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Ch. 7 Solutions

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Basic

Medium

Advanced

Basic

B-1. Commutation Relations of Ladder Operators

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Problem:

Expand the products \(\hat{a}\hat{a}^\dagger\) and \(\hat{a}^\dagger\hat{a}\) of the ladder operators

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

using only the canonical commutation relation \([\hat{x}, \hat{p}] = i\hbar\), and verify by direct calculation that

\([\hat{a}, \hat{a}^\dagger] = 1, \qquad \hat{H} = \hbar\omega\left(\hat{a}^\dagger\hat{a} + \frac{1}{2}\right)\)

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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Problem:

Using the commutator identity \([\hat{A}\hat{B}, \hat{C}] = \hat{A}[\hat{B}, \hat{C}] + [\hat{A}, \hat{C}]\hat{B}\) and the result from Problem 7.1, \([\hat{a}, \hat{a}^\dagger] = 1\), derive

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

From this, show that if \(\hat{H}|n\rangle = E_n|n\rangle\), then \(\hat{a}^\dagger|n\rangle\) is an eigenstate with energy \(E_n + \hbar\omega\), and \(\hat{a}|n\rangle\) is an eigenstate with energy \(E_n - \hbar\omega\).

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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Problem:

Starting from only \(\hat{a}|0\rangle = 0\) and \([\hat{a}, \hat{a}^\dagger] = 1\), derive the following normalized ladder operations:

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

Furthermore, starting from the ground state \(|0\rangle\), verify that the normalization of

\(|n\rangle = \frac{(\hat{a}^\dagger)^n}{\sqrt{n!}}|0\rangle\)

is correct.

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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Problem:

(a) Confirm that the free-particle Schrödinger equation \(i\hbar\,\partial_t \psi = -\frac{\hbar^2}{2m}\,\partial_x^2\, \psi\) contains a first-order time derivative on the left-hand side and a second-order spatial derivative on the right-hand side.

(b) Explain that under the Minkowski metric \(\eta_{\mu\nu} = \mathrm{diag}(-1, +1, +1, +1)\), the differential operator that forms a Lorentz scalar is \(\partial^\mu\partial_\mu = -c^{-2}\partial_t^2 + \nabla^2\) (the d'Alembertian), and that the orders of the time and spatial derivatives must match.

(c) As a straightforward candidate for making the Schrödinger equation Lorentz covariant, write down the Klein-Gordon equation obtained by promoting \(E^2 = p^2c^2 + m^2c^4\) to operators:

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

Confirm that this contains a second-order time derivative. State qualitatively why this "second-order" nature directly leads to the appearance of negative-energy solutions.

(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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Problem:

In string theory, each vibrational mode of a string (mode number \(n = 1, 2, 3, \ldots\)) behaves as an independent harmonic oscillator, and the angular frequency of each mode is an integer multiple \(\omega_n = n\,\omega_1\) (where \(\omega_1\) is the angular frequency of the fundamental vibration).

(a) Using the results from Section 7.4, explain that the zero-point energy of each mode is \(\hbar\omega_n/2 = n\hbar\omega_1/2\).

(b) Show that the total zero-point energy is formally given by

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

and confirm that this sum diverges.

(c) In zeta function regularization, \(\sum_{n = 1}^{\infty} n\) is replaced by \(\zeta(-1) = -\frac{1}{12}\). Evaluate \(E_{\text{zero}}\) under this prescription and show that the result is negative. Note that this is one of the origins of the critical dimension \(D = 26\) for the bosonic string discussed in Ch. 14.

(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.