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Problem (the following is the exercise problem from Neilson and Chuang)

Suppose $V$ is a Hilbert space with a subspace $W$. Suppose $U: W\rightarrow V$ is a linear operator which preserves inner products, that is, for any $\left|w_1\right>$ and $\left|w_2\right>$ in $W$, $$\left <w_1|U^ {\dagger} U|w_2\right> = \left<w_1|w_2\right> $$

Prove that there exists a unitary operator $U':V\rightarrow V$ which extends $U$. That is, $U'\left|w\right>=U\left|w\right>$ for all $\left|w\right>$ in $W$, but $U'$ is defined on the entire space $V$. Usually we omit the prime symbol $'$ and just write $U$ to denote the extension.


Solution
Following is the solution I found online from here Solution. It says that the desired operator can be defined as $U'=U\otimes I$, where the identity is defined only on $V \perp W$.


My approach and doubt
In the solution I found above the operator $U'=U\otimes I$, will act on a space with dimension $m*n$ where $m$ is dimension of space $W$ and $n$ is dimension of space $V \perp W$, but we wanted an operator which acted on space $V$ which has dimension $m+n$. I know that the operator $U$ will have range as some space $W^{'}$ ( some subspace of space $V$ ) of the same dimension as space $W$. What I came up as a solution was let $\{i\}$ and $\{j\}$ for $1<=i<=m$, be the orthonormal basis for spaces $W$ and $W^{'}$ respectively such that $U=\sum | j \rangle \langle i| $, then we can extend the orthonormal basis of space $W$ to an orthonormal basis of space $V$ ie. $\{i\}$ for $1<=i<=m+n$ and the define $U^{'}=\sum | j \rangle \langle i| $.

But I was still doubtful. Am I correct that the solution I found has a flaw and is my approach correct ?

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1 Answer 1

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Your construction is essentially correct. It should read $U=\sum_{i=1}^m |i\rangle\langle i|$, and $U'=\sum_{i=1}^{m+n}|i\rangle\langle i|$. Since $W$ is spanned by the first $m$ vectors, $U'$ agrees with $U$ on $W$, and will be the identity on the span of $|m+1\rangle,\dots,|m+n\rangle$.

As an aside, this notation is terribly confusing for mathematicians who don't work either in physics or quantum mechanics in some fashion. I happen to be grading for a course using this textbook right now, and I have seen all of the problems before, but the notation makes it difficult.

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  • $\begingroup$ Incidentally, I think the thing that makes the tensor "solution" incorrect is the fact that you can write $V$ as the sum of $W$ and its orthogonal complement, but not the tensor product of those. The dimensions don't agree because $W\otimes H$ has dimension $dim(W)dim(H)$ not $dim(W)+dim(H)$. $\endgroup$
    – Keaton
    Jun 26, 2015 at 19:52

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