Upper and lower bounds of trace

I’d like to know if my understanding is sound and reasonable.
if i have a system of one mode with 10 dims “fock space”
so the upper bound of the trace is always equal to 1
but the lower bound will be = 1 - (1/10) = 0.9
so if the trace is within 1 or 0.9 then the quantum system is still healthy?
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@josh can you please help me?

HI @kareem_essafty,

The cutoff of the fock space tells you how much “breathing room” your state has.The trace of the state tells you how much your state has support inside vs outside the cutoff.

A valid general lower bound for the trace is 0.

It is quite possible that a state could have zero support on the subspace defined by your cutoff (e.g., the Fock state |11>), leading to trace of zero.

On the other hand, some states will safely have full support inside the 10-dimensional cutoff subspace (e.g., the Fock state |5>).

Put another way, there is no general connection between the cutoff dimension and the trace (it all depends on which state you have).

Separately, usually a trace around 0.98~0.99 is a good sign that you are not missing any strong info by using a cutoff.

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thank you so much @nathan

what about this equation

does the cutoff have anything with the hilbert space dimension?

Unfortunately, the identity operator is not normalizable in the infinite-dimensional case, so the formula above (about completely mixed states) does not hold for the photonics case.

You could picture making a completely mixed state on a finite-dim subspace, e.g., up to the cutoff dimension, but it wouldn’t exactly be the restriction of some larger state to a subspace.

then how can we formulate it?
or is there any reference i can use to deepen my understanding?
thanks <3 again

What would you like to formulate?

the identity operator in infinite-dimensional space
and please pardon me if i’m wrong i’m trying study this area of CV model on my own right now

The natural way the identity operator appears in CV model is via the commutation relation [x,p]=i *hbar * Identity.

If you want to represent that operator in the Fock basis, it is exactly what you would expect: a matrix with all ones on the diagonal (n=0 to infinity). The projection of this operator onto a finite-dimensional subspace is also what you’d expect: a matrix with all ones on the diagonal (n=0 to D).

The main stopping point is if you want to make the “completely mixed state” in the Fock basis. Since the Identity operator is not normalizable (it’s trace is infinite), the usual route to making the maximally mixed state falls apart.

It is much more common in practice to encounter CV states with non-uniform distributions over the Fock basis