I have the $N$-fold tensor product of a convex combination of a pure state, i.e. $|\psi\rangle\langle\psi|$ with $|\psi\rangle$ a unit vector in a complex Hilbert space of dimension two, and the completely mixed state $\frac{\mathrm{I}_2}{2}$, with $\mathrm{I}_2$ the identity operator in two dimensions:

$$\rho(\lambda) = \left(\lambda|\psi\rangle\langle\psi|+(1-\lambda)\frac{\mathrm{I}_2}{2}\right)^{\otimes N},\\ \mathrm{with}~\lambda \in \left(0,1\right) $$

Writing w.l.o.g. $|\psi\rangle = \begin{bmatrix} \alpha\\ \beta \end{bmatrix}$ with $\alpha,\beta \in \mathbb{C}, ~|\alpha|^2+|\beta|^2 = 1$,

$$\rho(\lambda) = \left(\begin{bmatrix} \lambda|\alpha|^2+\frac{1-\lambda}{2} & \lambda\alpha\beta^*\\ \lambda\alpha^*\beta &\lambda|\beta|^2+\frac{1-\lambda}{2} \end{bmatrix}\right)^{\otimes N}$$

For $\lambda = 0$, the dimension of all the states in the form of $\rho$ with varying $|\psi\rangle$ is $2^N$, while for $\lambda = 1$ the space of all possible $\rho$ is the totally symmetric subspace, so that the dimension is given by $N+1$. A way to formalize the dimension of some space is to calculate the (lowest possible) trace of the identity operator working on that space. For example, $N = 2, \lambda = 1$ we can write our identity operator on all states of those form as

$$ \begin{bmatrix} 1 & 0 & 0 & 0\\ 1 & \frac{1}{2} & \frac{1}{2} & 0\\ 1 & \frac{1}{2} & \frac{1}{2} & 0\\ 0 & 0 & 0 & 1\\ \end{bmatrix} $$

The trace over this operator gives $2+1 = 3$, and doing the same for $N>2$ and $\lambda = 1$ you would be always able to find some operator that acts as the identity operator on those states, where $N+1$ will be the trace over that operator. We can assign the trace of that operator to the dimension of that space (see https://en.wikipedia.org/wiki/Dimension_(vector_space)#Trace).

What can you say about the dimension for intermediate values of $\lambda$? Do you immediately lose the ability to reduce the dimension for $\lambda<1$?

  • $\begingroup$ What do you mean by "dimension" here? The dimension of the range? $\endgroup$ – Martin Argerami Jul 31 '15 at 16:44
  • $\begingroup$ I updated the original post a bit to clarify what I mean by the dimension. $\endgroup$ – Kenneth Goodenough Jul 31 '15 at 17:18

For $\lambda\in [0,1) $, you are taking the positive (rank-one) operator $\lambda |\psi\rangle\langle\psi|$ and you are adding a positive multiple of the identity, so the resulting operator is invertible. Thus your dimension (the range) will be $2^N $.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.