2020 Mathematics Subject Classification: 11B65, 05A10

# Solvability of $\binom{2k}{k}=\binom{2a}{a}\binom{x+2b}{b}$

Meaghan Allen  Meaghan Allen
Department of Mathematics and Statistics
University of New Hampshire
Durham, NH 03824, US; Email<EMAIL_ADDRESS>
###### Abstract.

Suppose $k,x,$ and $b$ are positive integers, and $a$ is a nonnegative integer
such that $k=a+b$. In this paper, we will prove
$\binom{2k}{k}=\binom{2a}{a}\binom{x+2b}{b}$ if and only if $x=a=1$. We do
this by looking at different cases depending on the values of $x$ and $k$. We
use varying techniques to prove the cases, such as direct proof, verification
through Maple software, and a proof technique found in Moser’s paper. Previous
results from Hanson, Stănică, Shanta, Shorey and Nair are also used.

###### Key words and phrases:

[

###### Key words and phrases:

binomial coefficient, prime number

##  1\. Introduction

It was discovered by Moser in [4] that the equation

$\binom{2n}{n}=\binom{2a}{a}\binom{2b}{b}$

has no solutions. This result was further extended by P. Erdos in [1], where
he proved that

$\binom{2m}{m}\nmid\binom{2n}{n}$

for $2m>n$. Following the line investigated by Moser in [4] and Erdos in [1],
the purpose of this paper is to prove the following result:

######  Theorem 1.1.

Let $k,x,b$ be positive integers and $a$ be a nonnegative integer such that
$k=a+b$. Then

$\binom{2k}{k}=\binom{2a}{a}\binom{x+2b}{b},\qquad\text{ if and only if
}x=a=1.$

To prove this result, we need to overcome the difficulty that integers $a$ and
$b$ are no longer symmetric in the equation
$\binom{2k}{k}=\binom{2a}{a}\binom{x+2b}{b}$, unlike the case discussed in [4]
and [1]. The key tool used in our proof is an analysis of the existence of
prime numbers in the product of consecutive integers, which was extensively
investigated in [3] and [5].

##  2\. Proof of Main Result

To prove our main result, we will break the proof up into different cases. We
will state these cases as propositions, and prove them throughout the paper by
proving smaller lemmas. First, we will prove the result when $x\geq a$.

######  Proposition 2.1.

Let $k,x,b$ be positive integers and $a$ be a nonnegative integer such that
$k=a+b$. Assume that $x\geq a$. Then

$\binom{2k}{k}\leq\binom{2a}{a}\binom{x+2b}{b}.$

Moreover, the equality holds if and only if $x=a=1$.

###### Proof.

Note that

$\displaystyle\frac{\binom{2a}{a}\binom{a+2b}{b}}{\binom{2k}{k}}$
$\displaystyle=\frac{(2a)!}{a!\cdot
a!}\cdot\frac{(a+2b)!}{b!(a+b)!}\cdot\frac{k!\cdot k!}{(2k)!}$
$\displaystyle=\frac{(2a)(2a-1)\cdots(a+1)}{a(a-1)\cdots
1}\cdot\frac{k(k-1)\cdots(b+1)}{(2k)(2k-1)\cdots(a+2b+1)}$
$\displaystyle=2\cdot\frac{2a-1}{a-1}\frac{2a-2}{a-2}\cdots\frac{a+1}{1}\cdot\frac{1}{2}\cdot\frac{k-1}{2k-1}\frac{k-2}{2k-2}\cdots\frac{k-a+1}{2k-a+1}$
$\displaystyle=\Big{(}\frac{2a-1}{a-1}\frac{k-1}{2k-1}\Big{)}\Big{(}\frac{2a-2}{a-2}\frac{k-2}{2k-2}\Big{)}\cdots\Big{(}\frac{a+1}{1}\frac{k-a+1}{2k-a+1}\Big{)}$
$\displaystyle\left\\{\begin{aligned} &>1\quad\text{ for }1<a<k\\\
&=1\quad\text{ for }a=0,1\end{aligned}\right.$

Now, if $x>a$, then

$\binom{2a}{a}\binom{x+2b}{b}>\binom{2a}{a}\binom{a+2b}{b}\geq\binom{2k}{k}.$

Furthermore, if $x=a>1$, then

$\binom{2a}{a}\binom{x+2b}{b}=\binom{2a}{a}\binom{a+2b}{b}>\binom{2k}{k}.$

When $x=a=1$, we have

$\binom{2a}{a}\binom{x+2b}{b}=\binom{2a}{a}\binom{a+2b}{b}=\binom{2}{1}\binom{1+2b}{b}=\binom{2k}{k}.$

∎

Next, we will prove the result for $1\leq k\leq 10$.

######  Lemma 2.2.

Let $k,x,a,b$ be positive integers such that $k=a+b$. Assume that $1\leq k\leq
10$. Then

$\binom{2k}{k}=\binom{2a}{a}\binom{x+2b}{b}.$

if and only if $x=a=1$.

###### Proof.

By Proposition 2.1, we might assume that $x\leq a$. All cases for $1\leq k\leq
10$ and $x\leq a$ are verified by a direct computation with Maple. ∎

In the following three lemmas, we will prove the result for $a-x=1,2,3$ when
$4<2a\leq k$. We will fully prove Lemma 2.3, and omit the proofs of Lemma 2.4
and Lemma 2.5 since they are similar.

######  Lemma 2.3.

Let $k,x,a,b$ be positive integers such that $k=a+b$. Then

$\binom{2k}{k}\neq\binom{2a}{a}\binom{a+2b-1}{b},\qquad\text{ for }\ 4<2a\leq
k.$

###### Proof.

Let $k,x,a,b$ be positive integers such that $k=a+b$. Assume that

$\binom{2k}{k}=\binom{2a}{a}\binom{a+2b-1}{b}\ \ \text{ for }4<2a\leq k.$
(2.1)

So,

$\frac{(2k)!}{(k!)^{2}}=\frac{(2a)!}{(a!)^{2}}\cdot\frac{(2k-a-1)!}{(k-a)!(k-1)!}.$

Thus,

$\frac{2k(2k-1)(2k-2)\cdots(2k-a)}{k\cdot
k\cdot(k-1)\cdots(k-a+1)}=\frac{(2a)(2a-1)\cdots(a+2)(a+1)}{a(a-1)\cdots
2\cdot 1}.$

We have, for $k>a$,

$\frac{2k-1}{k}<2,\ \frac{2k-2}{k-1}<\frac{2a-1}{a-1},\
\frac{2k-3}{k-2}<\frac{2a-2}{a-2},\cdots,\frac{2k-a+1}{k-a+2}<\frac{a+2}{2}$

and

$2\cdot\frac{2k-a}{k-a+1}=2\Big{(}2+\frac{a-2}{k-a+1}\Big{)}<6\leq
a+1\qquad\text{ for }a\geq 5.$

Thus, if $a\geq 5$, then

$\binom{2k}{k}<\binom{2a}{a}\binom{a+2b-1}{b}\qquad\text{ for }k=a+b.$

When $a=4$, the assumption (2.1) becomes

$\binom{2k}{k}=\binom{8}{4}\binom{2k-5}{k-4},\qquad\text{ for }k\geq 8,$

or

$\frac{(2k)(2k-1)(2k-2)(2k-3)(2k-4)}{k\cdot k(k-1)(k-2)(k-3)}=\binom{8}{4}.$

A simple computation shows that

$8(2k-1)(2k-3)=70k(k-3),$

which has a unique integer solution $k=4$. Thus

$\binom{2k}{k}\neq\binom{8}{4}\binom{2k-5}{k-4},\qquad\text{ for }k\geq 8.$

When $a=3$, the assumption (2.1) becomes

$\binom{2k}{k}=\binom{6}{3}\binom{2k-4}{k-3},\qquad\text{ for }k\geq 6.$

Equivalently, we have,

$\frac{(2k)(2k-1)(2k-2)(2k-3)}{k\cdot k(k-1)(k-2)}=\binom{6}{3}.$

It induces that

$4(2k-1)(2k-3)=20k(k-2),$

where $k=-1,3$ are integer solutions. Thus,

$\binom{2k}{k}\neq\binom{6}{3}\binom{2k-4}{k-3},\qquad\text{ for }k\geq 6.$

The proof is complete. ∎

######  Lemma 2.4.

Let $k,x,a,b$ be positive integers such that $k=a+b$. Then

$\binom{2k}{k}\neq\binom{2a}{a}\binom{a+2b-2}{b},\qquad\text{ for }4<2a\leq
k.$

######  Lemma 2.5.

Let $k,x,a,b$ be positive integers such that $k=a+b$. Then

$\binom{2k}{k}\neq\binom{2a}{a}\binom{a+2b-3}{b},\qquad\text{ for }\ 4<2a\leq
k.$

Next, we will prove the main result in the case that $a\geq\frac{1}{2}k$ and
$b\geq 3$. To do so, we will need the following results from Hanson [2] and
Stănică [6].

######  Lemma 2.6 (Hanson [2]).

The product of $m$ consecutive integers $n(n+1)\cdots(n+m-1)$ greater than $m$
contains a prime divisor greater that $\frac{3}{2}m$ with the exceptions
$3\cdot 4$, $8\cdot 9$, and $6\cdot 7\cdot 8\cdot 9\cdot 10$.

######  Lemma 2.7 (Stănică [6]).

Let $m,n,r$ be positive integers, with $m>r\geq 1$ and $n\geq 1$. Then

$\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{8n}}n^{-\frac{1}{2}}\frac{m^{mn+\frac{1}{2}}}{(m-r)^{(m-r)n+\frac{1}{2}}r^{rn+\frac{1}{2}}}<\binom{mn}{rn}<\frac{1}{\sqrt{2\pi}}n^{-\frac{1}{2}}\frac{m^{mn+\frac{1}{2}}}{(m-r)^{(m-r)n+\frac{1}{2}}r^{rn+\frac{1}{2}}}$

Using the previous two results, we can prove the following Lemma.

######  Lemma 2.8.

Let $k,x,a,b$ be positive integers such that $k=a+b$. If $a\geq\frac{1}{2}k$,
$b\geq 3$, then

$\binom{2k}{k}\neq\binom{2a}{a}\binom{x+2b}{b}.$

###### Proof.

We follows the strategy used in [4]. Suppose that $a\geq\frac{1}{2}k$. Since
$b\geq 3$, we know $k\geq 6$. Assume
$\binom{2k}{k}=\binom{2a}{a}\binom{x+2b}{b}.$ Then

$\frac{\binom{2k}{k}}{\binom{2a}{a}}=\frac{(2k)(2k-1)\cdots(2a+1)}{(k(k-1)\cdots(a+1))^{2}}=\binom{x+2b}{b}$

is an integer. By Lemma 2.6, there exists a prime divisor $p$ of the product
$(2k)(2k-1)\cdots(2a+1)$ such that $p>\frac{3}{2}(2k-2a)=3b>2.$ We claim that
$p$ is not a divisor of $k(k-1)\cdots(a+1)$. In fact assume that $p$ divides
$k-i$ for some $0\leq i\leq b-1$ and $\alpha$ is the largest positive integer
such that $p^{\alpha}$ divides $k-i$. Then $\alpha$ is also the largest
positive integer such that $p^{\alpha}$ divides $2(k-i)$, and $p$ is not a
divisor of other terms $2k-j$, with $j\neq 2i$, in the numerator. In other
words, $\alpha$ is the largest positive integer such that $p^{\alpha}$ divides
$(2k)(2k-1)\cdots(2a+1)$. However, $p^{2\alpha}$ divides
$(k(k-1)\cdots(a+1))^{2}$. This contradicts with the assumption that
$\frac{\binom{2k}{k}}{\binom{2a}{a}}$ is an integer. Hence $p$ doesn’t divide
the denominator $k(k-1)\cdots(a+1)$.

So, we have found that $p$ divides $\binom{x+2b}{b}$. Since $p$ is prime, we
have $x+2b\geq p>3b$. By Lemma 2.7, we obtain, for $b\geq 3$,

$\binom{x+2b}{b}>\binom{3b}{b}\geq\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{8b}}b^{-\frac{1}{2}}\frac{3^{3b+\frac{1}{2}}}{2^{2b+\frac{1}{2}}}=\left(\sqrt{\frac{3}{4\pi}}\cdot\frac{1}{\sqrt{b}e^{\frac{1}{8b}}}\cdot\Big{(}\frac{27}{16}\Big{)}^{b}\right)\cdot
4^{b}>4^{b}.$

However,

$\frac{\binom{2k}{k}}{\binom{2a}{a}}=\frac{(2k)(2k-1)\cdots(2a+1)}{(k(k-1)\cdots(a+1))^{2}}=2^{b}\cdot\frac{(2k-1)(2k-3)\cdots(2a+1)}{k(k-1)\cdots(a+1)}<4^{b}.$

This is a contradiction. Thus, for $b\geq 3$,

$\binom{2k}{k}\neq\binom{2a}{a}\binom{x+2b}{b}.$

∎

Using the previous lemma, we can now prove our main result when
$a\geq\frac{1}{2}k$ and $x\geq 2$.

######  Lemma 2.9.

Let $k,x,a,b$ be positive integers such that $k=a+b$. If $a\geq\frac{1}{2}k$
and $x\geq 2$, then

$\binom{2k}{k}\neq\binom{2a}{a}\binom{x+2b}{b}.$

###### Proof.

By Lemma 2.8, we might assume that $1\leq b<3$. When $b=1$, for $x\geq 2$,

$\displaystyle\binom{2(k-1)}{k-1}\binom{x+2}{1}$
$\displaystyle=\frac{(2(k-1))!}{((k-1)!)^{2}}(x+2)$
$\displaystyle=\binom{2k}{k}(x+2)\cdot\frac{k^{2}}{2k(2k-1)}$
$\displaystyle>\binom{2k}{k}.$

When $b=2$, we have

$\frac{\binom{2(k-2)}{k-2}\binom{x+4}{2}}{\binom{2k}{k}}=\frac{(x+4)(x+3)}{2}\frac{k^{2}(k-1)^{2}}{(2k)(2k-1)(2k-2)(2k-3)}.$

If $x\geq 3$,

$\displaystyle\frac{(x+4)(x+3)}{2}\cdot\frac{k^{2}(k-1)^{2}}{2k(2k-1)(2k-2)(2k-3)}$
$\displaystyle=\frac{(x+4)(x+3)}{8}\cdot\frac{k(k-1)}{(2k-1)(2k-3)}$
$\displaystyle>\frac{(x+4)(x+3)}{32}$ $\displaystyle>1.$

If $x=2$,

$\frac{(2+4)(2+3)}{2}\cdot\frac{k^{2}(k-1)^{2}}{2k(2k-1)(2k-2)(2k-3)}=1$

has no integer solution. Thus, when $b=2$, for all $x\geq 2$,

$\binom{2(k-2)}{k-2}\binom{x+4}{2}\neq\binom{2k}{k}.$

∎

We will now prove our main result when $x=1$.

######  Proposition 2.10.

Let $k,a,b$ be positive integers such that $k=a+b$. Then

$\binom{2k}{k}=\binom{2a}{a}\binom{1+2b}{b}\quad\text{ if and only if }a=1.$

###### Proof.

First, assume $a=1$. Then,

$\displaystyle\frac{\binom{2k}{k}}{\binom{1+2b}{b}}$
$\displaystyle=\frac{(2k)!\cdot b!\cdot(b+1)!}{(1+2b)!\cdot k!\cdot k!}$
$\displaystyle=\frac{(2+2b)(1+2b)\dots(b+2)}{(1+2b)(2b)\dots(b+2)}\cdot\frac{b(b-1)\dots
1}{(b+1)b(b-1)\dots 1}$ $\displaystyle=\frac{2(1+b)}{1+b}$ $\displaystyle=2$
$\displaystyle=\binom{2a}{a}$

Next, assume $\binom{2k}{k}=\binom{2a}{a}\binom{1+2b}{b}$. By Lemma 2.2, we
might assume that $k>10$. Lemma 2.8 implies that, if $a\geq\frac{1}{2}k$ and
$b\geq 3$, then $\binom{2k}{k}\neq\binom{2a}{a}\binom{1+2b}{b}$. Moreover,
when $b=1,2$, the equation

$\binom{2k}{k}=\binom{2a}{a}\binom{1+2b}{b}$

has no integer solution for $k>10$. Therefore, we might assume that
$a<\frac{1}{2}k$, so $b>\frac{1}{2}k$. For the purpose of the contradiction,
we further assume that $a>1$.

Then

$\displaystyle\frac{\binom{2k}{k}}{\binom{1+2b}{b}}$
$\displaystyle=\frac{(2k)(2k-1)\cdots(2b+2)}{(k(k-1)\cdots(b+1))(k(k-1)\cdots(b+2))}$
$\displaystyle=\frac{(2k)(2k-1)\cdots(2b+3)\cdot 2}{(k(k-1)\cdots(b+2))^{2}}$
$\displaystyle=\binom{2a}{a}$

is an integer. By Lemma 2.6, there exists a prime divisor $p$ of the product
$(2k)(2k-1)\cdots(2b+3)$ such that $p>\frac{3}{2}(2k-2b-2)=3a-3>2.$ We claim
that $p$ is not a divisor of $k(k-1)\cdots(b+2)$. In fact assume that $p$
divides $k-i$ for some $0\leq i\leq a-1$ and $\alpha$ is the largest positive
integer such that $p^{\alpha}$ divides $k-i$. Then $\alpha$ is also the
largest positive integer such that $p^{\alpha}$ divides $2(k-i)$, and $p$ is
not a divisor of other terms $2k-j$, with $j\neq 2i$, in the numerator. In
other words, $\alpha$ is the largest positive integer such that $p^{\alpha}$
divides $(2k)(2k-1)\cdots(2b+3)\cdot 2$. However, $p^{2\alpha}$ divides
$(k(k-1)\cdots(b+2))^{2}$. This contradicts with the assumption that
$\frac{\binom{2k}{k}}{\binom{1+2b}{b}}$ is an integer. Hence $p$ doesn’t
divide the denominator $k(k-1)\cdots(b+2)$.

So, we must have $p$ divides $\binom{2a}{a}$, whence $2a\geq p>3a-3$. So
$a<3$. From the assumption that $a>1$, we have $a=2$. Now that equation
$\binom{2k}{k}=\binom{2a}{a}\binom{1+2b}{b}$ becomes

$\binom{2k}{k}=\binom{4}{2}\binom{2k-3}{k-2},\qquad\text{ when }a=2.$

Equivalently, we get,

$\frac{(2k)(2k-1)(2k-2)}{k\cdot k(k-1)}=6,$

which has no integer solution when $k>10$, a contradiction.

Therefore, $a\leq 1$. Since $a$ is a positive integer, it follows that $a=1$.
This completes the proof of the result. ∎

For the next Lemma we will prove, we need the following results from [3] and
[5].

######  Lemma 2.11 (Shanta & Shorey [3]).

The product of $m$ consecutive integers $n(n+1)\cdots(n+m-1)$ contains a prime
divisor greater than $1.8m$ if $n>m>2$ and $n+m\geq 150$.

######  Lemma 2.12 (Nair & Shorey [5]).

The product of $m$ consecutive integers $n(n+1)\cdots(n+m-1)$ contains a prime
divisor greater than $4.42m$ if $n>4m$, $m>3$ and $n+m\geq 150$.

######  Lemma 2.13.

Let $k,x,a,b$ be positive integers such that $k=a+b\geq 75$ and $x\geq 2$. If
$a\leq 0.9k$ and $1\leq b\leq 0.8k$, then

$\binom{2k}{k}\neq\binom{2a}{a}\binom{x+2b}{b}.$

###### Proof.

Assume that $a\leq 0.9k$ and $1\leq b\leq 0.8k$. Assume, by means of
contradiction, that $\binom{2k}{k}=\binom{2a}{a}\binom{x+2b}{b}.$ By Lemma
2.11, there exists a prime divisor of the product $(2k)(2k-1)\cdots(k+1)$ such
that $p>1.8k$. Obviously, $p$ is also a divisor of $\binom{2k}{k}$, thus a
divisor of $\binom{2a}{a}\binom{x+2b}{b}$. It follows that $2a>1.8k$ or
$x+2b>1.8k$. By Proposition 2.1, we might assume that $x<a$. Thus, if
$x+2b>1.8k$, then $b=a+2b-(a+b)>x+2b-k>0.8k$. Therefore, we conclude that
either $a>0.9k$ or $b>0.8k$, a contradiction to $a\leq 0.9k$ and $1\leq b\leq
0.8k$. Thus,

$\binom{2k}{k}\neq\binom{2a}{a}\binom{x+2b}{b}.$

∎

######  Lemma 2.14.

Let $k,x,a,b$ be positive integers such that $x\geq 2$ and $k=a+b\geq 150$. If
$a\geq\frac{171}{121}x$, then

$\binom{2k}{k}\neq\binom{2a}{a}\binom{x+2b}{b}.$

###### Proof.

Assume that $x\geq 2$, $a\geq\frac{171}{121}x$ and $k\geq 150$. Assume, by
means of contradiction, that $\binom{2k}{k}=\binom{2a}{a}\binom{x+2b}{b}$ for
some $b\geq 1.$ By Lemma 2.9, we can assume that $a<\frac{1}{2}k$. By Lemma
2.13, we can assume that $b>0.8k$, thus $a=k-b<0.2k$. In particular, $b>4a$.
By Lemma 2.3, Lemma 2.4, and Lemma 2.5 we can assume that $a-x>3$. So,

$\binom{2a}{a}=\frac{\binom{2k}{k}}{\binom{x+2b}{b}}=\frac{(2k)(2k-1)\cdots(x+2b+1)}{(k\cdot(k-1)\cdots(b+1))(k\cdot(k-1)\cdots(x+b+1))}.$

Then

$Q=\frac{(2k)(2k-1)\cdots(x+2b+1)}{(k\cdot(k-1)\cdots(x+b+1))^{2}}=\binom{2a}{a}\cdot(b+1)\cdots(b+x)$

is an integer. Let $n=b+x+1$ and $m=a-x$. Since $k-(x+b)=m>3$, we have,
$n>b>4a>4m$, and $n+m=k+1\geq 150$. So, by Lemma 2.12, there exists a prime
divisor $p$ of $k\cdot(k-1)\cdots(x+b+1)$ such that $p>4.42(a-x)$. Thus, for
some $0\leq i<a-x$, $p$ divides $k-i$ in the denominator
$k\cdot(k-1)\cdots(x+b+1)$ of $Q$. Let $\alpha$ be the largest positive
integer such that $p^{\alpha}$ divides $k-i$. Then $\alpha$ is also the
largest positive integer such that $p^{\alpha}$ divides $2(k-i)$, and no other
term $(2k-t)$, for $t\neq 2i$, in the numerator $(2k)(2k-1)\cdots(x+2b+1)$ of
$Q$ is divisible by $p$ as $a\geq\frac{171}{121}x$ and
$p>4.42(a-x)\geq(2a-x)$. However $p^{2\alpha}$ is a divisor of the denominator
$(k\cdot(k-1)\cdots(x+b+1))^{2}$ of $Q$. That contradicts with the fact that
$Q$ is an integer. Therefore, if $x\geq 2$, $a\geq\frac{171}{121}x$ and $k\geq
150$, then

$\binom{2k}{k}\neq\binom{2a}{a}\binom{x+2b}{b}.$

∎

Using the previous lemmas, we can prove our main result when $x\geq 2$ and
$k\geq 150$.

######  Proposition 2.15.

Let $k,x,a,b$ be positive integers such $x\geq 2$ and $k=a+b\geq 150$. Then

$\binom{2k}{k}\neq\binom{2a}{a}\binom{x+2b}{b}.$

###### Proof.

Assume, by means of contradiction, that
$\binom{2k}{k}=\binom{2a}{a}\binom{x+2b}{b}$ for some $b\geq 1.$ By
Proposition 2.1 and Lemma 2.14, we might assume that
$x<a<\frac{171}{121}x<\frac{3}{2}x$. By Lemma 2.13, we assume that $b>0.8k$,
which gives us $a<0.2k$. Note that

$\binom{x+2b}{b}=\frac{(x+2b)!}{b!\cdot(x+b)!}=\frac{(2b)!}{b!\cdot
b!}\cdot\frac{(2b+1)(2b+2)\cdots(2b+x)}{(b+1)(b+2)\cdots(b+x)}.$

By Lemma 2.7 and the assumption that $a<\frac{3}{2}x$,

$\displaystyle\frac{\binom{2k}{k}}{\binom{2a}{a}\binom{2b}{b}}$
$\displaystyle\leq\frac{\frac{1}{\sqrt{2\pi}}k^{-\frac{1}{2}}2^{2k+\frac{1}{2}}}{\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{8a}}a^{-\frac{1}{2}}2^{2a+\frac{1}{2}}\cdot\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{8b}}b^{-\frac{1}{2}}2^{2b+\frac{1}{2}}}$
$\displaystyle=e^{\frac{1}{8a}+\frac{1}{8b}}\sqrt{\frac{\pi ab}{k}}$
$\displaystyle<e^{1/4}\sqrt{\frac{3\pi x}{2}}.$

On the other hand,

$\displaystyle\frac{(2b+1)(2b+2)\cdots(2b+x)}{(b+1)(b+2)\cdots(b+x)}$
$\displaystyle=\Big{(}2-\frac{1}{b+1}\Big{)}\Big{(}2-\frac{2}{b+2}\Big{)}\cdots\Big{(}2-\frac{x}{b+x}\Big{)}$
$\displaystyle>\Big{(}2-\frac{x}{b+x}\Big{)}^{x}>\Big{(}2-\frac{a}{b+a}\Big{)}^{x}\geq\Big{(}2-\frac{1}{5}\Big{)}^{x}.$

Note that $\big{(}\frac{9}{5}\big{)}^{x}-e^{1/4}\sqrt{3\pi x/2}$ is an
increasing function for $x\geq 2$ and
$\big{(}\frac{9}{5}\big{)}^{3}>e^{1/4}\sqrt{9\pi/2}$. So, for $x\geq 3$ and
$k\geq 150$,

$\frac{(2b+1)(2b+2)\cdots(2b+x)}{(b+1)(b+2)\cdots(b+x)}>\Big{(}\frac{9}{5}\Big{)}^{x}>e^{1/4}\sqrt{\frac{3\pi
x}{2}}>\frac{\binom{2k}{k}}{\binom{2a}{a}\binom{2b}{b}}$

which gives us

$\binom{x+2b}{b}=\binom{2b}{b}\cdot\frac{(2b+1)(2b+2)\cdots(2b+x)}{(b+1)(b+2)\cdots(b+x)}>\frac{\binom{2k}{k}}{\binom{2a}{a}}$

and therefore,

$\binom{2a}{a}\binom{x+2b}{b}>\binom{2k}{k}$

which is a contradiction to our assumption. Also, we know that
$x<a<\frac{3}{2}x$, so when $x=2$, we must have $2<a<3$, a contradiction to
$a$ an integer. Therefore, for $x\geq 2$ and $k\geq 150$,

$\binom{2k}{k}\neq\binom{2a}{a}\binom{x+2b}{b}.$

∎

Finally, we will prove our main result when $x\geq 2$ and $x<k<150$.

######  Proposition 2.16.

Let $k,x,a,b$ be positive integers such $x\geq 2$, $k=a+b$, and $x<k<150$.
Then

$\binom{2k}{k}\neq\binom{2a}{a}\binom{x+2b}{b}.$

###### Proof.

Using Maple, we verified that, for $x\geq 2$ and $x<k<150$,

$\binom{2k}{k}\neq\binom{2a}{a}\binom{x+2k}{k}.$

∎

Now, we can prove the following, main theorem of the paper.

######  Theorem 1.1.

Let $k,x,b$ be positive integers and $a$ be a nonnegative integer such that
$k=a+b$. Then

$\binom{2k}{k}=\binom{2a}{a}\binom{x+2b}{b},\qquad\text{ if and only if
}x=a=1.$

###### Proof.

When $x=1$, we obtain the result from Proposition 2.10.

When $x\geq 2$, and $k\geq 150$, we obtain the result from Proposition 2.15.

When $x\geq 2$, and $x<k\leq 150$, we obtain the result from Proposition 2.16.

When $x\geq 2$, and $x\geq k$, we obtain the result from Proposition 2.1. ∎

## References

  * [1] P. Erdos, On some divisibility properties of $\binom{2n}{n}$, Canad. Math.Bull. 7 (1964), 513-518.
  * [2] D. Hanson, On a theorem of Sylvester and Schur, Canad. Math. Bull., 16 (1973), 195-199.
  * [3] S. Laishram; T.N. Shorey, The greatest prime divisor of a product of consecutive integers Acta Arith. 120 (2005), no. 3, 299–306.
  * [4] L. Moser, Insolvability of $\binom{2n}{n}=\binom{2a}{a}\binom{2b}{b}$, Canad. Math. Bull. (2) 6 (1963), 167-169.
  * [5] S. Nair; T.N. Shorey, Lower bounds for the greatest prime factor of product of consecutive positive integers J. Number Theory 159 (2016), 307–328.
  * [6] P. Stănică, Good lower and upper bounds on binormail coefficients, Journal of Inequalities in Pure and Applied Mathematics, Vol 2, Issue 3, Article 30, 2001.


# Passive dynamical decoupling of trapped ion qubits and qudits

R. T. Sutherland<EMAIL_ADDRESS>Quantinuum, 303 S
Technology Ct, Broomfield, CO 80021, USA  Department of Electrical and
Computer Engineering, University of Texas at San Antonio, San Antonio, Texas
78249, USA  S. D. Erickson  Quantinuum, 303 S Technology Ct, Broomfield, CO
80021, USA

###### Abstract

We propose a method to dynamically decouple every magnetically sensitive
hyperfine sublevel of a trapped ion from magnetic field noise, simultaneously,
using integrated circuits to adiabatically rotate its local quantization
field. These integrated circuits allow passive adjustment of the effective
polarization of any external (control or noise) field. By rotating the ion’s
quantization direction relative to this field’s polarization, we can perform
‘passive’ dynamical decoupling (PDD), inverting the linear Zeeman sensitivity
of every hyperfine sublevel. This dynamically decouples the entire ion, rather
than just a qubit subspace. Fundamentally, PDD drives the transition
$m_{F}\rightarrow-m_{F}$ for every magnetic quantum number $m_{F}$ in the
system—with only one operation—indicating it applies to qudits with constant
overhead in the dimensionality of the qudit. We show how to perform pulsed and
continuous PDD, weighing each technique’s insensitivity to external magnetic
fields versus their sensitivity to diabaticity and control errors. Finally, we
show that we can tune the sinusoidal oscillation of the quantization axis to a
motional mode of the crystal in order to perform a laser-free two qubit gate
that is insensitive to magnetic field noise.

##  I Introduction

Trapped ions offer high fidelity one- and two-qubit gates, long memory times,
and the potential to reduce circuit depths with the all-to-all connectivity
enabled by ion transport and reordering Harty  _et al._ (2014); Ballance  _et
al._ (2016); Gaebler  _et al._ (2016); Srinivas  _et al._ (2021); Clark  _et
al._ (2021); Moses  _et al._ (2023); Malinowski  _et al._ (2023). Regardless,
many challenges remain when integrating the capabilities promised in isolated
academic demonstrations into larger systems. One reason for this is that
large-scale computers must run many distinct operations that, sometimes, have
conflicting requirements. For example, many two qubit gating schemes Mintert
and Wunderlich (2001); Ballance  _et al._ (2016); Weidt  _et al._ (2016);
Sutherland (2019); Srinivas  _et al._ (2021); Clark  _et al._ (2021) require
shelving each ion to a magnetic field (B-field from here on) sensitive
(Zeeman) qubit before implementation, leaving it vulnerable to memory errors.
This is typically ameliorated with a spin-echo (dynamical decoupling) sequence
Viola and Lloyd (1998); Viola  _et al._ (1999) which exchanges a qubit’s
states to invert it’s B-field sensitivity. If a transition between our choice
of qubit states cannot be driven, the need for shelving or dressing pulses
will complicate any scheme. Further, since exchanging two states only works on
a qubit, extending the scheme to qudits ultimately adds control
complexity/errors Ringbauer  _et al._ ; Hrmo  _et al._ (2023).

Figure 1: (a) Example circuit design capable of arbitrarily changing the
effective quantization field experienced by a ‘target’ ion. (b) Representation
of a spin-echo sequence via adiabatic rotation of the local quantization field
$\vec{B}_{\text{t}}$ on a target qubit. The ion experiences a quantization
field initially pointing up on the page (i), that is then rotated into a
direction orthogonal to its original direction, continuing its rotation until
it points down (ii). Finally, the system is returned to its initial
$\vec{B}_{\text{t}}$ without rotating $\vec{B}_{\text{t}}$. This causes the
ion to remain in the state defined by the quantization field in (ii), inducing
a $\ket{F,m_{F}}\rightarrow\ket{F,-m_{F}}$ transition for every Zeeman state,
i.e. inverting the ion’s energy shifts from external B-fields.

In this work, we describe a method for dynamical decoupling that inverts the
(linear) magnetic sensitivity of every $m_{F}\neq 0$ state of a target ion.
Since it affects all states separately, it works equally well on qudits and
qubit states that cannot be driven directly. Using trap integrated circuits
Ospelkaus  _et al._ (2008, 2011); Warring  _et al._ (2013); Harty  _et al._
(2014, 2016); Srinivas  _et al._ (2019, 2021); Malinowski  _et al._ (2023)
(see Fig. 1a) we can locally manipulate the quantization field direction
experienced by a target ion. As we will show, adiabatically inverting the
quantization field direction also inverts the B-field sensitivity of the
entire ion, letting us increase memory times via dynamical decoupling.
Specifically, in Sec. II we discuss how one can temporarily alter the
effective size and direction of a quantization field local to an ion, perform
a specific task in the customized environment, then return to the ion’s
permanent quantization field by ramping the circuits off. Then, in Sec. III,
we show how the technique can be used to passively dynamically decouple (PDD)
the ion from magnetic field noise. By adiabatically rotating the local
quantization field until it is anti-parallel to its original direction, we can
invert the ion’s B-field sensitivity (see Fig. 1). In other words, we drive
the transition $\ket{F,m_{F}}\rightarrow\ket{F,-m_{F}}$ for every state in the
ion. This allows us to dynamically decouple Viola and Lloyd (1998); Viola  _et
al._ (1999) all internal states of the ion from B-field noise with no need to
directly drive a specific transition. The fact that PDD acts on an entire ion,
rather than a qubit subspace of that ion, extends dynamical decoupling to
qudit systems with constant overhead in the dimensionality of the qudit. In
Secs. III.1 and III.2, we discuss how to perform pulsed- and continuous-PDD.
Extending from the latter, in Sec. III.3 we propose a new scheme for laser-
free two qubit gates where the rotation frequency of the quantization field is
tuned near the motional sideband frequency of a multi-ion crystal in a static
magnetic field gradient. The gating scheme promises some of the advantages of
those based on oscillating gradients Ospelkaus  _et al._ (2008, 2011); Harty
_et al._ (2016); Sutherland  _et al._ (2019); Sutherland (2019); Srinivas  _et
al._ (2021), while requiring only a static gradient and remaining insensitive
to B-field noise. Finally, in Sec. IV we discuss the impact of diabaticity,
cross-talk, and anticipated control errors.

##  II Theory

We consider a set of ‘target’ ions experiencing two magnetic fields. The first
is the permanent quantization field $\vec{B}_{0}=(0,0,B_{0})$, identical for
every ion in the computer, and the second is a temporary/local B-field from
the near-field of the trap circuits $\vec{B}_{\text{c}}$. We consider the
effect these two fields have on the Zeeman states $\ket{F,m_{F}}$ of a system,
where $F$ is the total angular momentum of the state, and $m_{F}$ is its
magnetic quantum number. This makes the system’s ‘permanent’ Hamiltonian:

$\displaystyle\hat{H}_{0}=\frac{\hbar
A}{2}\vec{I}\cdot\vec{J}+\mu_{\mathrm{B}}B_{0}(g_{\mathrm{J}}\hat{J}_{\mathrm{z}}+g_{\mathrm{I}}\hat{I}_{\mathrm{z}}),$
(1)

where $A$ is the hyperfine splitting and
$\vec{L}\equiv(\hat{L}_{x},\hat{L}_{y},\hat{L}_{z})$ the angular momentum
operators $\vec{L}\in\\{\vec{I},\vec{J}\\}$. While the following results are
general, for clarity we consider only $S_{1/2}$ ground state manifolds,
meaning the system has two possible values for a state’s total angular
momentum $F^{+}=I+J$ and $F^{-}=I-J$ Edmonds (1996). In the following we will
write the magnitude of vectors $|\vec{V}|$ as $V$, and their unit vectors
$\vec{V}/V$ as $\hat{V}$. We consider only magnetic field magnitudes of
$\lesssim 10$ Gauss, as used in current commercial trapped ion processors Pino
_et al._ (2021); Moses  _et al._ (2023). This makes the $\sim~{}\text{MHz}$
transition Rabi frequencies small relative to the $A/2\pi\sim~{}\text{GHz}$
frequency separation of typical $S_{1/2}$ hyperfine manifolds Langer (2006).
This allows a perturbative treatment of these off-diagonal elements, resulting
in a repeatable AC Zeeman shift. Therefore, we simplify our analysis by making
the rotating wave approximation with respect to these terms, i.e. we drop
matrix elements between states with different values for $F$. After this
approximation, we are free to neglect the $\propto\vec{I}\cdot\vec{J}$ term in
Eq. (1). This reduces Eq. (1) to:

$\displaystyle\hat{H}_{0}\simeq\mu_{\mathrm{B}}B_{0}\Big{(}g_{\mathrm{J}}\hat{J}_{\mathrm{z}}+g_{\mathrm{I}}\hat{I}_{\text{z}}\Big{)}.$
(2)

We explore values of $B_{0}$ to be $\lesssim 10$ Gauss, so that a nearby
integrated circuit should be capable of generating a B-field
$\vec{B}_{\text{c}}$ larger than $B_{0}$ (with an experimentally feasible
current); for example, a wire carrying $0.25~{}$Amps of current generates
$10~{}$Gauss at a point $50~{}\mu\text{m}$ away from the trap, much less than
the $\sim 1~{}$Amp surface currents that have were used in recent high-
fidelity gate operations Srinivas  _et al._ (2021). This redefines the total
quantization field for the target ion as
$\vec{B}_{\text{t}}\equiv\vec{B}_{0}+\vec{B}_{\text{c}}$ for a user-specified
duration. Using 1-3 spatially separated circuits integrated into the plane of
the trap, we obtain 1-3 degrees of freedom (respectively) to control the
magnitude and direction of the total magnetic field
$\vec{B}_{\mathrm{t}}=(B_{\text{c},x},~{}B_{\text{c},y},~{}B_{\text{c},z}+B_{0})$
experienced by the ion (see Fig. 1a). In this work, we will define
$\vec{B}_{\mathrm{c}}$ such that it has no projection along the $y$-direction,
making $\vec{B}_{\mathrm{t}}=(B_{\text{c},x},~{}0,~{}B_{\text{c},z}+B_{0})$.
Since $g_{\text{J}}\gg g_{\text{I}}$, it is straightforward to rotate
$\vec{B}_{\text{t}}$ at a rate that is slow compared to the electron spin
interaction ($\mu_{B}B_{\text{t}}g_{\text{J}}/2\pi\hbar$) but (nearly)
instantaneous compared to the nuclear spin interaction
($\mu_{B}B_{\text{t}}g_{\text{I}}/2\pi\hbar$). This means we can drop the
$\propto\hat{I}$ term in Eq. (2) for simplicity, noting the following scheme
does not invert the shift from the nucleus because inverting that shift would
require operation timescales $\sim g_{\text{J}}/g_{\text{I}}$ longer than
those we discuss here. Considering this, when $\vec{B}_{\text{c}}$ has been
ramped on, the total Hamiltonian becomes:

$\displaystyle\hat{H}_{\mathrm{t}}=\mu_{\mathrm{B}}g_{\mathrm{J}}\vec{B}_{\mathrm{t}}\cdot\vec{J},$
(3)

which we will use in the numerical examples below.

The operator $\hat{U}_{\mathrm{t}}$ that diagonalizes $\hat{H}_{\mathrm{t}}$
can be represented as a rotation of the system such that its quantization
field direction is redefined to be along $\hat{B}_{\text{t}}$. This
‘redefinition’ can be encapsulated by a single rotation about an axis
orthogonal to both $\vec{B}_{0}$ and $\vec{B}_{\mathrm{t}}$—here taken to be
the $y$-direction. We can represent such a rotation by:

$\displaystyle\hat{U}_{\mathrm{t}}=e^{-i\phi\hat{J}_{y}},$ (4)

where $\tan(\phi[t])=-B_{\mathrm{t},x}/B_{\mathrm{t},z}$. Using this operator
to transform Eq. (3) according to
$\tilde{H}=\hat{U}^{\dagger}_{\text{t}}\hat{H}_{\text{t}}\hat{U}_{\text{t}}+i\hbar\dot{\hat{U}}^{\dagger}_{\text{t}}\hat{U}_{\text{t}}$
gives:

$\displaystyle\tilde{H}$ $\displaystyle=$
$\displaystyle\mu_{\mathrm{B}}B_{\text{t}}g_{\mathrm{J}}\hat{J}_{z}+\hbar\dot{\phi}(t)\hat{J}_{y}.$
(5)

In the limit that we change the quantization field slowly compared to the
ion’s Zeeman splitting, we can let $\dot{\phi}\rightarrow 0$ and ignore the
latter term in the above equation—its largest effect being an added,
repeatable AC contribution to the effective value of $B_{\text{t}}$, which can
be calibrated out (see Sec. IV).

### Passive Field Rotations

When the ion experiences a second external field after its quantization
direction has been rotated, the second field’s effective polarization will be
defined by its relation to $\vec{B}_{\text{t}}$, not $\vec{B}_{0}$. In other
words, if an ion would have experienced an operator
$\vec{F}=(\hat{F}_{x},\hat{F}_{y},\hat{F}_{z})$ when the ion’s quantization
field was $\vec{B}_{0}$, the ion will experience the operator
$\tilde{F}\equiv\hat{U}_{\text{t}}^{\dagger}\vec{F}\hat{U}_{\text{t}}$ after
we rotate $\vec{B}_{0}\rightarrow\vec{B}_{\text{t}}$. For the rotation
described by Eq. (4), this would give:

$\displaystyle\tilde{F}=(\hat{F}_{x}\cos[\phi]+\hat{F}_{z}\sin[\phi],\hat{F}_{y},\hat{F}_{z}\cos[\phi]-\hat{F}_{y}\sin[\phi]).$
(6)

In a sense, this “passively” rotates the polarization of the external field
relative to the target ion—independent of our ability to change its lab-frame
polarization. It is often difficult to actively change the polarization of a
control field, so the ability to control this polarization electronically
could simplify many experiments with conflicting polarization requirements.
Interestingly, rotating $\hat{B}_{\text{t}}$ rotates the effective
polarization of stray magnetic fields as well, providing a unique way of
mitigating their harm.

##  III Passive Dynamical Decoupling

For any ion interacting with a stray magnetic field, its $1^{\text{st}}$-order
Zeeman shifts are proportional to the field’s projection onto its local
quantization direction $\hat{B}_{\text{t}}$. Therefore, if we adiabatically
rotate $\hat{B}_{\text{t}}$ into $-\hat{z}$, we invert shift from the field.
In other words, we drive a $m_{F}\rightarrow-m_{F}$ transition for every
Zeeman state in the ion. This is a unique benefit of ‘passive’ dynamical
decoupling PDD: there is no requirement whatsoever on the ability to directly
drive transitions between a system’s information carrying states, or even how
many information carrying states there are.

Traditional methods for dynamically decoupling a quantum system from magnetic
field noise involve inverting the magnetic sensitivity of only a qubit
subspace, not the whole quantum system; this typically requires driving a
transition (directly or indirectly) between the two states. There is no such
requirement for PDD, since it affects every Zeeman state in the system. For
example, if we perform a gate on 137Ba+ with one qubit state defined as
$\ket{S_{1/2},F,m_{F}}$ and one as $\ket{D_{5/2},F^{\prime},m_{F}^{\prime}}$,
any dynamical decoupling sequence via traditional means would require a laser
beam to drive a transition between the $S_{1/2}$ and $D_{5/2}$ manifolds. With
passive dynamical decoupling there is no such requirement since the protocol
simply maps the qubit onto
$\\{\ket{S_{1/2},F,-m_{F}},\ket{D_{5/2},F^{\prime},-m_{F}^{\prime}}\\}$. The
applicability of PDD is more general than this, however, since it is a single
operation that dynamically decouples entire atoms, rather than qubit
subspaces; while we will focus on qubits below, the control sequences we
describe would similarly dynamically decouple qudit systems with constant
overhead.

Figure 2: Probabilities versus time of a 137Ba+ ion initialized to
$\ket{\psi(0)}=\sqrt{1/3}\ket{F=1,m_{F}=-1}+\sqrt{2/3}\ket{F=1,m_{F}=1}$ of
the $S_{1/2}$ manifold of 137Ba+, undergoing an inversion of it’s magnetic
field sensitivity. Up to a calibratable phase shift, every hyperfine sublevel
of the ion undergoes the transformation $m_{F}\rightarrow-m_{F}$ at the end of
the sequence. Here, $\ket{F=1,m_{F}=-1}$ is red, $\ket{F=1,m_{F}=0}$ is grey,
and $\ket{F=1,m_{F}=1}$ is green. (Inset) The quantization field components
experienced by the ion versus time, where $B_{x}=\sin^{2}(\pi t/\tau)$ is the
orange line and $B_{z}=1-\cos(\pi t/\tau)$ is the blue line.

###  III.1 Pulsed PDD

After we rotate $\hat{B}_{\text{t}}$ into $-\hat{z}$, we can return
$\hat{B}_{\text{t}}$ to $+\hat{z}$ without reversing the first operation via
ensuring the field remains aligned with $\hat{z}$ during the return; if we
ensure $|\hat{B}_{\text{t}}\cdot\hat{z}|=1$, Eq. (3) will remain diagonal and
the system will not return to its original state. The final states of the ion
will have undergone an $m_{F}\rightarrow-m_{F}$ transition (see Fig. 1).
Writing down the Hamiltonian for an ion experiencing an extraneous B-field
oscillating at $\omega_{e}$ we get:

$\displaystyle\hat{H}_{e}=\mu_{\text{B}}B_{e}\cos(\omega_{e}t)g_{\text{J}}\hat{J}_{z},$
(7)

the effect of which we can analyze using Eq. (6), inserting $\vec{J}$ for
$\vec{F}$. When setting $\phi=\pi$, the scheme inverts the shift from the
extraneous field on every Zeeman state in the target ion, transforming
$\hat{H}_{e}$ into $-\hat{H}_{e}$. If needed, this operation could be repeated
in a pattern to perform higher-order pulsed PDD sequences Hayes  _et al._
(2012). Importantly, this does not dynamically decouple the quadratic shift
due to $B_{0}$ mixing the two hyperfine manifolds; this means, for example,
that pulsed PDD could not be used to increase the memory time of the
$\\{\ket{F^{+},0},\ket{F^{-},0}\\}$ ‘clock’ qubit.

We provide a numerical example of the dynamics of such a transition in Fig. 2.
Here we show a spin-echo sequence for a system initialized to
$\ket{\psi(0)}=\sqrt{1/3}\ket{F=1,m_{F}=-1}+\sqrt{2/3}\ket{F=1,m_{F}=1}$ of
the $S_{1/2}$ ground-state manifold of 137Ba+, giving our qubit a B-field
sensitivity of $\frac{\partial\omega_{q}}{\partial B}\simeq 2\pi\times
1.4~{}\text{MHz/Gauss}$. As shown in the inset, we set the time dependence of
the temporary quantization field to be
$\vec{B}_{\text{t}}=|{B}_{\text{t}}|(1-\cos[\pi t/\tau],0,\sin^{2}[\pi
t/\tau])$, where $\tau=2~{}\mu\text{s}$ and $B_{\text{t}}=10$ Gauss (see inset
of Fig. 2. As will be discussed in Sec. IV, choosing functions for
$\vec{B}_{\text{t}}$ such that
$\vec{B}_{\text{t}}(0)=\vec{B}_{\text{t}}(t_{f})$ and
$\dot{\vec{B}}_{\text{t}}(0)=\dot{\vec{B}}_{\text{t}}(t_{f})=0$ significantly
reduces the value of $B_{\text{t}}$ needed to reduce state leakage to a given
degree. After the rotation, we see that the system has undergone the desired
$m_{F}\rightarrow-m_{F}$ transition discussed above. Importantly, while we are
driving a transition between $\ket{F=1,m_{F}=1}$ and $\ket{F=1,m_{F}=-1}$, we
are not directly coupling them, which would violate selection rules. In the
figure we can see this in the fact that $\ket{F=1,m_{F}=0}$ acts as a bus
during the operation, being populated only transiently.

Figure 3: Filter function $S_{\omega_{e}}$ corresponding to the memory error
for a 137Ba+ ion initialized to the
$\ket{\psi(0)}=\sqrt{2/3}\ket{F=1,m_{F}=1}+\sqrt{1/3}\ket{F=1,m_{F}=-1}$ state
of the $S_{1/2}$ ground state manifold experiencing B-field noise for
$t_{f}=100~{}\mu\text{s}$. Each rotation assumes the magnitude of the rotated
quantization field to be $B_{\text{t}}=10$ Gauss. The grey (dashed dotted)
line corresponds to an ion undergoing no dynamical decoupling, the blue
(solid) line corresponds to an ion undergoing a ‘passive’ spin-echo rotation
(pulsed PDD) lasting $\tau=10~{}\mu\text{s}$ centered at $t_{f}/2$. The green
(dashed) line corresponds the scenario where we continuously rotate the
quantization field in the $xz$-plane at a rate $\omega_{\text{r}}=2\pi/\tau$
(continuous PDD).

In Fig. 3 we show the increased resistance to B-field noise that results from
a ‘passive’ spin-echo, using the same initial state shown in Fig. 2; the
control sequence is the same, except we set the rotation time to be
$\tau=10~{}\mu\text{s}$. There we plot the filter function $F(\omega)$,
defined as the function that, when integrated against the noise spectral
density $S(\omega)$, gives the total infidelity of the system
$\mathcal{I}=\int^{\infty}_{0}d\omega^{\prime}F(\omega^{\prime})S(\omega^{\prime})=1-|\braket{T}{\psi(0)}|^{2}$
where $\ket{T}$ is the target state. In this example we calculate the memory
error of an ion after $t_{f}=100~{}\mu\text{s}$, so
$\ket{T}\equiv\ket{\psi(0)}$ up to a repeatable AC Zeeman phase shift that can
be calibrated out (see Sec. IV) of the error budget. We determine $F(\omega)$
numerically by calculating the infidelity of a system experiencing a small
B-field, representing a noise term:

$\displaystyle\hat{H}_{e}=\mu_{\text{B}}B_{e}\cos(\omega_{e}t+\theta)g_{\text{J}}\hat{J}_{z},$
(8)

where we ensure that $B_{e}$ is small enough that $\mathcal{I}\propto
B_{e}^{2}$, but large compared to the infidelity due to state leakage.
Averaging over $\theta$ and dividing the result by $B_{e}^{2}$ gives
$S(\omega_{e})$ for the operation. Comparing $S(\omega_{e})$ for the ion with
and without a spin-echo rotation, we can see as $\omega_{e}\rightarrow 0$ that
$S(\omega_{e})\rightarrow 0$ for the echoed sequence, while, for the sequence
with no PDD, $S({\omega_{e}})$ approaches maximum value as
$\omega_{e}\rightarrow 0$; since magnetic field NSDs are typically largest at
low frequencies, Fig. 3 shows that passive PDD should lead to significantly
longer qubit and qudit memory times relative to their non-echoed parallels
Sepiol  _et al._ (2019).

###  III.2 Continuous PDD

We can perform continuous PDD by rotating the direction of
$\hat{B}_{\text{t}}$ about a vector in the $xy$-plane, here taken to be
$\hat{y}$. If we keep the value of $B_{\text{t}}$ constant, this gives:

$\displaystyle\vec{B}_{\text{t}}=B_{\text{t}}(\sin[\omega_{r}t],0,\cos[\omega_{r}t]),$
(9)

rotating $\hat{B}_{\text{t}}$ in a circle at a rate $\omega_{r}\equiv
2\pi/\tau$. As a result, projections of any extraneous B-field
$\vec{B}_{e}\cdot\hat{B}_{\text{t}}$ will sinusoidally oscillate at
$\omega_{r}$. In other words:

$\displaystyle\tilde{H}_{e}\\!=\\!\mu_{\text{B}}B_{e}\\!\cos(\omega_{e}t\\!+\\!\theta)g_{\text{J}}\Big{(}\\!\hat{J}_{x}\sin[\omega_{r}t],0,\hat{J}_{z}\cos[\omega_{r}t]\\!\Big{)},$
(10)

rendering B-field noise where $\omega_{e}\ll\omega_{r}$ off-resonant. We show
this in Fig. 3 where we plot $S(\omega_{e})$ for the same system described
previously, only undergoing continuous PDD. We can see that, for this value of
$\tau=10~{}\mu\text{s}$, continuous PDD suppresses magnetic field noise by
several orders-of-magnitude relative to a spin-echo sequence. In the
simulation, we suppress state leakage by adiabatically ramping
$\vec{B}_{\text{t},x}$ on/off via setting
$\vec{B}_{\text{t},x}=\sin^{2}(\omega_{r}t)$ when $\omega_{r}t<\pi/2$ and
$(t_{f}-t)\omega_{r}<\pi/2$, ensuring that
$\dot{\vec{B}}_{\text{t}}(0)=\dot{\vec{B}}_{\text{t}}(t_{f})=0$. Although we
gain insensitivity to ambient B-field noise, continuous PDD requires leaving
the ions exposed to uncertainties in $\vec{B}_{\text{t}}$ for significantly
longer than pulsed PDD, since it requires the control fields to be on for all
$t_{f}$.

###  III.3 PDD-assisted laser-free entangling gate

In the presence of a static B-field gradient, we can rotate
$\hat{B}_{\text{t}}$ at a rate tuned near the frequency of a motional mode,
driving a spin-dependent force for every state in the ion. This reduces to the
the spin-dependent force typically associated with
$\hat{\sigma}_{z}\otimes\hat{\sigma}_{z}$ two-qubit gates Mølmer and Sørensen
(1999); Sørensen and Mølmer (2000); Leibfried  _et al._ (2002) when applied to
a qubit subspace. If an ion is at rest in a magnetic field with a non-
trivially large gradient pointing in the $z$-direction, its Hamiltonian can be
represented as:

$\displaystyle\hat{H}_{zz}=\mu_{\text{B}}\Big{(}B_{z}+\frac{\partial
B_{z}}{\partial z}\hat{z}\Big{)}g_{\text{J}}\hat{J}_{z}.$ (11)

The impact of $B_{z}$ is repeatable and can be calibrated out, so we ignore
its effect for simplicity. Projecting the position operator $\hat{z}$ onto a
specified motional mode with ladder operators $\hat{a}(\hat{a}^{\dagger})$ and
frequency $\omega_{a}$, we can make the rotating wave approximation with
respect to the other modes in the system and write down:

$\displaystyle\hat{H}_{zz}=\hbar\Omega_{zz}\hat{J}_{z}\Big{(}\hat{a}^{\dagger}e^{i\omega_{a}t}+\hat{a}e^{-i\omega_{a}t}\Big{)},$
(12)

where
$\mu_{\text{B}}g_{\text{J}}B_{z}^{\prime}\beta_{a}\equiv\hbar\Omega_{zz}$, and
$\beta_{a}$ is the projection of the $\hat{z}$-operator onto mode $a$. If this
system undergoes continuous PDD, i.e. $\hat{B}_{\text{t}}$ is rotated in a
circle, the transformed Hamiltonian will be:

$\displaystyle\tilde{H}_{zz}$ $\displaystyle\simeq$
$\displaystyle\hbar\Omega_{zz}\Big{(}\hat{a}^{\dagger}e^{i\omega_{a}t}+\hat{a}e^{-i\omega_{a}t}\Big{)}\Big{(}\hat{J}_{z}\cos[\omega_{r}t]+\hat{J}_{x}\sin[\omega_{r}t]\Big{)}$
(13) $\displaystyle+\mu_{\mathrm{B}}B_{\text{t}}g_{\mathrm{J}}\hat{J}_{z},$

assuming the diabatic term is negligible. Transforming into the interaction
picture with respect to the $\propto B_{\text{t}}$ control field term, we make
the rotating wave approximation with respect to the $\propto\hat{J}_{x}$
terms—assuming that $\omega_{a}\pm\omega_{r}$ is far-detuned from the Zeeman
splitting. Finally, if we tune $\omega_{r}\sim\omega_{a}$, we get:

$\displaystyle\tilde{H}_{zz}^{\prime}=\frac{\hbar\Omega_{zz}}{2}\Big{(}\hat{a}^{\dagger}e^{i\delta
t}+\hat{a}e^{-i\delta t}\Big{)}\hat{J}_{z},$ (14)

where $\delta\equiv\omega_{a}-\omega_{r}$. Projecting this Hamiltonian onto a
qubit subspace, we get the typical Hamiltonian for a
$\hat{\sigma}_{z}\otimes\hat{\sigma}_{z}$ entangling gate.

A key advantage of this gate scheme is that it is continuously dynamically
decoupled from B-field noise in a similar way as the laser-free gates in Refs.
Harty  _et al._ (2016); Srinivas  _et al._ (2021), suppressing qubit dephasing
during the gate. Another advantage this scheme has over previous laser-free
schemes, however, is that there is (again) no requirement on the ability to
directly drive a transition between the qubit states. Importantly, this new
scheme allows experimentalists to tune near a motional mode of the system in a
static gradient—without the need for $\sim\text{GHz}$ frequency fields to
drive the qubit transition, as is the case in Refs. Mintert and Wunderlich
(2001); Weidt  _et al._ (2016). This means it could be useful in gates that
use permanent gradients.

##  IV Sources of error

We have shown how to use PDD to render a system less sensitive to ambient
B-field noise during a given operation time $t_{f}$, but have not discussed
potential sources of error. In the limit that we can manipulate the currents
in trap integrated circuits with errors much smaller than memory errors from
B-field noise, PDD offers a clear advantage. If it is not clear that this is
the case we must consider the main sources of error intrinsic to the specific
PDD scheme we use. In the following we will examine the (rough) extent to
which errors from uncertainties in the control field operations, and from
diabaticity, should affect the usefulness of PDD. We note that, while we do
not examine crosstalk in detail since this will likely be device specific,
idle qubits far way from the source of $\vec{B}_{\text{c}}$ will only
experience small repeatable perturbations to $\vec{B}_{0}$, resulting in phase
shifts that can be tracked.

###  IV.1 Diabaticity

Figure 4: State leakage probability
$1-(|\braket{0}{\psi(t)}|^{2}+|\braket{1}{\psi(t)}|^{2})$ versus temporary
quantization field magnitude $B_{t}$ used to implement dynamical decoupling.
For each figure we show pulsed a) and continuous b) dynamical decoupling
schemes for a system initialized to the state
$\ket{\psi(t)}=\sqrt{2/3}\ket{F=1,m_{F}=1}+\sqrt{1/3}\ket{F=1,m_{F}=-1}$ of
the $S_{1/2}$ manifold of 137Ba+. Each set of simulations is for a
$t_{f}=100~{}\mu\text{s}$ sequence. In a) we show a ‘passive’ spin-echo, where
we rotate $\vec{B}_{t}\rightarrow-\vec{B}_{t}$ over $\tau=10~{}\mu\text{s}$,
and in b) we rotate $\vec{B}_{t}$ continuously at $\omega_{r}=2\pi/\tau$.

So far, we have not discussed the role of the $\propto\dot{\phi}(t)$ term in
Eq. (5), which represents off-resonant diabatic transitions. If we rotate the
system slowly enough, it will be fully adiabatic and $\dot{\phi}(t)\rightarrow
0$. Since we cannot guarantee adiabaticity for every system of interest, it is
crucial to understand the general behaviour of PDD when $\dot{\phi}(t)\neq 0$.
To examine this, we first transform this term into the interaction picture
with respect to the $\propto B_{\text{t}}(t)$ quantization field term in Eq.
(5) using:

$\displaystyle\hat{U}_{I}=\exp\Big{(}-i\varepsilon[t]\hat{J}_{z}\Big{)},$ (15)

where
$\varepsilon(t)\equiv(\mu_{\text{B}}g_{\text{J}}/\hbar)\int^{t}_{0}dt^{\prime}B_{\text{t}}(t^{\prime})$
is the time integral of the Zeeman splitting induced by $\vec{B}_{\text{t}}$.
This gives:

$\displaystyle\hat{H}_{\text{t},I}=\hbar\dot{\phi}(t)\Big{(}\hat{J}_{y}\cos[\varepsilon(t)]+\hat{J}_{x}\sin[\varepsilon(t)]\Big{)},$
(16)

showing the ‘diabatic’ term in Eq. (5) can be represented with an angular
momentum operator $\vec{J}$ that rotates in the $xy$-plane. We can approximate
the time propagator for Eq. (16), using the Magnus expansion Magnus (1954):

$\displaystyle\hat{U}$ $\displaystyle=$
$\displaystyle\exp\Big{(}\frac{-i}{\hbar}\int^{t}_{0}dt^{\prime}\\!\hat{H}_{\text{t},I}(t^{\prime})$
$\displaystyle-\frac{1}{2\hbar^{2}}\int^{t}_{0}\\!\\!\int^{t^{\prime}}_{0}dt^{\prime}dt^{\prime\prime}\Big{[}\hat{H}_{\text{t},I}(t^{\prime}),\hat{H}_{\text{t},I}(t^{\prime\prime})\Big{]}\Big{)}.$

Plugging Eq. (16) into this equation simplifies to:

$\displaystyle\hat{U}$ $\displaystyle=$
$\displaystyle\exp\Big{(}-i\int^{t}_{0}dt^{\prime}\dot{\phi}(t^{\prime})\Big{(}\hat{J}_{y}\cos[\varepsilon(t^{\prime})]+\hat{J}_{x}\sin[\varepsilon(t^{\prime})]\Big{)}$
(18)
$\displaystyle-\frac{\hat{J}_{z}}{2}\\!\int^{t}_{0}\int^{t^{\prime}}_{0}\\!\\!\\!\\!dt^{\prime}dt^{\prime\prime}\dot{\phi}(t^{\prime})\dot{\phi}(t^{\prime\prime})\sin\Big{(}\varepsilon[t^{\prime\prime}]-\varepsilon[t^{\prime}],\Big{)}$

which we can separate into two distinct effects: state leakage, represented by
the $1^{\text{st}}$-order terms, and an additional $\propto\hat{J}_{z}$ shift,
represented by the $2^{\text{nd}}$-order terms. If we approximate
$\dot{\phi}(t)\rightarrow\phi_{0}$ and
$\varepsilon(t)\rightarrow\varepsilon_{0}t$ to be constant over $t_{f}$, we
can see that the $1^{\text{st}}$-order terms generally represent off-resonant
transitions inducing leakage that scales
$\propto(\phi_{0}/\varepsilon_{0})^{2}$. We can further suppress off-resonant
leakage errors by “pulse shaping” $\dot{\phi}(t)$. For our examples of
continuous PDD we did this by setting
$B_{\text{t},x}=B_{\text{t}}\sin^{2}(\omega_{r}t)$ at the start and end of the
control sequence. This ensures $\ddot{\phi}(0)=\ddot{\phi}(t_{f})=0$, which
further suppresses the probability of an off-resonant transition in the same
way it does for carrier Roos (2008); Sutherland (2019) and spin-motion
Sutherland  _et al._ (2023) interactions. In Fig. 4 we show increasing the
value of $B_{\text{t}}$ also suppresses state leakage, making these
transitions more off-resonant by increasing the Zeeman splitting of the atom.
We demonstrate this in Fig. 4 for the example case used in Figs. 2 and 3,
plotting the leakage error probability
($1-|\braket{1,-1}{\psi(t_{f})}|^{2}-|\braket{1,1}{\psi(t_{f})}|^{2}$) versus
$B_{t}$ for both pulsed and continuous PDD. This shows leakage can be
suppressed below $10^{-4}$ at $\sim 1$ Gauss for our pulsed and $\sim 2.5$
Gauss for the continuous PDD examples; for reference, a $0.1$ Amp current in
an infinitely thin wire produces a $4$ Gauss B-field at distance of
$50~{}\mu\text{m}$. This corresponds to resistive heat loads $\sim 100$ times
lower than the $\sim~{}1$ Amp currents used in Ref Srinivas  _et al._ (2021).
The $2^{\text{nd}}$-order terms in Eq. (18) correspond to a
$\propto\hat{J}_{z}$ AC shift, that, in general, grows like
$\propto\dot{\phi}^{2}/\varepsilon$. If we can consistently repeat both
$\dot{\phi}$ and $\varepsilon$ over many operations, then we can track this
shift and eliminate its effect.

###  IV.2 Control field uncertainties

In any experiment, a control field will drift from its ‘ideal’ value, i.e.
$\vec{B}_{\text{c}}(t)\rightarrow\vec{B}_{\text{c}}(t)+\delta{\vec{B}_{\text{c}}}(t)$.
The biggest source of infidelity from $\delta\vec{B}_{\text{c}}$ will likely
be due to its projection onto $\hat{B}_{\text{t}}$, since this will give a
$1^{\text{st}}$-order shift that can be described by the time propagator:

$\displaystyle\hat{U}_{\delta
B_{\text{c},\text{t}}}\simeq\exp\Big{(}-\frac{i}{\hbar}\frac{\partial\omega_{q}}{\partial
B}\hat{J}_{z}\int^{t_{r}}_{0}dt^{\prime}\hat{B}_{\text{t}}[t^{\prime}]\cdot\delta\vec{B}_{\text{c}}[t^{\prime}]\Big{)},$
(19)

where $t_{r}$ is here the time the control fields are on. Assuming
$\delta\vec{B}_{\text{c}}$ drifts slowly over a calibration cycle, we can take
$\delta\vec{B}_{\text{c}}$ to be constant and write down an approximate
equation for $\mathcal{I}$. We can Taylor expand Eq. (19) if $\mathcal{I}\ll
1$, letting us obtain:

$\displaystyle\mathcal{I}=\Big{(}\delta
B_{\text{c}}\frac{\partial\omega_{q}}{\partial
B}t_{r}\Big{)}^{2}\lambda^{2}_{\sigma_{z}},$ (20)

where
$\lambda_{\hat{\sigma}_{z}}^{2}\equiv\braket{\hat{\sigma}_{z}^{2}}-\braket{\hat{\sigma}_{z}}^{2}$
is the variance of the $\hat{\sigma}_{z}$ Pauli-z operator of the qubit
subsystem. We here take $\lambda_{\hat{\sigma}_{z}}^{2}=1/3$ upon averaging
over SU(2). If we choose $B_{\text{c}}\simeq 2.5~{}\text{Gauss}$ and assume an
uncertainty of $10^{-4}$ for
$\delta\vec{B}_{\text{c}}\cdot\hat{B}_{\text{t}}$, we get an infidelity of
$\mathcal{I}\simeq 1\times 10^{-5}$ for the pulsed spin-echo case (using
$t_{r}\simeq 3~{}\mu\text{s}$) and $\mathcal{I}\simeq 2\times 10^{-2}$ for
continuous PDD (using $t_{r}\simeq 100~{}\mu\text{s}$). We can see continuous
PDD is significantly more sensitive to control field uncertainties because
$t_{r}$ is larger, making pulsed PDD more appealing as a near-term tool.

##  V Conclusion

In this work, we discussed how trap integrated circuits can be used to
implement arbitrary passive rotations of the quantization axis temporarily
experienced by a target ion. This lets us implement ‘passive’ dynamical
decoupling by inverting the ion’s quantization field’s projection onto
external B-fields, inverting the magnetic susceptibility of its hyperfine
sublevels. We showed how to do ‘pulsed’ PDD, where the energy dependence of
every Zeeman state in the ion is reversed over a short timescale, and how to
do ‘continuous’ PDD, where the ion’s quantization direction is rotated
sinusoidally. We also proposed a new way to perform laser-free two-qubit gates
by showing we can tune this continuous rotation to the frequency of a motional
mode sideband while the ion is in a static gradient. Finally, we discussed
potential sources of error for the scheme.

## Acknowledgements

We would like to thank M. Foss-Feig, J. Gaebler, B. J. Bjork, C. Langer, M.
Schecter, C. N. Gilbreth, J. Bartolotta, E. Hudson and P. J. Lee for helpful
discussions.

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