I'd like to get an answer to this question from someone who knows his density matrix theory. I want to compare two different systems and ask how their density matrix representation looks.

First look at an ensemble of 100 hydrogen atoms, and let us suppose that one of those atoms is excited atoms to 2p state. Let's call this Case 1.

Now for Case 2, let us consider an alternate system where all 100 atoms are in the mixed state, each of them excited to 1% of the same 2p state as Case 1.

Two questions, really: first what is the density matrix representation of these two states? and second, how would we distinguish these two states experimentally?

EDIT: There is a thoughtful and well-reasoned answer by Mikael Kuisma which I believe is incorrect. However, it made me think about the following alternative question which I might have asked, that appears to have the same form as the question I actually did ask: yet for this modified question, Mikael's answer appeaars to be correct. Here is the alternative question:

"A beam of 100 silver atoms is shot throught a Stern-Gerlach magnet. In Case 1, there are 99 atoms spin-up and one atom spin-down. In Case 2, all the atoms are in a superposition 0.995|up> + 0.1|down>. Are these two beams the same or different; and if so, how could you distinguish them?

So the new question is: are these cases perfectly analogous...the bottle of hydrogen atoms versus the beam of silver atoms...and does the same reasoning apply to both?

  • $\begingroup$ Can you explain what you mean by "excited to 1%"? Do you mean being in a state that is a superposition of 1s and 2p, with 1% of the amplitude (squared) in 2p? $\endgroup$
    – garyp
    May 2 '16 at 20:00
  • $\begingroup$ Yes. So roughly speaking, 0.995|s> + 0.1|p> $\endgroup$ May 2 '16 at 22:44

Let's first write the density matrices. Assuming that the first atom is the excited one, we can write $$ \rho_{total}=\left|2p\right>\left<2p\right| \otimes\left(\bigotimes^{100}_{i=2}\left|1s\right>\left<1s\right| \right) $$ while in the second case we have $$ \rho_{total}=\bigotimes^{100}_{i=1}\left[{\frac{1}{ 100}}\left|2p\right>\left<2p\right| + {\frac{99}{100}}\left|1s\right>\left<1s\right| \right] $$ where $\otimes$ is the tensor product.

One experiment able to tell in which configuration we are is the measure of the energy of one atom in system. The hamiltonian of a single atom is $$ h_i = E_{1s}\left|1s\right>_i\left<1s\right|_i + E_{1s}\left|2p\right>_i\left<2p\right|_i $$ In the many particles Hilbert space, this operator becomes $$ H_i = \left(\bigotimes^{i-1}_{j=1}\mathbf{1}\right)\otimes \left( E_{1s}\left|1s\right>_i\left<1s\right|_i + E_{1s}\left|2p\right>_i\left<2p\right|_i\right) \otimes \left(\bigotimes^{100}_{i=j+1}\mathbf{1}\right) $$ where $\mathbf 1$ is the single particle identity matrix.

Measuring the energy of the $i$ particles means $$ E_i = Tr\left(\rho_{total}H_i\right)=\left<2p\right|\rho_{total}H_i\left|2p\right> + \left<1s\right|\rho_{total}H_i\left|1s\right> $$ In the first case we have $$ E_i= E_{2p}$$ if $i = 1$ and $$ E_i= E_{1s} $$ otherwise.

In the second case we have $$ E_i={\frac{1}{ 100}} E_{2p} + {\frac{99}{ 100}}E_{1s} $$ for every $i$.

Notice that, if we measure the total energy of the system instead, we obtain the same value for both the cases.

I hope this helps.

  • $\begingroup$ I am not sure that it is in the true spirit of the density matrix theory to require that one of the atoms in an ensemble be identifiable from the others, as it is when you say "let us assume it is the first atom which is in the excitied state". And the same goes for distintuishing the two systems experimentally. If you know which atom is excited, yes you can presumably measure its energy. But how do you look at a bottle of 1o0 hydrogen atoms and say THIS one is in the p state and that's the one I'm going to measure? $\endgroup$ May 3 '16 at 14:37
  • $\begingroup$ Density matrix of whole system is NOT a direct product of density matrices of its components (in following formula I put only two atoms A and B) $\rho^{AB}\neq\sum_i p_i \rho_i^A\otimes\rho_i^B$. Your derivation is plausible, but subject to this fundamental mistake. Furthermore if you create a system with such density matrix $\sum_i p_i \rho_i^A\otimes\rho_i^B$, and run on it a time evolution operator with proper hamiltonian for time $\Delta t$, then it will no longer satisfy the direct product rule, since atoms are getting entangled due to the fact that they are interacting with each other. $\endgroup$
    – cosurgi
    May 8 '16 at 20:33
  • $\begingroup$ On a separete note to ensure full anti-symmetry of a fermionic system you should for example use Slater determinant, which mixes states of all atoms to obtain complete anti-symmetry. $\endgroup$
    – cosurgi
    May 8 '16 at 20:34
  • $\begingroup$ Thanks Marty and Cosurgi, you both have good comments, and I think my explanation was a bit misleading. I thought we were talking about a conceptual experiment, in that case, if we could ignore the fact that these particles are fermions, we could write $\left | \psi\right> = \otimes \left| \phi_i\right>$ and thus, the density matrix is the tensor product of the density matrices, being $\rho= \left |\psi \right >\left <\psi\right |$. However as Cosurgi mentioned, in real world, these particles are actually fermions, which means that the full density matrix can't be written as a tensor product. $\endgroup$ May 9 '16 at 2:53

This answer is written from perspective that the atoms are indistinguisable and that you mean the 100-density matrix, not a reduced one.

The main conclusion is that these systems can be distinguished if multiple atoms can be measured simultaneously.

Assuming indistinguisable atoms, there is something in the definition of your system 2, which can lead to too hasty conclusions. However, let's proceed anyway.

There are $2^N$ distinguisable states manybody states. Assuming no interaction (i.e. they are not in a lattice, where they could be labeled), one is left with 101 states (from 0 to 100 occupation in the 2p level). The first system has one atom at 2p. Thus it has wave function $|2p^1 1s^{99}>$, which is a 100-body symmetric or antisymmetric wave function.

Now for the second system, how do we specify that each atom is 0.01 2p, since atoms are indistinguisable. We can try $\widehat A [ (a|1s>+b|2p>)^N]$, where $\widehat A$ is (anti)symmetrization operator.

(Since I'm writing this with mobile) I will not derive the result in full, but I believe that the end result will be roughly (omitting binomial factors arising from $(a+b)^n$ ):

$$\sqrt{0.99}^{100} |1s^{100}> + \sqrt{0.99}^{99} \sqrt{0.01} |1s^{99} 2p> + \sqrt{0.99}^{98} \sqrt{0.01}^2 |1s^{98} 2p^2> + \ldots \sqrt{0.01}^{N} |2p^{100}>$$

In other words. If you have each atom with 0.01 probability at 2p, you will have a probability of $0.01^{100}$ to find them all in 2p. This is clearly different from case 1, where there is at most 1 atom. If one can measure multiple atoms at same time, the systems are clearly distinguisable.

Also, the expectation value of a binomial distribution is np, so this approach kept the probabilities consistent.

You can constrict the density matrices,

$$\rho = |\psi> <\psi|$$

and nothing will change.


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