Timeline for Finding the trace of a system explicitly
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 30, 2020 at 19:42 | vote | accept | DJA | ||
| Nov 29, 2020 at 22:26 | comment | added | J. Murray | Let us continue this discussion in chat. | |
| Nov 29, 2020 at 21:45 | comment | added | DJA | So the above is the way I do it since that is how I have been taught but I suppose the link below outlines how you would take the partial trace (it actually is a far more straight forward method) and therefore it makes sense the be particular about the position of the states and being consistent with their Hilbert spaces. I am still however trying to see how the method I have been taught is the same thing. thphy.uni-duesseldorf.de/~ls3/teaching/1515-QOQI/Additional/… | |
| Nov 29, 2020 at 17:35 | comment | added | J. Murray | @DJA The thing to understand is that bras are what you get when you plug a Hilbert space vector into the first slot of the inner product and leave the second slot vacant - that is, $\langle \psi|$ is a different way to write $\langle \psi, \bullet \rangle$, which acts on vectors to produce complex numbers. As long as you understand which ket vector a bra is the adjoint of, then there will ultimately be no problem. | |
| Nov 29, 2020 at 17:22 | comment | added | J. Murray | @DJA It looks like your author uses a convention whereby they reverse the order of the states in the tensor product when they take the adjoint, so $(|\alpha\rangle|\beta\rangle)^\dagger = \langle \beta|\langle\alpha|$. I think that is a bad idea for several reasons, but my opinion is not universal. In any case, if you perform an actual calculation you will find that as long as you're consistent, you'll get the right answer. | |
| Nov 29, 2020 at 17:11 | comment | added | DJA | Consider a system made up of a pair of two level atoms whereby $|g_{\alpha}\rangle $ and $|e_{\alpha}\rangle $ are the ground and excited states of atom $\alpha$ and $|g_{\beta}\rangle $ and $|e_{\beta}\rangle $ are the ground and excited states of atom $\beta$. How is it that we can denote the reduced density of $\beta$ as $\rho_{\beta} = \sum_{\nu, \mu = g_{\beta} ,e\beta} \bigg(\langle\nu| g_{\alpha} |\rho |g_{\alpha}\rangle |\mu\rangle + \langle\nu| e_{\alpha} |\rho |e_{\alpha}\rangle |\mu\rangle \bigg)|\nu \rangle \langle \mu| $ | |
| Nov 29, 2020 at 17:11 | comment | added | DJA | However my confusion arises from the below worked example and perhaps you can reconcile this with the fact that $\rho_{j,l,k,l} = \langle\alpha_j| \langle\beta_l |\rho |\alpha_k\rangle |\beta_l\rangle$. | |
| Nov 29, 2020 at 16:57 | comment | added | J. Murray | @DJA Yes. I assume you are following the notation of whatever source you're learning from, but my preference (as stated in my answer to your previous question) is to write $|\alpha_j\rangle \otimes |\beta_k\rangle \equiv |\alpha_j ,\beta_k\rangle$ rather than $|\alpha_j\rangle |\beta_k\rangle$ because the object in question is a single ket belonging to the tensor product space, and the latter naively seems to suggest otherwise. | |
| Nov 29, 2020 at 16:45 | comment | added | DJA | Oh okay so I see what your saying, and so this would suggest that $\rho_{j,l,k,l} = \langle\alpha_j| \langle\beta_l |\rho |\alpha_k\rangle |\beta_l\rangle$. | |
| Nov 29, 2020 at 15:46 | history | answered | J. Murray | CC BY-SA 4.0 |