\left(\hat{x} + \frac{i\hat{p}}{m\omega}\right)\left(\hat{x} - \frac{i\hat{p}}{m\omega}\right)\)

Expanding the brackets:

\(= \alpha^2\left(\hat{x}^2 - \frac{i\hat{x}\hat{p}}{m\omega} + \frac{i\hat{p}\hat{x}}{m\omega} + \frac{\hat{p}^2}{m^2\omega^2}\right) = \alpha^2\left(\hat{x}^2 + \frac{\hat{p}^2}{m^2\omega^2} + \frac{i}{m\omega}[\hat{p}, \hat{x}]\right)\)

Using \([\hat{p}, \hat{x}] = -[\hat{x}, \hat{p}] = -i\hbar\):

\(\hat{a}\hat{a}^\dagger = \alpha^2\left(\hat{x}^2 + \frac{\hat{p}^2}{m^2\omega^2} + \frac{\hbar}{m\omega}\right) \tag{1}\)

Similarly, calculating \(\hat{a}^\dagger\hat{a}\), where the sign from \([\hat{x}, \hat{p}] = +i\hbar\) is used:

\(\hat{a}^\dagger\hat{a} = \alpha^2\left(\hat{x}^2 + \frac{\hat{p}^2}{m^2\omega^2} - \frac{\hbar}{m\omega}\right) \tag{2}\)

Taking (1) - (2):

\([\hat{a}, \hat{a}^\dagger] = \hat{a}\hat{a}^\dagger - \hat{a}^\dagger\hat{a} = \alpha^2 \cdot \frac{2\hbar}{m\omega} = \frac{m\omega}{2\hbar} \cdot \frac{2\hbar}{m\omega} = 1 \quad\blacksquare\)

Rewriting the Hamiltonian

We rewrite the right-hand side of (2) in terms of \(\hat{H}\). Since \(\hat{H} = \hat{p}^2/(2m) + (1/2)m\omega^2\hat{x}^2\):

\(\frac{m\omega}{2\hbar}\left(\hat{x}^2 + \frac{\hat{p}^2}{m^2\omega^2}\right) = \frac{1}{\hbar\omega}\left(\frac{1}{2}m\omega^2\hat{x}^2 + \frac{\hat{p}^2}{2m}\right) = \frac{\hat{H}}{\hbar\omega}\)

Therefore, (2) becomes:

\(\hat{a}^\dagger\hat{a} = \frac{\hat{H}}{\hbar\omega} - \frac{1}{2} \implies \hat{H} = \hbar\omega\left(\hat{a}^\dagger\hat{a} + \frac{1}{2}\right) \quad\blacksquare\)

Medium

M-1. Commutation Relations of the Hamiltonian with Ladder Operators

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Calculating \([\hat{H}, \hat{a}^\dagger]\)

For \(\hat{H} = \hbar\omega(\hat{a}^\dagger\hat{a} + 1/2)\), the \(1/2\) is a constant and does not contribute to the commutator:

\([\hat{H}, \hat{a}^\dagger] = \hbar\omega\,[\hat{a}^\dagger\hat{a}, \hat{a}^\dagger]\)

Using the identity \([\hat{A}\hat{B}, \hat{C}] = \hat{A}[\hat{B}, \hat{C}] + [\hat{A}, \hat{C}]\hat{B}\) with \(\hat{A} = \hat{a}^\dagger\), \(\hat{B} = \hat{a}\), \(\hat{C} = \hat{a}^\dagger\):

\([\hat{a}^\dagger\hat{a}, \hat{a}^\dagger] = \hat{a}^\dagger[\hat{a}, \hat{a}^\dagger] + [\hat{a}^\dagger, \hat{a}^\dagger]\hat{a} = \hat{a}^\dagger \cdot 1 + 0 \cdot \hat{a} = \hat{a}^\dagger\)

Therefore

\([\hat{H}, \hat{a}^\dagger] = +\hbar\omega\,\hat{a}^\dagger \quad\blacksquare\)

Calculating \([\hat{H}, \hat{a}]\)

Similarly, \([\hat{a}^\dagger\hat{a}, \hat{a}] = \hat{a}^\dagger[\hat{a}, \hat{a}] + [\hat{a}^\dagger, \hat{a}]\hat{a} = 0 + (-1)\hat{a} = -\hat{a}\), so

\([\hat{H}, \hat{a}] = -\hbar\omega\,\hat{a} \quad\blacksquare\)

Raising and lowering the energy

Let \(\hat{H}|n\rangle = E_n|n\rangle\). Acting \(\hat{H}\) on \(\hat{a}^\dagger|n\rangle\):

\(\hat{H}(\hat{a}^\dagger|n\rangle) = (\hat{a}^\dagger\hat{H} + [\hat{H}, \hat{a}^\dagger])|n\rangle = \hat{a}^\dagger E_n|n\rangle + \hbar\omega\,\hat{a}^\dagger|n\rangle = (E_n + \hbar\omega)(\hat{a}^\dagger|n\rangle)\)

Thus \(\hat{a}^\dagger|n\rangle\) is an eigenstate with eigenvalue \(E_n + \hbar\omega\). Similarly,

\(\hat{H}(\hat{a}|n\rangle) = (E_n - \hbar\omega)(\hat{a}|n\rangle) \quad\blacksquare\)

Physical interpretation

\(\hat{a}^\dagger\) is the operator that "raises the energy by one level" (= creates one quantum), and \(\hat{a}\) is the operator that "lowers the energy by one level" (= annihilates one quantum).

M-2. Normalized Ladder Operations

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Normalization of \(\hat{a}^\dagger|n\rangle\)

Let \(|n\rangle\) be a normalized eigenstate of \(\hat{N} = \hat{a}^\dagger\hat{a}\) (\(\hat{N}|n\rangle = n|n\rangle\), \(\langle n|n\rangle = 1\)). Since \(\hat{H} = \hbar\omega(\hat{N} + 1/2)\) gives \(\hat{H}|n\rangle = \hbar\omega(n + 1/2)|n\rangle\), the eigenstates of \(\hat{N}\) and the Hamiltonian are the same. From Problem 7.2, \(\hat{a}^\dagger|n\rangle\) is an eigenstate with energy \(\hbar\omega(n + 1 + 1/2)\), i.e., an eigenstate of \(\hat{N}\) with eigenvalue \(n + 1\), so \(\hat{a}^\dagger|n\rangle \propto |n+1\rangle\). To determine the proportionality constant from normalization, we compute the squared norm:

\(\|\hat{a}^\dagger|n\rangle\|^2 = \langle n|\hat{a}\hat{a}^\dagger|n\rangle\)

From Problem 7.1, \(\hat{a}\hat{a}^\dagger = \hat{a}^\dagger\hat{a} + 1 = \hat{N} + 1\), so

\(\|\hat{a}^\dagger|n\rangle\|^2 = \langle n|(\hat{N} + 1)|n\rangle = n + 1\)

Therefore

\(\hat{a}^\dagger|n\rangle = \sqrt{n+1}\,|n+1\rangle \quad\blacksquare\)

Normalization of \(\hat{a}|n\rangle\)

Similarly

\(\|\hat{a}|n\rangle\|^2 = \langle n|\hat{a}^\dagger\hat{a}|n\rangle = \langle n|\hat{N}|n\rangle = n\)

Therefore

\(\hat{a}|n\rangle = \sqrt{n}\,|n-1\rangle \quad\blacksquare\)

When \(n = 0\), the right-hand side becomes \(0\), which is consistent with the condition that one cannot go below the ground state: \(\hat{a}|0\rangle = 0\).

Construction of \(|n\rangle\)

\(\hat{a}^\dagger|0\rangle = \sqrt{1}\,|1\rangle = |1\rangle\), \(\hat{a}^\dagger|1\rangle = \sqrt{2}\,|2\rangle\), \(\hat{a}^\dagger|2\rangle = \sqrt{3}\,|3\rangle\), ...

Therefore

\(|n\rangle = \frac{1}{\sqrt{n}}\hat{a}^\dagger|n-1\rangle = \frac{1}{\sqrt{n}}\cdot\frac{1}{\sqrt{n-1}}(\hat{a}^\dagger)^2|n-2\rangle = \cdots = \frac{(\hat{a}^\dagger)^n}{\sqrt{n!}}|0\rangle \quad\blacksquare\)

Lemma (commutation formula): Provable by mathematical induction:

\([\hat{a}, (\hat{a}^\dagger)^n] = n\,(\hat{a}^\dagger)^{n-1}\)

(\(n = 1\): \([\hat{a}, \hat{a}^\dagger] = 1\) holds. Assuming it holds for \(n\), then \([\hat{a}, (\hat{a}^\dagger)^{n+1}] = [\hat{a}, \hat{a}^\dagger]\,(\hat{a}^\dagger)^n + \hat{a}^\dagger[\hat{a}, (\hat{a}^\dagger)^n] = (\hat{a}^\dagger)^n + \hat{a}^\dagger\cdot n(\hat{a}^\dagger)^{n-1} = (n+1)(\hat{a}^\dagger)^n\).)

From this follows \(\hat{a}(\hat{a}^\dagger)^n = (\hat{a}^\dagger)^n\hat{a} + n(\hat{a}^\dagger)^{n-1}\).

Verification of normalization: \(\langle n|n\rangle = \frac{1}{n!}\langle 0|\hat{a}^n(\hat{a}^\dagger)^n|0\rangle\). Using the above lemma to repeatedly simplify \(\hat{a}^n(\hat{a}^\dagger)^n|0\rangle\), terms where the rightmost \(\hat{a}\) annihilates \(|0\rangle\) successively drop out, and ultimately \(n!|0\rangle\) remains. Therefore \(\langle n|n\rangle = \frac{n!}{n!}\langle 0|0\rangle = 1\), confirming the normalization is correct.

M-3. Why the Schrödinger Equation Is Not Lorentz Covariant

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(a) Order of derivatives

The free-particle Schrödinger equation

\(i\hbar\,\partial_t \psi = -\frac{\hbar^2}{2m}\,\partial_x^2\, \psi\)

The left-hand side contains \(\partial_t\) (first-order time derivative), while the right-hand side contains \(\partial_x^2\) (second-order spatial derivative). The orders of differentiation do not match.

(b) Lorentz-scalar differential operator

Under the Minkowski metric \(\eta_{\mu\nu} = \mathrm{diag}(-1, +1, +1, +1)\), combining the 4-derivative \(\partial_\mu = (\partial_t/c, \nabla)\) to form a Lorentz scalar gives the d'Alembertian

\(\Box \equiv \partial^\mu\partial_\mu = -\frac{1}{c^2}\partial_t^2 + \nabla^2\)

The essential point is that the second-order time derivative and the second-order spatial derivative appear at the same order.

Under a Lorentz transformation \(x^\mu \to \Lambda^\mu{}_\nu x^\nu\), we have \(\partial_\mu \to \Lambda_\mu{}^\nu \partial_\nu\), so time and spatial derivatives mix. The first-order \(\partial_t\) will contain first-order \(\partial_x\) terms after transformation, but since the original equation has no first-order \(\partial_x\) term, new terms appear in the transformed frame. Therefore, the form of the equation is not preserved under Lorentz transformations.

(c) The Klein-Gordon equation and negative-energy solutions

Promoting the relativistic energy-momentum relation \(E^2 = p^2c^2 + m^2c^4\) to operators via \(E \to i\hbar\,\partial_t\), \(\vec{p} \to -i\hbar\nabla\):

\((i\hbar\,\partial_t)^2\phi = \left[(-i\hbar\nabla)^2c^2 + m^2c^4\right]\phi\)

Rearranging:

\(\left(-\frac{1}{c^2}\partial_t^2 + \nabla^2 - \frac{m^2c^2}{\hbar^2}\right)\phi = 0 \quad\blacksquare\)

The left-hand side is \(\Box\phi - (m^2c^2/\hbar^2)\phi\), which contains a second-order time derivative and is Lorentz covariant.

Why negative-energy solutions appear

Substituting the plane wave \(\phi = e^{-iEt/\hbar + ipx/\hbar}\), the equation gives

\(E^2 = p^2c^2 + m^2c^4\)

This is a quadratic equation in \(E\), so the solutions are

\(E = \pm\sqrt{p^2c^2 + m^2c^4}\)

Both the positive branch (ordinary particles) and the negative branch appear. The Schrödinger equation is first-order in \(E\) (since \(i\hbar\,\partial_t\) corresponds to \(E\)), so only positive solutions arise. The moment we make the theory relativistic, the "second-order time derivative" appears, which produces a quadratic equation in \(E\) and gives rise to negative-energy solutions—this was the starting point for Dirac's discovery of antiparticles (details in Quantum Mechanics Ch. 27).

Summary: How the solutions in this chapter connect directly to string theory

Problem Reappearance in string theory
Problems 7.1–7.3 (ladder operator algebra) Ch. 14 Mode expansion and quantization of the string
Problem 7.4 (infinite sum of zero-point energies) Ch. 14 Critical dimension of the bosonic string \(D = 26\)
Problem 7.5 (non-covariance of the Schrödinger equation) Ch. 8 Quantum field theory, Ch. 13 The string action is relativistic from the start

The harmonic oscillator algebra acquired in this chapter becomes the direct tool for quantizing the string. Starting from the next chapter, we trace the path of quantum field theory that resolves the conflict between relativity and quantum mechanics, and beyond that, string theory emerges.

Advanced

A-1. Infinite Harmonic Oscillators and the Zero-Point Energy of a String

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(a) Zero-point energy of each mode

In Section 7.4, we showed that the ground state energy of a harmonic oscillator with angular frequency \(\omega\) is \(E_0 = \hbar\omega/2\). Denoting the angular frequency of mode number \(n\) as \(\omega_n = n\omega_1\), the zero-point energy of this mode is

\(\frac{1}{2}\hbar\omega_n = \frac{n\hbar\omega_1}{2} \quad\blacksquare\)

(b) Total zero-point energy and divergence

Summing the zero-point energies over all modes:

\(E_{\text{zero}} = \sum_{n = 1}^{\infty}\frac{\hbar\omega_n}{2} = \frac{\hbar\omega_1}{2}\sum_{n = 1}^{\infty} n = \frac{\hbar\omega_1}{2}(1 + 2 + 3 + \cdots)\)

The sum \(1 + 2 + 3 + \cdots\) clearly diverges (it is monotonically increasing with no upper bound). Therefore \(E_{\text{zero}}\) diverges to infinity.

(c) Zeta function regularization

The Riemann zeta function \(\zeta(s) = \sum_{n=1}^{\infty} n^{-s}\) converges for \(\mathrm{Re}(s) > 1\) and can be analytically continued to other regions. Its analytically continued value gives

\(\zeta(-1) = -\frac{1}{12}\)

(The derivation relies on theorems of complex analysis. In physics, we accept this as a prescription.) Adopting the regularization that identifies \(\sum n = \zeta(-1)\), we obtain

\(E_{\text{zero}} = \frac{\hbar\omega_1}{2}\cdot\left(-\frac{1}{12}\right) = -\frac{\hbar\omega_1}{24} \quad\blacksquare\)

The fact that the sign becomes negative may seem strange at first glance, but physically it means "things are consistent when we choose this energy reference for the vacuum."

Remark: Connection to the critical dimension of strings

As discussed in Ch. 14, for the bosonic string with spacetime dimension \(D\), the zero-point energy is multiplied by \((D-2)\) as a sum over independent transverse oscillation directions (the number of physical degrees of freedom in light-cone quantization). The requirement of Lorentz invariance (the condition that a massless vector particle appears consistently in the spectrum) leads to

\(\frac{D-2}{24} = 1 \implies D = 26\)

This is the critical dimension of the bosonic string. The zero-point energy of harmonic oscillators obtained in this chapter is directly connected to the profound result that determines the dimensionality of spacetime in string theory.