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What two arbitrary states in the same Hilbert space can be connected through an unitary transformation? And how to construct the unitary transformation? Whether is there a general answers for these problems.

We know that any two spin 1/2 states can be connected through a su(2) transformation. However, there is not a unitary transformation that connecting two different Fock states |m> and |n>.

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2 Answers 2

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Given a Hilbert space $H$ and two vectors $u,v\in H$ there is a unitary $u\in B(H)$ that maps $u$ onto $v$. Take the rank-1 operator $\theta_{u,v}$ defined as $$\theta_{u,v}(z) = \frac{(v,z)}{(v,v)}u,\qquad\forall z\in H$$ which defines a partial isometry that can be extended to a unitary over the whole of $H$.

The problem you're having with the Fock space is that you are restricting to a unitary representation of a symmetry group, e.g. the Poincaré group. Then no unitary transformation from this representation can change the particle content of a state, which then gives a decomposition of the Fock space into invariant subspaces. On the other hand, if you consider the algebra generated by the field operator surely you can find unitaries that are able to create/annihilate particles, thus intertwining between the different subspaces.

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  • $\begingroup$ Yes, you are right. In this way, we can always find a unitary transformation to connect arbitrary two different states. In some sense, this unitary matrix is customized. And it is irrelevant with symmetry group. $\endgroup$
    – Chong Chen
    Jul 17, 2015 at 6:44
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However, there is not a unitary transformation that connecting two different Fock states $|m\rangle$ and $|n\rangle$.

That's not true. There is always a unitary transformation that maps between any two given normalized states. Indeed and infinite number of them. In your example, take the number states $|0\rangle,\,|1\rangle,\,|2\rangle,\,\cdots$ as the basis vectors for your quantum state space. Then the matrix with elements

$$U_{j\,k} = \delta_{j\,n} \delta_{k\,m} + \delta_{j\,m} \delta_{k\,n} + (1-\delta_{j\,n})(1-\delta_{j\,m})(1-\delta_{k\,n})(1-\delta_{k\,m})\delta_{j\,k}\tag{1}$$

will map any state:

$$\alpha_0\,|0\rangle+\alpha_1\,|1\rangle+\cdots +\alpha_n\,|n\rangle + \cdots +\alpha_m\,|m\rangle + \cdots$$

to:

$$\alpha_0\,|0\rangle+\alpha_1\,|1\rangle+\cdots +\alpha_m\,|n\rangle + \cdots +\alpha_n\,|m\rangle + \cdots$$

i.e. it swaps the superposition weights of base states $|n\rangle$ and $|m\rangle$ and, in particular, maps $|n\rangle$ to $|m\rangle$ and contrariwise.

If your having trouble visualizing my equation (1) it is simply this: it's the identity matrix aside from the $1$ at position $(m,\,m)$ in the identity is shifted to position $(m,\,n)$ in the matrix $U$; likewise the $1$ at position $(n,\,n)$ in the identity is shifted to position $(n,\,m)$.


To extend the partial isometry in Phoenix87's Answer, work conceptually as follows. Let $\hat{X}$ be the vector you wish to map to $Y$. Now calculate

$$\hat{Y}_\perp=\frac{\frac{Y}{\|Y\|}-\frac{\langle X,\,Y\rangle}{\|Y\|\,\|X\|}\,X}{\|\frac{Y}{\|Y\|}-\frac{\langle X,\,Y\rangle}{\|Y\|\,\|X\|}\,X\|}$$

so that $\hat{X}$ and $\hat{Y}_\perp$ are an orthogonal basis for the linear subspace spanned by $X$ and $Y$. Now use the Gram Schmidt Process to find an orthonormal basis $\hat{X}_3,\,\hat{X}_4\,\,\cdots$ for the orthogonal complement of this span (I mean imagine the algorithm working to show that the basis exists, not literally doing it by hand, which is impossible for an infinite dimension Hilbert space!). In this basis, your original $Y$ is $\langle Y,\,\hat{X}\rangle \hat{X} + \langle Y,\,\hat{Y}_\perp\rangle \hat{Y}_\perp =\stackrel{def}{=}y_x\,\hat{X}+y_y\,\hat{Y}_\perp$. The full new basis is $\hat{X},\,\hat{Y}_\perp,\,\hat{X}_3,\,\hat{X}_4\,\,\cdots$

Now the matrix for your mapping with respect to this new basis is the identity matrix aside from the $2\times2$ upper right submatrix. This $2\times2$ unitary upper right submatrix is:

$$\left(\begin{array}{cc}y_x&-y_y^*\\y_y&y_x^*\end{array}\right)$$

and the rest of your operator is the same as for the identity matrix. Now transform this matrix $M$ back to the original basis by $M\mapsto T\,M\,T^\dagger$ where $T$ is the transformation from the original basis to the one found by the Gram-Schmidt process.

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