Definition Of Operator Exponential?

1. Sep 18, 2014

ostrich2

I've been reading Mermin's book on Quantum Computer Science, and in the section in which he discusses the construction of a QFT using 1-Qbit and 2-Qbit gates, he makes reference to some expressions involving linear operators that I'm not familiar with (at least if I've seen them before I've forgotten). I'm wondering if someone can help understand/interpret these expressions.

In particular, he defines the following linear operator,

$\textbf{Z}|y\rangle_{n} = e^{2\pi i / 2^{n}}|y\rangle_{n}$

Where $|y\rangle_{n}$ represents the vector for the basis state corresponding to the n-qbit number whose binary expansion is the integer y.

From there, the following expression for Z is defined as follows (for a specific case in which n=4):

$$\textbf{Z} = exp\big(\frac{i \pi}{8}(8\textbf{n}_3 + 4\textbf{n}_2 + 2\textbf{n}_1 + \textbf{n}_0 ) \big)$$

where $\textbf{n}$ represents the 1-qbit number operator $\textbf{n}|x\rangle = x|x\rangle , x \in \{ 0,1 \}$ (subscript denotes the number operator referring to the ith qbit)

This expression refers to an operator in the argument of an exponential, which I've never seen before, so I'm not sure how to interpret this. I did find a definition of a matrix exponential in terms of an infinite series:

$$e^{\textbf{Z}} = \sum_{k=0}^{\infty}\frac{1}{k!}\textbf{Z}^{k}$$

The number operator can be expressed in terms of the matrix $\begin{pmatrix}0 & 0 \\ 0 & 1 \end{pmatrix}$, so I'm not sure if the corresponding operator expression can be understood in a similar way? It's a little hard to tell because Mermin seems to present the definitions and then make derivations from them without a lot of detailed explanation, the implication perhaps being that things follow straightforwardly from the basic definitions (which they might if I understood them properly).

I suspect I don't, because just taking as an example another simple definition of the same kind, it is stated that the following operator identity holds:

$$exp(2\pi i \textbf{n}) = \textbf{1}$$

$\textbf{1}$ being the identity operator. I tried to see if I could derive this identity interpreting it in terms of the corresponding matrix exponential, but for this to be valid, it seems to me that the corresponding infinite series had better converge to the identity matrix. But based on the definition of the number operator, and multiplying the matrix representation by the complex scalar $2\pi i$, I get:

$$e^{(2\pi i \textbf{n})} = \sum_{k=0}^{\infty}\frac{1}{k!}(2\pi i \textbf{n})^{k}$$.

In terms of matrices,

$$(2\pi i \textbf{n})^{k} = \begin{pmatrix}0 & 0 \\ 0 & (2\pi i)^{k} \end{pmatrix}$$

I don't see how the infinite series could possibly ever converge to the identity matrix, not least of which because there is only one element of any power of this matrix that will ever be nonzero? Or am I way off here?

2. Sep 19, 2014

Blazejr

Remember that n^0 is identity matrix. For positive , n^k=n. Therefore we have $$\exp (2\pi i n)= \sum _{k=0}^{\infty}\frac{1}{k!}(2 \pi i n)^k=id + \sum _{k=1}^{\infty}\frac{1}{k!}(2 \pi i n)^k=id + \sum _{k=1}^{\infty}\frac{1}{k!}(2 \pi i)^kn$$
$$=id +\left( \sum _{k=1}^{\infty}\frac{1}{k!}(2 \pi i)^k\right)n=id + (exp(2 \pi i)-1)n=id$$
Where id is identity matrix.

Last edited by a moderator: Sep 22, 2014
3. Sep 19, 2014

Fredrik

Staff Emeritus
The series definition of $e^A$ works when A is a bounded linear operator on a Hilbert space. Bounded means that the set $\{\|Ax\|:x\in\mathcal H,\,\|x\|=1\}$ (a subset of $\mathbb R$) is bounded from above. $\mathcal H$ is of course the Hilbert space on which A is a linear operator.

It also works when A is an element of an arbitrary Banach algebra. But it doesn't work for unbounded operators like position and momentum.

4. Sep 21, 2014

ostrich2

Thanks, that does make more sense now. So now I'm looking further at the original operator definition:

$$\textbf{Z}|y\rangle_n = e^{\frac{2 \pi i y}{2^{n}}} |y\rangle_{n}$$

In terms of number operators (again for the 4 bit case) this is again expressed as:

$$\textbf{Z} = exp\big(\frac{i \pi}{8}(8\textbf{n}_3 + 4\textbf{n}_2 + 2\textbf{n}_1 + \textbf{n}_0 ) \big)$$

It isn't obvious to me at a glance that the above operator would produce this state, so I'm trying to work this out to convince myself this works. Firstly, it seems to me that the same argument provided above would work more generally (as an aside, did something funny happen with the post formatting after the upgrade, it seems to me part of what Blazejr posted isn't showing up for me anymore?) , so we should have

\begin{align} exp(x \pi i \textbf{n})&= I + \sum_{k=1}^{\infty}\frac{1}{k!}(x \pi)^{k}\textbf{n} \\ &= I + (e^{x \pi} - 1)\textbf{n} \end{align}

Now we want an expression for $\textbf{Z}|y\rangle$. Using this, and of course the fact that the individual number operators are supposed to act as the identity on the other qbits other than the one they're associated with,

\begin{align} \textbf{Z} &= \prod_{j=0}^{3} [\textbf{I}_{j} + (e^{\frac{2 j \pi i}{8}} - 1)\textbf{n}_{j}]~|y_{j}\rangle \\ &= \left( \prod_{j=0}^{3}[1 + (e^{\frac{2 j \pi i}{8}} - 1)y_{j}]\right) |y_{3}\rangle |y_{2}\rangle |y_{1}\rangle |y_{0}\rangle \end{align}

(In the first case the product being the tensor product of the corresponding vector expressions for each qbit, and in the second case the scalar factor pulled out from the tensor product of the basis vectors).

Does this look like it's on the right track? It's not evident to me from this that I can end up with $e^{\frac{2 \pi i y}{2^{n}}}$ from this product, but I may be fairly rusty with these types of periodic functions