Normalize states & calculate a matrix element

Knowing an eigenstate’s energy does not tell us how strongly an operator connects it to another state. That strength depends on normalized vectors and their matrix element. Here a regular five-site Bethe state gives an exact example: four nonzero momentum transfers have equal strength, while the zero-momentum transition vanishes. You will derive the state, calculate the amplitudes, and check their sum against a local spin identity. Everything takes place in a finite periodic chain.

Required background. Construct Bethe vectors algebraically supplies the regular creation-block state and its monodromy convention. You need complex inner products, orthonormal spin configurations and spin-raising operators. Helpful background. The linear-algebra bridge reviews normalization and complete bases; one-magnon waves explains the Fourier eigenstates used below.

A regular five-site state

Use the spin-$1/2$ XXX ring with $N=5$, $J\gt0$, $\hbar=1$, unit lattice spacing and

$$H=\frac J2\sum_{x=1}^{5}(I-P_{x,x+1}),\qquad 6\equiv1.$$

Here $S_x^a=\sigma_x^a/2$, $S_x^\pm=S_x^x\pm iS_x^y$, and $P$ exchanges two spins. In the orthonormal configuration basis,

$$|x\rangle=S_x^-|0\rangle,\qquad |x,y\rangle=S_x^-S_y^-|0\rangle\quad(x\lt y),$$

where $|0\rangle$ is all up. The active right translation sends each site to its successor: $U|x\rangle=|x+1\rangle$, with cyclic labels.

Keep the creation block $B(v)$ of

$$T_a(v)=L_{a5}(v)\cdots L_{a1}(v),\qquad L_{ax}(v)=(v-i/2)I+iP_{ax}.$$

The regular two-root exercise establishes that

$$|\Phi\rangle=B(1/2)B(-1/2)|0\rangle$$

is a nonzero eigenvector. These are creation parameters $\lambda_B$; the coordinate convention has $\lambda_{\rm coord}=-\lambda_B$. The set $\{1/2,-1/2\}$ is unchanged by that sign reversal, so this particular root set cannot diagnose a momentum-convention error by itself.

Entry check: a vector is not yet normalized

Suppose an operator calculation produces $\langle f|O|\Phi\rangle$, where $|f\rangle$ is normalized but the norm of $|\Phi\rangle$ has not been calculated. Can its squared modulus already be used as the normalized transition strength? Would knowing its energy settle the question?

Check and repair

No. Calculate $\|\Phi\|$ and divide the amplitude by it, or divide the squared modulus by $\|\Phi\|^2$. An eigenvalue equation cannot detect this missing factor: rescaling an eigenvector leaves its energy unchanged. Review inner products and normalization if this distinction is unfamiliar.

For any nonzero raw initial and final vectors, the normalized matrix element is

$$M_{fi}=\frac{\langle\Phi_f|O|\Phi_i\rangle} {\|\Phi_f\|\,\|\Phi_i\|}.$$

Replacing normalized states by $e^{i\alpha_i}|\Psi_i\rangle$ and $e^{i\alpha_f}|\Psi_f\rangle$ multiplies $M_{fi}$ by $e^{i(\alpha_i-\alpha_f)}$. Its weight $|M_{fi}|^2$ is unchanged. A matrix element retains phase information; a weight does not.

Calculate the state and its norm

Separate the ten unordered pairs into the five cyclic neighbors and the other five pairs:

$$\begin{aligned} \mathcal A&=\{12,23,34,45,15\},\\ \mathcal D&=\{13,14,24,25,35\}. \end{aligned}$$

The exact vector is

$$|\Phi\rangle=\frac18\left( \sum_{xy\in\mathcal D}|x,y\rangle -\sum_{xy\in\mathcal A}|x,y\rangle\right).$$

In particular, sites $1$ and $5$ are adjacent. Treating them as endpoints of an open chain would change the vector’s interpretation and its Hamiltonian check.

Derive the raw amplitudes from the creation block

The one-block formula is

$$B(w)|0\rangle =i\sum_{x=1}^{5}(w+i/2)^{5-x}(w-i/2)^{x-1}|x\rangle.$$

At $w=-1/2$, the coefficients in site order $1,\ldots,5$ are

$$\frac14(-i,\ 1,\ i,\ -1,\ -i).$$

Multiplying the local blocks for $B(1/2)$ gives these entries:

Input site $x$	Coefficient in $B(-1/2)\lvert0\rangle$	$\langle1,2\rvert B(1/2)\lvert x\rangle$	$\langle1,3\rvert B(1/2)\lvert x\rangle$
$1$	$-i/4$	$-i/4$	$-1/4$
$2$	$1/4$	$-1/4$	$1/2$
$3$	$i/4$	$0$	$-1/4$

The columns for either target vanish on inputs $4,5$. To see the sign of the less immediate $1/2$ entry, put $\alpha=(1+i)/2$. An input down spin at site $2$ becomes the pair $1,3$ through three auxiliary spin exchanges. Their factor is $i^3$, followed by two diagonal factors $\alpha$, giving $i^3\alpha^2=1/2$.

Taking the indicated products gives

$$\begin{aligned} \langle1,2|\Phi\rangle &=\left(-\frac i4\right)^2-\frac1{16} =-\frac18,\\ \langle1,3|\Phi\rangle &=\frac i{16}+\frac18-\frac i{16} =\frac18. \end{aligned}$$

The regular Bethe calculation gives $U|\Phi\rangle=|\Phi\rangle$. The translation orbits of $12$ and $13$ exhaust the ten configurations, so these two components determine the whole vector. The displayed vector also verifies that translation property directly.

The configuration states are orthonormal, hence

$$\langle\Phi|\Phi\rangle=\frac{10}{64}=\frac5{32}, \qquad \|\Phi\|=\frac{\sqrt{10}}8.$$

We choose the normalized state

$$|\Psi\rangle=\frac{|\Phi\rangle}{\|\Phi\|} =\frac1{\sqrt{10}}\left( \sum_{\mathcal D}|x,y\rangle-\sum_{\mathcal A}|x,y\rangle\right).$$

This fixes a global phase by making every adjacent-pair amplitude negative real. Other phase choices describe the same physical state.

There is a short Hamiltonian check independent of rapidity formulas. Write $|A\rangle=\sum_{\mathcal A}|x,y\rangle$ and $|D\rangle=\sum_{\mathcal D}|x,y\rangle$. Each adjacent configuration has two unequal bonds; each nonadjacent configuration has four. Applying the bond exchanges and collecting equal configurations gives

$$H|A\rangle=J(|A\rangle-|D\rangle),\qquad H|D\rangle=J(|D\rangle-|A\rangle).$$

Therefore

$$H|\Psi\rangle=2J|\Psi\rangle,\qquad U|\Psi\rangle=|\Psi\rangle.$$

The state has coordinate momentum $K_i=0$ modulo $2\pi$ and total spin component $S^z_{\rm tot}=5/2-2=1/2$.

Spin and momentum select the final state

For the five allowed momenta, define

$$q_m=\frac{2\pi m}{5},\qquad O_q^+=\frac1{\sqrt5}\sum_{x=1}^{5}e^{-iqx}S_x^+, \qquad m=0,\ldots,4.$$

Our spin operators and their matrix elements are dimensionless with $\hbar=1$. The Fourier factor $1/\sqrt5$ is part of the operator definition; changing it changes every weight.

Since $[S^z_{\rm tot},O_q^+]=O_q^+$, this operator turns the two-down-spin state into the one-down-spin sector. A normalized basis of that final sector is

$$|k\rangle=\frac1{\sqrt5}\sum_{y=1}^{5}e^{iky}|y\rangle, \qquad k=\frac{2\pi\ell}{5},\quad \ell=0,\ldots,4.$$

The positive exponent in the state and the negative exponent in the operator are deliberate. Active translation gives

$$U|k\rangle=e^{-ik}|k\rangle, \qquad US_x^+U^{-1}=S_{x+1}^+.$$

Relabeling the finite sum, with $e^{5iq}=1$ at the seam, yields

$$UO_q^+U^{-1}=e^{iq}O_q^+.$$

Thus an initial state with $U$ eigenvalue $e^{-iK_i}$ can have a nonzero matrix element only to a state with

$$K_f=K_i-q\pmod{2\pi}.$$

Here $K_i=0$, so the only possible final Fourier state is $|-q\rangle$. This is a necessary selection rule. It does not guarantee a nonzero amplitude.

Calculate the amplitude

After a raising operator removes one down spin, the remaining down spin can be at any site $y$. The removed spin was at either $y\pm1$ or $y\pm2$. The former pairs have negative amplitudes; the latter have positive amplitudes. Consequently,

$$\begin{aligned} \langle y|O_q^+|\Psi\rangle &=\frac1{\sqrt{50}}\bigl[ -e^{-iq(y-1)}-e^{-iq(y+1)}\\ &\hspace{4.8em}+e^{-iq(y-2)}+e^{-iq(y+2)}\bigr]\\ &=\frac{2[\cos(2q)-\cos q]}{\sqrt{50}}e^{-iqy}. \end{aligned}$$

Cyclic site labels are legitimate because $q$ is an allowed ring momentum. Comparing with the normalized Fourier vector gives the entire output:

$$O_q^+|\Psi\rangle=M_q|-q\rangle,\qquad M_q=\frac{2[\cos(2q)-\cos q]}{\sqrt{10}}.$$

The fixed state phases make these amplitudes real:

Momentum label $m$	$M_{q_m}$	Weight $\lvert M_{q_m}\rvert^2$
$0$	$0$	$0$
$1,4$	$-1/\sqrt2$	$1/2$
$2,3$	$1/\sqrt2$	$1/2$

For example, $\cos(2\pi/5)=(\sqrt5-1)/4$ and $\cos(4\pi/5)=-(\sqrt5+1)/4$ give the $m=1$ row. At $q=0$, the two negative and two positive contributions cancel at every remaining site. The momentum-zero final state exists and satisfies the selection rule, yet its transition amplitude is zero.

The one-magnon energies are $E_k=J(1-\cos k)$. They identify possible energy differences, but do not supply the table’s weights or signs. In particular, replacing $q$ by $-q$ preserves energy and weight while reversing the selected momentum. The independent exercise makes that convention check explicit.

These weights are transition strengths, not a probability distribution over the five different operators $O_q^+$. There is no requirement that their sum be one. For a fixed nonzero output, normalizing $O_q^+|\Psi\rangle$ produces a final state; at $q=0$ the output is zero and cannot be normalized.

Check completeness and the total strength

The discrete Fourier identity

$$\frac15\sum_{m=0}^{4}e^{iq_m(x-y)}=\delta_{xy} \qquad (x,y=1,\ldots,5)$$

proves that the five Fourier vectors form a complete orthonormal basis of the one-magnon sector. If $\Pi_1$ projects onto that sector, then

$$\sum_k|k\rangle\langle k|=\Pi_1, \qquad \sum_k|\langle k|O_q^+|\Psi\rangle|^2 =\|O_q^+|\Psi\rangle\|^2.$$

The last equality holds because the output belongs to that sector. It requires no claim about completeness of arbitrary Bethe roots in other sectors.

There is also a sum over operators. Using the same Fourier identity,

$$\begin{aligned} \sum_q(O_q^+)^\dagger O_q^+ &=\sum_x S_x^-S_x^+\\ &=\sum_x\left(\frac12-S_x^z\right). \end{aligned}$$

The final expression counts down spins. Since the initial state has exactly two,

$$\sum_q\|O_q^+|\Psi\rangle\|^2 =0+4\left(\frac12\right)=2.$$

For a fixed site $y$, translation invariance gives the related local check

$$\|S_y^+|\Psi\rangle\|^2 =\left\langle\frac12-S_y^z\right\rangle =\frac25.$$

Directly, precisely four of the ten pairs contain $y$; their surviving amplitudes each have squared modulus $1/10$.

Normalized matrix elements and a complete intermediate-state sum are the ingredients of the spectral framework in Caux 2009, § II, printed pp. 3–4, equations (1)–(3), PDF. That source defines a ground-state correlation and uses an unnormalized spatial Fourier operator. Here the initial state is an excited eigenstate and the operator includes $1/\sqrt5$; our finite spin calculation establishes the corresponding weights directly. The next lesson, Build a spectral correlation sum, uses them to determine time dependence and frequency signs.

Exercises

Guided practice: change a raw vector’s scale

Replace $|\Phi\rangle$ by $|\Phi'\rangle=(3+4i)|\Phi\rangle$, keeping final states and operators fixed. Find its norm, normalized state and matrix elements. Then calculate the total strength one would incorrectly obtain by using the original unnormalized $|\Phi\rangle$ in place of $|\Psi\rangle$.

Hint

A complex scalar changes a norm by its modulus. Its remaining phase changes amplitudes, but cancels from squared moduli.

Solution

Since $|3+4i|=5$,

$$\|\Phi'\|=\frac{5\sqrt{10}}8,\qquad |\Psi'\rangle=\frac{3+4i}{5}|\Psi\rangle.$$

Every amplitude becomes $M_q'=(3+4i)M_q/5$, while every weight stays the same. Using the original raw state instead multiplies each correct weight by $\|\Phi\|^2=5/32$. The resulting incorrect total is $(5/32)\times2=5/16$. An energy eigenvalue check would still pass.

Independent practice: reverse the Fourier sign

Define $\overline O_q^+=5^{-1/2}\sum_x e^{+iqx}S_x^+$. Determine its translation phase and selected final momentum. At $q=2\pi/5$, compare the actual output with an erroneous prediction that still selects $|-q\rangle$. Can the energy and output norm distinguish them?

Hint

Relabel the operator sum under $U$. The cosine coefficient is even in $q$, but $|q\rangle$ and $|-q\rangle$ are distinct orthogonal Fourier states for this momentum.

Solution

Now $U\overline O_q^+U^{-1}=e^{-iq}\overline O_q^+$. Therefore

$$\overline O_q^+|\Psi\rangle=M_q|q\rangle, \qquad \langle-q|\overline O_q^+|\Psi\rangle=0 \quad(q=2\pi/5).$$

The actual coefficient of $|q\rangle$ is $-1/\sqrt2$. Predicting the same coefficient of $|-q\rangle$ would give the same norm and energy because $E_q=E_{-q}$. The active translation phases are instead $e^{-iq}$ and $e^{+iq}$, which differ. Testing translation, or projecting onto the full Fourier basis, detects the error that energy alone misses.

Transfer: change the operator’s spin selection

Replace $O_q^+$ by the uniform longitudinal operator

$$Z_0=\frac1{\sqrt5}\sum_x S_x^z =\frac{S^z_{\rm tot}}{\sqrt5}.$$

Find $Z_0|\Psi\rangle$ and its squared norm. If every one-magnon matrix element vanishes, does that imply the output is zero? Identify exactly which completeness step must change.

Hint

The initial state’s total spin component is $1/2$. A longitudinal operator preserves the number of down spins.

Solution

The output is

$$Z_0|\Psi\rangle=\frac1{2\sqrt5}|\Psi\rangle, \qquad \|Z_0|\Psi\rangle\|^2=\frac1{20}.$$

It remains in the two-magnon sector. Every one-magnon overlap vanishes by orthogonality of sectors, so summing those overlaps computes $\|\Pi_1 Z_0|\Psi\rangle\|^2=0$, not the full output norm. A resolution of the identity in the correct sector recovers $1/20$. In this special example the output is already parallel to $|\Psi\rangle$, so that single projection exhausts its weight; it does not make one vector a complete basis of the ten-dimensional sector.

References

• Caux, Jean-Sébastien. “Correlation functions of integrable models: a description of the ABACUS algorithm.” Journal of Mathematical Physics 50, 095214 (2009). DOI: 10.1063/1.3216474. Author version arXiv:0908.1660v1, 12 August 2009; open PDF. Section II, printed pp. 3–4, equations (1)–(3), supports the matrix-element and spectral-sum method. The explicit five-site excited-state amplitudes are derived here.