PSet 3 Discussion

Conventions used below: for spins we use Pauli operators \(\sigma_{x,z}\) with eigenvalues \(\pm 1\); the physical spin components are \(S_{x,z}=(\hbar/2)\sigma_{x,z}\). For coupled light–matter states we write \(\ket{n,\pm}\) where \(n\) is the cavity photon number and \(\sigma_z\ket{\pm}=\pm\ket{\pm}\).

Problem 3-1: Generalizing EPR-style arguments to three-particle entangled states¶

The three-spin state is

\[ \ket{\psi}=\frac{1}{2}\Big(\ket{+++}-\ket{+--}-\ket{-+-}-\ket{--+}\Big), \]

where \(\ket{\pm}\) denotes an \(S_z\) eigenstate (equivalently, a \(\sigma_z\) eigenstate) for each particle \(A,B,C\) in that order.

We will repeatedly use the standard basis change

\[ \ket{+z}=\frac{1}{\sqrt{2}}\left(\ket{+x}+\ket{-x}\right),\qquad \ket{-z}=\frac{1}{\sqrt{2}}\left(\ket{+x}-\ket{-x}\right). \]

(a)¶

If all three observers measure along \(z\), the only possible outcomes are exactly the four computational-basis strings that appear in \(\ket{\psi}\): \(+++\), \(+--\), \(-+-\), and \(--+\). Each has probability \(1/4\) (since each basis ket has amplitude magnitude \(1/2\)).

To diagnose correlations cleanly, work with Pauli outcomes \(s_z^A,s_z^B,s_z^C\in\{\pm 1\}\).

First look at two-body correlations, e.g. between \(A\) and \(B\). The possible \((s_z^A,s_z^B,s_z^C)\) triples are

\[ (+,+,+),\quad (+,-,-),\quad (-,+,-),\quad (-,-,+), \]

each occurring with probability \(1/4\). The corresponding products \(s_z^A s_z^B\) across these four outcomes are

\[ (+1),\quad (-1),\quad (-1),\quad (+1), \]

so they average to zero:

\[ \langle\sigma_z^A\sigma_z^B\rangle=\frac{1}{4}\left(1-1-1+1\right)=0. \]

By the same check (or by symmetry), all pairwise correlations vanish:

\[ \langle\sigma_z^A\sigma_z^B\rangle=\langle\sigma_z^A\sigma_z^C\rangle=\langle\sigma_z^B\sigma_z^C\rangle=0. \]

workout

However, the three-body product \(s_z^A s_z^B s_z^C\) equals \(+1\) for every allowed outcome above, hence

\[ \langle\sigma_z^A\sigma_z^B\sigma_z^C\rangle=1. \]

Equivalently, in terms of physical spins,

\[ \left\langle S_z^A S_z^B S_z^C\right\rangle=\left(\frac{\hbar}{2}\right)^3. \]

So: there are no two-body correlations in \(z\), but there is a perfect three-body correlation.

(b)¶

Here Alice and Bob measure along \(x\) while Carl measures along \(z\).

workout

A compact way to see the claimed parity constraint is to show that \(\ket{\psi}\) is an eigenstate of the product observable \(\sigma_x^A\sigma_x^B\sigma_z^C\). Acting on each \(z\)-basis product ket, \(\sigma_x\) flips \(\ket{+z}\leftrightarrow\ket{-z}\) and \(\sigma_z\) contributes a sign on the third qubit. One finds

\[ \sigma_x^A\sigma_x^B\sigma_z^C\,\ket{\psi}=-\ket{\psi}. \]

Therefore, if \((\sigma_x^A,\sigma_x^B,\sigma_z^C)\) are measured in the same run with outcomes \(s_x^A,s_x^B,s_z^C\in\{\pm 1\}\), their product is deterministic:

\[ s_x^A s_x^B s_z^C=-1. \]

Now interpret this in terms of the physical-spin outcomes \(\pm \hbar/2\). A “\(+1/2\) outcome” corresponds to the Pauli eigenvalue \(+1\). For three \(\pm 1\) numbers, having product \(-1\) is equivalent to having an even number of \(+1\)’s (since the number of \(-1\)’s must be odd). Thus, among the three outcomes \((S_x^A,S_x^B,S_z^C)\), there must be an even number of \(+\hbar/2\) results.

We can also expand \(\ket{\psi}\) into the \((xxz)\) basis, and can see that exactly four outcome triples occur (each with probability \(1/4\)):

\[ \ket{\psi}=\frac12\Big( \ket{+x,-x,+z} +\ket{-x,+x,+z} -\ket{+x,+x,-z} +\ket{-x,-x,-z}\Big), \]

which gives rise to

\[ (s_x^A,s_x^B,s_z^C)=(+,-,+),\;(-,+,+),\;(+,+,-),\;(-,-,-), \]

and these are precisely the ones satisfying \(s_x^A s_x^B s_z^C=-1\).

(c)¶

Here Alice measures along \(x\) while Bob and Carl measure along \(z\).

Projecting onto definite \((s_z^B,s_z^C)\) outcomes, note that for each of the four allowed \((B,C)\) strings in \(\ket{\psi}\), the corresponding \(A\) spin is actually a definite \(z\) eigenstate:

workout

\[ (B,C)=(+,+)\Rightarrow A=+,\quad (B,C)=(-,-)\Rightarrow A=+,\quad (B,C)=(+,-)\Rightarrow A=-,\quad (B,C)=(-,+)\Rightarrow A=-. \]

Thus, conditioned on any \((s_z^B,s_z^C)\), Alice’s state is either \(\ket{+z}\) or \(\ket{-z}\). Measuring \(\sigma_x\) on either \(\ket{+z}\) or \(\ket{-z}\) gives a completely random outcome:

\[ \Pr(s_x^A=+1\mid A=\pm z)=\Pr(s_x^A=-1\mid A=\pm z)=\frac{1}{2}. \]

Since the four \((s_z^B,s_z^C)\) outcomes each occur with probability \(1/4\), the full joint distribution is uniform:

\[ \Pr(s_x^A,s_z^B,s_z^C)=\frac{1}{8}\qquad\text{for all 8 possible triples.} \]

So the measurement outcomes in the \((xzz)\) setting are completely random: knowing two outcomes does not constrain the third.

(d)¶

workout

Under cyclic permutation of the particles \(A\to B\to C\to A\), the set of four \(z\)-basis kets appearing in \(\ket{\psi}\) maps into itself with the same coefficients, so the state is invariant under cyclic relabeling.

Therefore:

The conclusion of part (c) (uniform/random statistics) holds for any measurement setting with exactly one \(x\) and two \(z\) measurements: \((xzz)\), \((zxz)\), and \((zzx)\).
The conclusion of part (b) (the deterministic parity rule) holds for any measurement setting with exactly one \(z\) and two \(x\) measurements: \((xxz)\), \((xzx)\), and \((zxx)\).

(e)¶

Consider runs where Bob and Carl measure along \(z\). From part (a) we know the allowed outcomes satisfy

\[ s_z^A s_z^B s_z^C=+1. \]

Thus, once Bob and Carl later compare notes, they can predict with certainty what Alice would have obtained had she measured along \(z\) in that same run:

workout

\[ s_z^A=s_z^B s_z^C. \]

EPR-style reasoning then goes as follows: Alice’s particle is spacelike separated from Bob and Carl, so (assuming locality) Bob and Carl’s choices/outcomes cannot influence what would happen at Alice. But Bob and Carl can nonetheless predict Alice’s hypothetical \(S_z^A\) outcome with certainty. Therefore (in the EPR sense) \(S_z^A\) must have had a predetermined value in that run: it is an “element of reality”.

(f)¶

Now consider runs where Bob measures along \(z\) and Carl measures along \(x\). By part (d), the state also satisfies a deterministic constraint for the \((xzx)\) setting:

workout

\[ \sigma_x^A\sigma_z^B\sigma_x^C\,\ket{\psi}=-\ket{\psi}. \]

Hence the outcomes obey

\[ s_x^A s_z^B s_x^C=-1, \]

so once Bob and Carl compare their outcomes, they can predict with certainty what Alice would have obtained had she measured along \(x\):

\[ s_x^A=-s_z^B s_x^C. \]

Applying the same locality logic as in part (e), this implies (again in the EPR sense) that \(S_x^A\) also has a predetermined value for that run.

(g)¶

EPR reasoning from (e) and (f) suggests that in each run particle \(A\) has definite predetermined values for both \(\sigma_z^A\) and \(\sigma_x^A\). By symmetry (part (d)), the same should hold for particles \(B\) and \(C\) as well. A local hidden-variable model would then assign, for each run, fixed numbers

\[ a_x,a_z,b_x,b_z,c_x,c_z\in\{\pm 1\}, \]

independent of which measurements are actually chosen.

workout

However, the deterministic quantum predictions from parts (a) and (d) impose the following constraints on any such assignment:

From the \((zzz)\) setting in part (a),

\[ a_z b_z c_z=+1. \]
From the “two \(x\), one \(z\)” settings in part (d),

\[ a_x b_x c_z=-1,\qquad a_x b_z c_x=-1,\qquad a_z b_x c_x=-1. \]

Now multiply the last three equations. The left-hand side becomes

\[ (a_x b_x c_z)(a_x b_z c_x)(a_z b_x c_x) =(a_x^2)(b_x^2)(c_x^2)\,(a_z b_z c_z) =a_z b_z c_z, \]

because \(a_x^2=b_x^2=c_x^2=1\). But the right-hand side gives \((-1)^3=-1\). Thus any hidden-variable assignment would imply

\[ a_z b_z c_z=-1, \]

which contradicts the required \(a_z b_z c_z=+1\) from the \((zzz)\) statistics. Therefore it is impossible to construct a deterministic local hidden-variable theory reproducing the correlations of \(\ket{\psi}\).

Problem 3-2: Superdense coding and security¶

In superdense coding, Alice and Bob initially share a maximally entangled Bell state, e.g.

\[ \ket{\Phi^+}=\frac{1}{\sqrt{2}}\left(\ket{00}+\ket{11}\right). \]

Alice encodes a classical 2-bit message \(m\in\{00,01,10,11\}\) by applying one of four single-qubit unitaries \(U_m\) on her half of the pair, then sends her qubit to Bob. Bob then performs a Bell-basis measurement to decode \(m\) perfectly.

Suppose an eavesdropper Eve intercepts only Alice’s transmitted (single) qubit. We want to show that, regardless of what Eve measures, she learns nothing about \(m\).

workout

Let

\[ \rho_{AB}^{(m)}=(U_m\otimes I)\ket{\Phi^+}\!\bra{\Phi^+}(U_m^\dagger\otimes I) \]

be the joint two-qubit state after Alice’s encoding. Eve has access only to the reduced state on Alice’s subsystem:

\[ \rho_A^{(m)}=\operatorname{Tr}_B\!\left[\rho_{AB}^{(m)}\right]. \]

Taking the partial trace and using cyclicity with respect to operators on \(A\),

\[ \rho_A^{(m)} =\operatorname{Tr}_B\!\left[(U_m\otimes I)\ket{\Phi^+}\!\bra{\Phi^+}(U_m^\dagger\otimes I)\right] =U_m\left(\operatorname{Tr}_B\!\left[\ket{\Phi^+}\!\bra{\Phi^+}\right]\right)U_m^\dagger. \]

But for a maximally entangled Bell state the reduced state is maximally mixed:

\[ \operatorname{Tr}_B\!\left[\ket{\Phi^+}\!\bra{\Phi^+}\right]=\frac{I}{2}. \]

Therefore, for every message \(m\),

\[ \rho_A^{(m)}=U_m\left(\frac{I}{2}\right)U_m^\dagger=\frac{I}{2}. \]

This already proves security. Indeed, for any measurement Eve might do, described as a POVM \(\{M_y\}\) on the intercepted qubit, the probability of outcome \(y\) conditioned on \(m\) is

\[ \Pr(y\mid m)=\operatorname{Tr}\!\left[M_y\rho_A^{(m)}\right] =\operatorname{Tr}\!\left[M_y\frac{I}{2}\right], \]

which is independent of \(m\). Hence Eve’s measurement results have zero information about Alice’s message.

Problem 3-3: Perturbative treatment of a light–matter interaction¶

The Hamiltonian is

\[ \hat H=\omega_{\mathrm{cav}}\,\hat a^\dagger\hat a+\frac{\omega_{\mathrm{at}}}{2}\,\sigma_z+g\left(\hat a^\dagger\sigma_-+\hat a\,\sigma_+\right), \]

with \(\omega_{\mathrm{cav}}>\omega_{\mathrm{at}}\) and where \(\sigma_-=\ket{-}\!\bra{+}\) and \(\sigma_+=\ket{+}\!\bra{-}\). We treat the interaction term (proportional to \(g\)) as a perturbation.

(a)¶

At \(g=0\), the Hamiltonian is

\[ \hat H_0=\omega_{\mathrm{cav}}\,\hat a^\dagger\hat a+\frac{\omega_{\mathrm{at}}}{2}\,\sigma_z. \]

workout

Its ground state is clearly the no-photon, spin-down state \(\ket{0,-}\), with energy

\[ E_0^{(0)}=-\frac{\omega_{\mathrm{at}}}{2}. \]

Moreover, \(\ket{0,-}\) remains an eigenstate for any value of \(g\), because the interaction term annihilates it:

\[ \hat a\,\ket{0}=0,\qquad \sigma_-\,\ket{-}=0 \]

imply

\[ \left(\hat a^\dagger\sigma_-+\hat a\,\sigma_+\right)\ket{0,-}=0. \]

Thus \(\hat H\ket{0,-}=-(\omega_{\mathrm{at}}/2)\ket{0,-}\) for all \(g\).

(b)¶

The candidates for first excited eigenstates of \(\hat H_0\) are the two states in the one-excitation sector:

\[ \ket{0,+}\quad\text{with}\quad E_{0,+}^{(0)}=\frac{\omega_{\mathrm{at}}}{2}, \]

and

\[ \ket{1,-}\quad\text{with}\quad E_{1,-}^{(0)}=\omega_{\mathrm{cav}}-\frac{\omega_{\mathrm{at}}}{2}. \]

Since \(\omega_{\mathrm{cav}}>\omega_{\mathrm{at}}\), we have \(E_{1,-}^{(0)}-E_{0,+}^{(0)}=\omega_{\mathrm{cav}}-\omega_{\mathrm{at}}>0\), so the first excited state of \(\hat H_0\) is \(\ket{0,+}\).

The first-order energy shift is

workout

\[ E_{0,+}^{(1)}=\bra{0,+}\,g\left(\hat a^\dagger\sigma_-+\hat a\,\sigma_+\right)\ket{0,+}. \]

Both terms vanish: \(\sigma_-\ket{+}=\ket{-}\) but then \(\bra{+}\ket{-}=0\) for the diagonal matrix element, and \(\hat a\ket{0}=0\). Hence

\[ E_{0,+}^{(1)}=0. \]

(c)¶

The first-order correction to the eigenstate \(\ket{0,+}\) is

\[ \ket{0,+}^{(1)}=\sum_{m\neq (0,+)}\frac{\bra{m}V\ket{0,+}}{E_{0,+}^{(0)}-E_m^{(0)}}\ket{m}, \]

where \(V=g(\hat a^\dagger\sigma_-+\hat a\,\sigma_+)\). The only state connected to \(\ket{0,+}\) by \(V\) is \(\ket{1,-}\), because

\[ \hat a^\dagger\sigma_-\,\ket{0,+}=\ket{1,-},\qquad \hat a\,\sigma_+\,\ket{0,+}=0. \]

workout

The matrix element is

\[ \bra{1,-}V\ket{0,+}=g. \]

The energy denominator is

\[ E_{0,+}^{(0)}-E_{1,-}^{(0)}=\frac{\omega_{\mathrm{at}}}{2}-\left(\omega_{\mathrm{cav}}-\frac{\omega_{\mathrm{at}}}{2}\right)=\omega_{\mathrm{at}}-\omega_{\mathrm{cav}}=-(\omega_{\mathrm{cav}}-\omega_{\mathrm{at}}). \]

Therefore

\[ \ket{0,+}^{(1)}=-\frac{g}{\omega_{\mathrm{cav}}-\omega_{\mathrm{at}}}\ket{1,-}. \]

So, to first order in \(g\),

\[ \ket{\psi_{\mathrm{1st}}}\propto \ket{0,+}-\frac{g}{\omega_{\mathrm{cav}}-\omega_{\mathrm{at}}}\ket{1,-}. \]

(d)¶

The second-order energy shift is

\[ E_{0,+}^{(2)}=\sum_{m\neq (0,+)}\frac{|\bra{m}V\ket{0,+}|^2}{E_{0,+}^{(0)}-E_m^{(0)}}. \]

Again only \(\ket{1,-}\) contributes, yielding

\[ E_{0,+}^{(2)}=\frac{|g|^2}{\omega_{\mathrm{at}}-\omega_{\mathrm{cav}}} =-\frac{g^2}{\omega_{\mathrm{cav}}-\omega_{\mathrm{at}}}. \]

(e)¶

Define the total excitation number operator

\[ \hat N\equiv \hat a^\dagger\hat a+\sigma_+\sigma_-. \]

Here \(\hat a^\dagger\hat a\) counts cavity photons, while \(\sigma_+\sigma_-=\ket{+}\!\bra{+}\) is the projector onto the atomic excited state (so it counts atomic excitations). Using \(\sigma_z\ket{\pm}=\pm\ket{\pm}\), we can rewrite

\[ \sigma_+\sigma_-=\ket{+}\!\bra{+}=\frac{1+\sigma_z}{2}, \]

so equivalently

\[ \hat N=\hat a^\dagger\hat a+\frac{1+\sigma_z}{2}. \]

It is helpful to see explicitly that \(\hat N\) really does “count total excitations” on the product basis \(\ket{n,\pm}\):

\[ \hat N\ket{n,-}=n\ket{n,-},\qquad \hat N\ket{n,+}=(n+1)\ket{n,+}. \]

Now we show \([\hat N,\hat H]=0\). First recall the basic bosonic commutators \([\hat a,\hat a^\dagger]=1\) and the number-operator identities

\[ [\hat a^\dagger\hat a,\hat a^\dagger]=\hat a^\dagger,\qquad [\hat a^\dagger\hat a,\hat a]=-\hat a. \]

For the atom, use the standard spin algebra

\[ [\sigma_z,\sigma_\pm]=\pm 2\sigma_\pm. \]

Since \(\sigma_+\sigma_-=(1+\sigma_z)/2\), this implies

\[ [\sigma_+\sigma_-,\sigma_\pm]=\frac{1}{2}[\sigma_z,\sigma_\pm]=\pm\sigma_\pm. \]

We will also use two facts that simplify the bookkeeping:

Cavity operators commute with atomic operators, e.g. \([\hat a,\sigma_\pm]=0\) and \([\hat a^\dagger\hat a,\sigma_z]=0\).
The Leibniz rule \([A,BC]=[A,B]C+B[A,C]\).

The Hamiltonian is

\[ \hat H=\omega_{\mathrm{cav}}\,\hat a^\dagger\hat a+\frac{\omega_{\mathrm{at}}}{2}\,\sigma_z+g\left(\hat a^\dagger\sigma_-+\hat a\,\sigma_+\right). \]

The commutator with the “free” part is immediate:

\[ [\hat N,\omega_{\mathrm{cav}}\,\hat a^\dagger\hat a]=0,\qquad [\hat N,\tfrac{\omega_{\mathrm{at}}}{2}\sigma_z]=0, \]

because \([\hat a^\dagger\hat a,\hat a^\dagger\hat a]=0\), \([\sigma_+\sigma_-,\sigma_z]=0\), and cavity/atomic operators commute.

The only nontrivial step is the interaction term. Consider the first piece \(\hat a^\dagger\sigma_-\). Using the Leibniz rule and the commuting-subsystem facts,

\[ \begin{aligned} [\hat N,\hat a^\dagger\sigma_-] &=[\hat a^\dagger\hat a,\hat a^\dagger]\sigma_-+\hat a^\dagger[\hat a^\dagger\hat a,\sigma_-]+[\sigma_+\sigma_-,\hat a^\dagger]\sigma_-+\hat a^\dagger[\sigma_+\sigma_-,\sigma_-] \\ &=(\hat a^\dagger)\sigma_-+0+0+\hat a^\dagger(-\sigma_-) \\ &=0. \end{aligned} \]

Similarly, for \(\hat a\,\sigma_+\),

\[ \begin{aligned} [\hat N,\hat a\,\sigma_+] &=[\hat a^\dagger\hat a,\hat a]\sigma_+ + \hat a[\sigma_+\sigma_-,\sigma_+] \\ &=(-\hat a)\sigma_+ + \hat a(+\sigma_+) \\ &=0. \end{aligned} \]

Putting this together, we have

\[ [\hat N,\hat H]=0. \]

This means \(\hat H\) conserves the total excitation number \(N\) and therefore block-diagonalizes into fixed-\(N\) subspaces.

\(N=0\) subspace: the only state is \(\ket{0,-}\), so it is an exact eigenstate (the ground state) for all \(g\).
\(N=1\) subspace: a convenient basis is \(\{\ket{0,+},\ket{1,-}\}\).

We now build the Hamiltonian matrix in the \(N=1\) subspace explicitly. The diagonal terms come from the free Hamiltonian:

\[ \hat H_0\ket{0,+}=\frac{\omega_{\mathrm{at}}}{2}\ket{0,+},\qquad \hat H_0\ket{1,-}=\left(\omega_{\mathrm{cav}}-\frac{\omega_{\mathrm{at}}}{2}\right)\ket{1,-}. \]

For the off-diagonal coupling, note

\[ \hat a^\dagger\sigma_-\,\ket{0,+}=\ket{1,-},\qquad \hat a\,\sigma_+\,\ket{1,-}=\ket{0,+}, \]

while the other combinations annihilate these basis states. Hence in the ordered basis \(\{\ket{0,+},\ket{1,-}\}\),

workout

\[ H_{N=1}= \begin{pmatrix} \omega_{\mathrm{at}}/2 & g \\ g & \omega_{\mathrm{cav}}-\omega_{\mathrm{at}}/2 \end{pmatrix}. \]

Let \(\Delta\equiv\omega_{\mathrm{cav}}-\omega_{\mathrm{at}}>0\). The eigenvalues of this \(2\times 2\) matrix are

\[ E_{\pm}=\frac{\omega_{\mathrm{cav}}}{2}\pm\frac{1}{2}\sqrt{\Delta^2+4g^2}. \]

At \(g\to 0\), \(E_-\to \omega_{\mathrm{at}}/2\) and \(E_+\to \omega_{\mathrm{cav}}-\omega_{\mathrm{at}}/2\). Therefore the exact first excited energy for any \(g\) is

\[ \boxed{E_{\mathrm{1st}}^{\mathrm{exact}}=E_-=\frac{\omega_{\mathrm{cav}}}{2}-\frac{1}{2}\sqrt{(\omega_{\mathrm{cav}}-\omega_{\mathrm{at}})^2+4g^2}.} \]

It is also nice to write the corresponding exact eigenstate. Solve

\[ \left(H_{N=1}-E_\pm I\right)\begin{pmatrix}c_\pm\\ d_\pm\end{pmatrix}=0, \]

which gives the ratio

\[ \frac{d_\pm}{c_\pm}=\frac{E_\pm-\omega_{\mathrm{at}}/2}{g}. \]

Equivalently, one can introduce a mixing angle \(\theta\) defined by

\[ \tan(2\theta)=\frac{2g}{\Delta},\qquad 0<\theta<\frac{\pi}{2}, \]

so that (up to an overall phase)

\[ \ket{E_-}=\cos\theta\,\ket{0,+}-\sin\theta\,\ket{1,-},\qquad \ket{E_+}=\sin\theta\,\ket{0,+}+\cos\theta\,\ket{1,-}. \]

Finally, to confirm agreement with perturbation theory to order \(g^2\), expand \(E_-\) for small \(g\):

\[ \sqrt{\Delta^2+4g^2}=\Delta\sqrt{1+\frac{4g^2}{\Delta^2}} =\Delta\left(1+\frac{2g^2}{\Delta^2}+O(g^4)\right) =\Delta+\frac{2g^2}{\Delta}+O(g^4), \]

so

\[ E_-=\frac{\omega_{\mathrm{cav}}}{2}-\frac{1}{2}\left(\Delta+\frac{2g^2}{\Delta}\right)+O(g^4) =\boxed{\frac{\omega_{\mathrm{at}}}{2}-\frac{g^2}{\omega_{\mathrm{cav}}-\omega_{\mathrm{at}}}+O(g^4)}, \]

matching parts (b) and (d).

Problem 3-4: Perturbing a harmonic oscillator¶

We take \(\hbar=1\) as in the problem statement. The Hamiltonian is

\[ \hat H=\frac{\hat p^2}{2m}+\frac{1}{2}m\omega^2\hat x^2+\lambda\,\omega\left(\frac{\hat x}{x_{\mathrm{zpt}}}\right)^4, \]

where

\[ x_{\mathrm{zpt}}=\sqrt{\frac{1}{2m\omega}},\qquad \lambda\ll 1. \]

Let \(H_0=\frac{\hat p^2}{2m}+\frac{1}{2}m\omega^2\hat x^2\) so that \(E_n^{(0)}=(n+\tfrac{1}{2})\omega\), and take the perturbation

\[ V=\lambda\,\omega\left(\frac{\hat x}{x_{\mathrm{zpt}}}\right)^4. \]

Using \(\hat x=x_{\mathrm{zpt}}(\hat a+\hat a^\dagger)\), we can write

\[ V=\lambda\,\omega(\hat a+\hat a^\dagger)^4. \]

(a)¶

The first-order shift is

\[ E_n^{(1)}=\bra{n}V\ket{n}=\lambda\,\omega\,\bra{n}(\hat a+\hat a^\dagger)^4\ket{n}. \]

A convenient normal-ordered expansion is

\[ (\hat a+\hat a^\dagger)^4 =\hat a^4+\hat a^{\dagger4}+(4\hat N+6)\hat a^2+(4\hat N-2)\hat a^{\dagger2}+(6\hat N^2+6\hat N+3), \]

where \(\hat N=\hat a^\dagger\hat a\). Acting on \(\ket{n}\), every term except the last changes the number state and therefore has zero diagonal matrix element. Hence

workout

\[ \bra{n}(\hat a+\hat a^\dagger)^4\ket{n}=6n^2+6n+3, \]

and

\[ E_n^{(1)}=\lambda\,\omega\,(6n^2+6n+3)=3\lambda\,\omega\,(2n^2+2n+1). \]

(b)¶

The second-order shift for a non-degenerate level is

\[ E_n^{(2)}=\sum_{k\neq n}\frac{|\bra{k}V\ket{n}|^2}{E_n^{(0)}-E_k^{(0)}}. \]

Because \((\hat a+\hat a^\dagger)^4\) only changes the number \(n\) by \(\Delta n=\pm 2,\pm 4\), only \(k=n\pm 2\) and \(k=n\pm 4\) contribute.

From the expansion above, the needed matrix elements of \((\hat a+\hat a^\dagger)^4\) are

\[ \bra{n+4}(\hat a+\hat a^\dagger)^4\ket{n}=\sqrt{(n+1)(n+2)(n+3)(n+4)}, \]

\[ \bra{n-4}(\hat a+\hat a^\dagger)^4\ket{n}=\sqrt{n(n-1)(n-2)(n-3)}, \]

\[ \bra{n+2}(\hat a+\hat a^\dagger)^4\ket{n}=(4n+6)\sqrt{(n+1)(n+2)}, \]

\[ \bra{n-2}(\hat a+\hat a^\dagger)^4\ket{n}=(4n-2)\sqrt{n(n-1)}. \]

Also,

\[ E_n^{(0)}-E_{n\pm 2}^{(0)}=\mp 2\omega,\qquad E_n^{(0)}-E_{n\pm 4}^{(0)}=\mp 4\omega. \]

Using \(\bra{k}V\ket{n}=\lambda\,\omega\,\bra{k}(\hat a+\hat a^\dagger)^4\ket{n}\), we can write the second-order shift as a sum over the four connected levels:

\[ \begin{aligned} E_n^{(2)} &=(\lambda\omega)^2\left[ \frac{\left|\bra{n+2}(\hat a+\hat a^\dagger)^4\ket{n}\right|^2}{-2\omega} +\frac{\left|\bra{n-2}(\hat a+\hat a^\dagger)^4\ket{n}\right|^2}{+2\omega}\right. \\ &\qquad\qquad\left. +\frac{\left|\bra{n+4}(\hat a+\hat a^\dagger)^4\ket{n}\right|^2}{-4\omega} +\frac{\left|\bra{n-4}(\hat a+\hat a^\dagger)^4\ket{n}\right|^2}{+4\omega} \right] \\ &=\lambda^2\omega\left[ -\frac{1}{2}\left|\bra{n+2}(\hat a+\hat a^\dagger)^4\ket{n}\right|^2 +\frac{1}{2}\left|\bra{n-2}(\hat a+\hat a^\dagger)^4\ket{n}\right|^2\right. \\ &\qquad\qquad\left. -\frac{1}{4}\left|\bra{n+4}(\hat a+\hat a^\dagger)^4\ket{n}\right|^2 +\frac{1}{4}\left|\bra{n-4}(\hat a+\hat a^\dagger)^4\ket{n}\right|^2 \right]. \end{aligned} \]

Squaring the matrix elements listed above gives

\[ \left|\bra{n+4}(\hat a+\hat a^\dagger)^4\ket{n}\right|^2=(n+1)(n+2)(n+3)(n+4), \]

\[ \left|\bra{n-4}(\hat a+\hat a^\dagger)^4\ket{n}\right|^2=n(n-1)(n-2)(n-3), \]

\[ \left|\bra{n+2}(\hat a+\hat a^\dagger)^4\ket{n}\right|^2=(4n+6)^2(n+1)(n+2), \]

\[ \left|\bra{n-2}(\hat a+\hat a^\dagger)^4\ket{n}\right|^2=(4n-2)^2\,n(n-1). \]

(For small \(n\), the “downward” terms automatically vanish because of the factors \(n(n-1)\) and \(n(n-1)(n-2)(n-3)\).)

Plugging in, we get

\[ \begin{aligned} E_n^{(2)} =\lambda^2\omega\Bigg[ &-\frac{1}{2}(4n+6)^2(n+1)(n+2) +\frac{1}{2}(4n-2)^2 n(n-1) \\ &-\frac{1}{4}(n+1)(n+2)(n+3)(n+4) +\frac{1}{4}n(n-1)(n-2)(n-3) \Bigg]. \end{aligned} \]

Now expand each polynomial factor:

\[ (4n+6)^2(n+1)(n+2)=16n^4+96n^3+212n^2+204n+72, \]

\[ (4n-2)^2 n(n-1)=16n^4-32n^3+20n^2-4n, \]

\[ (n+1)(n+2)(n+3)(n+4)=n^4+10n^3+35n^2+50n+24, \]

\[ n(n-1)(n-2)(n-3)=n^4-6n^3+11n^2-6n. \]

workout

Substituting these back in and collecting like powers of \(n\), the \(n^4\) terms cancel and we obtain

\[ E_n^{(2)}=\lambda^2\omega\left[-(68n^3+102n^2+118n+42)\right] =\boxed{-2\lambda^2\omega\left(34n^3+51n^2+59n+21\right)}. \]

(c)¶

workout

Yes, \(E_n^{(2)}\) is negative for all \(n\ge 0\) because the polynomial \(34n^3+51n^2+59n+21\) is strictly positive.

It is also useful to understand why the sign comes out negative, since in general second-order shifts can have either sign.

Start from the second-order formula

\[ E_n^{(2)}=\sum_{k\neq n}\frac{|V_{kn}|^2}{E_n^{(0)}-E_k^{(0)}}. \]

For the harmonic oscillator, \(E_k^{(0)}=(k+\tfrac{1}{2})\omega\), so:

If \(k>n\), then \(E_n^{(0)}-E_k^{(0)}<0\), so these terms contribute negatively.
If \(k<n\), then \(E_n^{(0)}-E_k^{(0)}>0\), so these terms contribute positively.

In our case, \(V=\lambda\omega(\hat a+\hat a^\dagger)^4\) only connects \(n\) to \(n\pm 2\) and \(n\pm 4\), so we can write schematically

\[ E_n^{(2)}=\lambda^2\omega\left[ -\frac{1}{2}|M_{n\to n+2}|^2 +\frac{1}{2}|M_{n\to n-2}|^2 -\frac{1}{4}|M_{n\to n+4}|^2 +\frac{1}{4}|M_{n\to n-4}|^2 \right], \]

where \(M_{n\to k}\equiv \bra{k}(\hat a+\hat a^\dagger)^4\ket{n}\).

The reason the negative terms win is that “upward” matrix elements (those involving more raising operators) carry slightly larger ladder factors than the corresponding “downward” ones. Concretely, compare the \(n\to n\pm 2\) pair:

\[ |M_{n\to n+2}|^2=(4n+6)^2(n+1)(n+2),\qquad |M_{n\to n-2}|^2=(4n-2)^2 n(n-1). \]

Both are \(O(n^4)\) at large \(n\), but for any finite \(n\) the “upward” factor is larger because it contains \((n+1)(n+2)\) rather than \(n(n-1)\) and also \((4n+6)^2\) rather than \((4n-2)^2\). In the same spirit, for the \(n\to n\pm 4\) pair,

\[ |M_{n\to n+4}|^2=(n+1)(n+2)(n+3)(n+4),\qquad |M_{n\to n-4}|^2=n(n-1)(n-2)(n-3), \]

and again the upward factor is larger.

We can view these ladder factors as ratios of factorials. For example,

\[ (n+1)(n+2)=\frac{(n+2)!}{n!},\qquad n(n-1)=\frac{n!}{(n-2)!}, \]

and similarly for the \(\pm 4\) terms. These ratios make it clear that, when comparing contributions with the same number of raising vs lowering operations, the raising (upward) transitions carry more “weight” in magnitude. Since the raising transitions are precisely the ones with negative denominators, the overall second-order sum becomes negative.

Finally, note that \(E_n^{(2)}<0\) does not mean the anharmonic potential lowers the energy overall: for \(\lambda>0\), the first-order shift \(E_n^{(1)}=3\lambda\omega(2n^2+2n+1)\) is positive and dominates at small \(\lambda\), while the negative \(E_n^{(2)}\) is a (smaller) correction arising from level mixing.

(d)¶

We want an order-of-magnitude estimate of the largest \(n\) for which perturbation theory is valid.

A common (and more robust than only looking at \(E_n^{(1)}\)) criterion is:

\[ \Delta V \ll \Delta E, \]

where \(\Delta E\) is the unperturbed level spacing and \(\Delta V\) is the uncertainty (RMS fluctuation) of the perturbation in the unperturbed state \(\ket{n}\):

\[ (\Delta V)^2\equiv \bra{n}V^2\ket{n}-\left(\bra{n}V\ket{n}\right)^2. \]

For the harmonic oscillator, the spacing is constant:

\[ \Delta E=E_n^{(0)}-E_{n-1}^{(0)}=\omega. \]

Now compute \(\Delta V\). Since \(V=\lambda\omega(\hat a+\hat a^\dagger)^4\), we have

\[ (\Delta V)^2=(\lambda\omega)^2\left(\bra{n}(\hat a+\hat a^\dagger)^8\ket{n}-\bra{n}(\hat a+\hat a^\dagger)^4\ket{n}^2\right). \]

We already found

\[ \bra{n}(\hat a+\hat a^\dagger)^4\ket{n}=6n^2+6n+3. \]

To evaluate \(\bra{n}(\hat a+\hat a^\dagger)^8\ket{n}\) efficiently, one can use the normal-ordered form of \((\hat a+\hat a^\dagger)^4\) from part (a) and square it. When taking \(\bra{n}(\cdot)\ket{n}\), only terms with an equal number of raising and lowering operators survive (because the final state must return to \(\ket{n}\)). Carrying this out gives

\[ \bra{n}(\hat a+\hat a^\dagger)^8\ket{n}=70n^4+140n^3+350n^2+280n+105. \]

Therefore,

\[ (\Delta V)^2 =(\lambda\omega)^2\left[\left(70n^4+140n^3+350n^2+280n+105\right)-\left(6n^2+6n+3\right)^2\right], \]

so

\[ (\Delta V)^2=(\lambda\omega)^2\left(34n^4+68n^3+278n^2+244n+96\right). \]

The validity criterion \(\Delta V\ll\Delta E=\omega\) becomes

workout

\[ \lambda\sqrt{34n^4+68n^3+278n^2+244n+96}\ll 1. \]

For large \(n\), the leading behavior is \(\Delta V\sim \lambda\omega\sqrt{34}\,n^2\), hence the breakdown scale is

\[ \lambda\sqrt{34}\,n^2\sim 1 \quad\Rightarrow\quad \boxed{n_{\max}\sim \frac{1}{34^{1/4}\sqrt{\lambda}}\approx \frac{1}{2.4\sqrt{\lambda}}}. \]

This reproduces the same scaling \(n_{\max}\propto 1/\sqrt{\lambda}\) as simpler estimates, but with a more concrete \(O(1)\) prefactor coming from the RMS-fluctuation criterion.

For comparison, a simpler (cruder) criterion is to demand that the mean perturbation (i.e. the first-order shift) be small compared to the spacing. From part (a), at large \(n\)

\[ E_n^{(1)}\sim \lambda\,\omega\,n^2\qquad (n\gg 1). \]

Requiring \(|E_n^{(1)}|\ll \omega\) again gives \(n_{\max}\sim 1/\sqrt{\lambda}\); the RMS criterion above just refines this with a more systematic measure of “typical perturbation size”.