User John Jack | chat.stackexchange.com

John Jack

Jul 12, 2017 20:43

@Danu
Expand using $e^{a \hat{A}} = \sum_{n= 0}^{\infty} \frac{a^n}{n !}\hat{A}^n$.?

John Jack

Jul 12, 2017 20:39

@DanielSank I didn't...I am trying to. This is in 'Quantum Optics' by Gerry Knight. Do you agree with it?

John Jack

Jul 12, 2017 20:37

If $| \alpha \rangle$ is a coherent state and $\hat{n} = \hat{a}^{\dagger}\hat{a}$ the number operator, would it be true that $e^{-i \omega t \hat{n}} | \alpha \rangle = |\alpha e^{-i \omega t} \rangle$ ?

John Jack

Jul 12, 2017 20:35

hi alll

John Jack

Jul 9, 2017 17:19

@ACuriousMind Understood thanks.

John Jack

Jul 9, 2017 17:18

@ACuriousMind Does it follow then as a property that for measurement operators, say $\{ A_m \}_m$, we always have that $Tr[A_m P A_m^{\dagger}] \geq 0$?

John Jack

Jul 9, 2017 17:13

@ACuriousMind Okay thanks for checking. Out of interest, what are you using as your definition of positive operator, I am using an "Hermitian operator with non-negative eigenvalues".

John Jack

Jul 9, 2017 17:11

@ACuriousMind They are both zero because $PQ = 0$ and $QP = 0$.

John Jack

Jul 9, 2017 17:06

@ACuriousMind Since $P$ and $Q$ are simultaneously diagonalised, the sum $P + Q$ is a diagonal matrix with the sum of eigenvalues of $P$ and $Q$ respectively on the diagonal. Since they are positive operators, they respectively have non-negative eigenvalues hence it follows that $|P + Q| = P + Q$...?

John Jack

Jul 9, 2017 17:00

@ACuriousMind
So if I assume that $P$ and $Q$ are simultaneously diagonalized then I can simply add the conclusion as the final step: $|X| = |\rho - \sigma| = |P-Q| = \sqrt{(P-Q)^2} = \sqrt{P^2 + Q^2} = |P + Q| = P + Q$?

John Jack

Jul 9, 2017 16:53

@ACuriousMind I can see the first part but not how the simultaneous diagonalizing with the definition implies the conclusion.
So what I have is The first part you mention is: $$PQ~~\text{orthogonal} \Leftrightarrow PQ = 0 \Leftrightarrow [P,Q] = 0 \Leftrightarrow P~\text{and }~Q~~\text{can be diagonalized simultaneously.}$$

Then from definition $|X| = |\rho - \sigma| = |P-Q| = \sqrt{(P-Q)^2} = \sqrt{P^2 + Q^2} = |P + Q|$

What am I missing?

John Jack

Jul 9, 2017 16:28

If we consider the operator $X = \rho - \sigma$ (where $\rho$ and $\sigma$ are density operators). Then $X$ is hermitian. We can also show $X$ can be written as the difference of two positive operators that have support on an orthogonal subspaces (support of an operator is the space spanned by eigenvectors with non-zero eigenvalues). Hence $X = P - Q$. Can you see from the definition of $| \cdot |$ why $|\rho - \sigma| = P + Q$?

John Jack

Jul 9, 2017 16:25

Will just take a second to type...

John Jack

Jul 9, 2017 16:22

@ACuriousMind I have just one more question if you have a second...@0celo7

John Jack

Jul 9, 2017 16:20

Finite dimensional yes there's no assumption of symmetric matrix. Yes I see since it is basis-independent it makes sense @ACuriousMind Yes I understand.

John Jack

Jul 9, 2017 16:18

@0celo7 Oh okay. It's stated as $|X| = \sqrt{X^2}$ but then says that "equivalently it is the matrix obtained from $X$ making all it's eigenvalues equal to their absolute values". Does this make sense to you?

John Jack

Jul 9, 2017 16:15

@0celo7 No to what?

John Jack

Jul 9, 2017 16:15

@0celo7 It's supposed to be a matrix.

John Jack

Jul 9, 2017 16:13

@ACuriousMind I'm just confirming that $Tr[|\rho|]$ is basis independent based on the definition of $|\cdot|$ that I gave.

John Jack

Jul 9, 2017 16:11

@0celo7 That's not how it is stated in the text but I would assume that's what is meant. Have you seen this kind of definition before?

John Jack

Jul 9, 2017 15:12

Does anyone know the definition of the absolute value of a matrix $X$ defined by $|X| = \sqrt{X^2}$? I just want to confirm that if I was interested in taking $tr[|X|]$, then this trace is independent of the basis we choose for $X$?

John Jack

Jul 6, 2017 17:23

@EmilioPisanty Yeah that's the kind of idea I am still working with.

John Jack

Jul 6, 2017 17:20

@EmilioPisanty Do you see it as photon particles with higher energy or is that a silly way of modelling it...do you just think of it as eigenstates?

John Jack

Jul 6, 2017 17:19

@EmilioPisanty Oh okay thanks.

John Jack

Jul 6, 2017 17:12

@EmilioPisanty What's so important about $n = 1$?

John Jack

Jul 6, 2017 17:11

@EmilioPisanty Is it what is commonly called a number state? The eigenstate of $\hat{N}$ where $\hat{H} = \hbar \omega(\hat{N} + \frac{1}{2})$?

John Jack

Jul 6, 2017 17:07

@EmilioPisanty I mean how does it enter into the description, what does a 'photon' refer to in this description?

John Jack

Jul 6, 2017 17:04

@ACuriousMind :) A few messages above my last message.

John Jack

Jul 6, 2017 17:03

@ACuriousMind What do the photons correspond to when you describe a quantized single mode field (as I described above)?

John Jack

Jul 6, 2017 16:54

Do we consider $\hat{p}$ and $\hat{q}$ as Hermitian operators, since we equate them with the position and momentum operators (as they are in the Harmonic oscillator case in QM)?

John Jack

Jul 6, 2017 16:53

In a text I am using it shows the quantization of a single mode electric field in a cavity by showing that the Hamiltonian is analagous to the Hamronic oscillator and then basically just shows that the energy of hamrnoic oscillator is quantized. It then considerd the electric field as an operator:$$\hat{E_x}(z,t) = (\frac{2 \omega^2}{V \epsilon_0})\hat{q}(t)\sin(kz)$$ using Maxwell's equation we find the corresponding magnetic field operator in terms of $\hat{p}(t)$.

John Jack

Jul 6, 2017 16:47

Hello all

John Jack

Jul 4, 2017 16:40

@ACuriousMind I know that two commuting matrices can be simultanously diagonalised...In this case, is the idea that since the $diag(0,0,D_i,0,)$ commute with each other they diagonalise the blocks seperately hence we can apply them all at once in any order since they only affect one block at a time?

John Jack

Jul 4, 2017 16:32

@ACuriousMind So your $D_{1}$ is some unitary matrix such that $D_{1}\rho_1 D_{1}^{\dagger}$ is diagonal?

John Jack

Jul 4, 2017 16:25

@ACuriousMind It states further that it is irrelevant that each of the $\rho_n$ will have a different eigenbasis since $V$ is block-diagonal where each $\rho_n$ is one of the blocks. I don't really see how that helps.

John Jack

Jul 4, 2017 16:10

Oh Damnit

John Jack

Jul 4, 2017 16:07

@ACuriousMind That would resolve it it seems. Where did you find that result? There are so many of these results but that one seems very important.

John Jack

Jul 4, 2017 15:48

@ACuriousMind I think they do because to diagonalize each of $\rho_n$ at a time you would need a $U_n$ for each of them right? And so we would still get $S(V) = S(
\sum_n p_n U_n\rho_n U_n^{\dagger} \otimes |n \rangle \langle n |)$ in order for the statement "you can diagonalize each $\rho_n$ without changing the total entropy of the joint state $V$" to make sense. We are diagonalizing $V$ so all $\rho_n$ need to be diagonalized.

John Jack

Jul 4, 2017 15:33

@ACuriousMind How else would you diagonalize each $\rho_n$ than to find a unitary $U_n$ for each $n$? The claim from the text is that "you can diagonalize each $\rho_n$ without changing the total entropy of the joint state $V$"....

John Jack

Jul 4, 2017 15:31

@ACuriousMind That's my claim. Each $\rho_n$ would need a different $U_n$ to diagonalize it (possibly need a different $U_n$).

John Jack

Jul 4, 2017 15:30

@ACuriousMind Yeah

John Jack

Jul 4, 2017 15:28

@ACuriousMind It doesn't show it for the case of a series of $n$. It shows it for the one case I mentioned where you have $\rho_n \otimes |n \rangle \langle n|$. That's what I'm asking. It seems you would need to invoke linearity to extend to the series case.

John Jack

Jul 4, 2017 15:23

@ACuriousMind From time to time it becomes easier to consider in the way I mentioned.
For each block $\rho_n$ I know that the entropy is invariant under unitary transformation $U_n$ i.e. $S(\rho_n) = S(U_n \rho_n U_n^{\dagger})$. Further if we define $U_n \otimes I$ then $S(U_n \rho_n U_n^{\dagger} \otimes |n \rangle \langle n|) = S(\rho_n) + S(|n \rangle \langle n |) = S(\rho_n)$.

But, do you see though how for $V = \sum_n p_n \rho_n \otimes |n \rangle \langle n |$ we have that $S(V) = S(
\sum_n p_n U_n\rho_n U_n^{\dagger} \otimes |n \rangle \langle n |)$. Since $S$ is not a linear operat…

John Jack

Jul 4, 2017 15:03

@ACuriousMind Yeah but it becomes trickier when working with tensor products so I'm just confirming. So you would have to diagonalize each block matrix of $V$ to find $\lambda_n$ ?

John Jack

Jul 4, 2017 14:58

@ACuriousMind So your answer is yes?

John Jack

Jul 4, 2017 14:55

@ACuriousMind Do you know if we simply define the von Neumann entropy of a joint system say $V = \sum_n \rho_n \otimes |n \rangle \langle n |$ as $-\sum_n\lambda_n \text{ln} \lambda_n$ where $\lambda_n$ are the diagonal elements of a dialgonalized $V$?

John Jack

Jul 4, 2017 12:41

Is it simply because if $U$ is unitary then $U \otimes I$ is unitary on tensor products and von Neuamnn entropy is invariant under unitary operators?

John Jack

Jul 4, 2017 12:38

Suppose you have the joint state of two systems averaged: $$\sum_{n}p_n \rho_n \otimes |n \rangle \langle n |$$ where $p_n$ is the probability of being in that joint state. Why would it follow that we can diagonalize the states $\rho_n$ without affecting the total von Neumann entropy of the joint state?

John Jack

Jul 3, 2017 18:47

@Mithrandir24601 Okay kewl no prob.

John Jack

Jul 3, 2017 18:02

@Mithrandir24601 Hey do you maybe know why it follows that if a measurement is made on a system in state $\rho$, which has final states $\{ \tilde{\rho}_n \}$ with corresponding probabilities $\{ p_n \}$ then we have that $$S(\rho) - \sum_n p_n S(\tilde{\rho}_n) \leq H[\{ p_n \}]$$
where $S$ is the von Neumann entropy and $H$ is the Shannon entropy.

John Jack

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