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PeterDonis
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That still doesn't answer the question I asked in post #29 about the Hamiltonian.deepalakshmi said:I evolved it with time evolution operator.
That still doesn't answer the question I asked in post #29 about the Hamiltonian.deepalakshmi said:I evolved it with time evolution operator.
a, b are annihilation operator. b is similar to a. just change in variablePeterDonis said:That still doesn't answer the question I asked in post #29 about the Hamiltonian.
I don't understand. You only have one set of number operator states ##\ket{n}##. That means you have only one annihilation operator and one creation operator. "Change in variable" makes no sense here.deepalakshmi said:b is similar to a. just change in variable
Not here.deepalakshmi said:Is there any other way to share my work.
Then you will need to decide how much of it to post here. First, however, you might try answering post #38, and also post #30.deepalakshmi said:Calculation is 3 pages
I wrote the number terms in terms of creation operator. Then I inserted UUdagger which is basically time evolution operator. Later I applied BadaggerBdagger formula( baker campbell formula). I got a dagger cos theta+ ib dagger sin theta. I substituted this in place of creation operator. Later I used binomial expression and simplifiedPeterDonis said:I don't understand. You only have one set of number operator states ##\ket{n}##. That means you have only one annihilation operator and one creation operator. "Change in variable" makes no sense here.
As I said earlier, the initial state and hamiltonian is given to me by my guide.PeterDonis said:Not here.Then you will need to decide how much of it to post here. First, however, you might try answering post #38, and also post #30.
deepalakshmi said:##e^{-\frac{|\alpha|^2}{2}}\frac{(\alpha)^n}{\sqrt{n!}}|n>##
You forgot the sum over ##n##.deepalakshmi said:##e^{-\frac{|\alpha|^2}{2}}\frac{(\alpha)^n}{\sqrt{n!}}|n>##
Yeah I forgotDemystifier said:You forgot the sum over ##n##.
Think harder. It poses no contradiction.deepalakshmi said:actually the fidelity is always between 0 to 1 right? But I got exponential terms with alpha
Ok. I understood. Since there is a exponential power minus, value always lies between 0 to 1. I also took an example and the value always lies between 0 to 1. ThankyouHaorong Wu said:Think harder. It poses no contradiction.
You are still not answering my question. Your formula for the Hamiltonian has two sets of ladder operations, the a's and the b's. But you only have one set of number eigenstates ##\ket{n}##, so you should only have one set of ladder operators. Why are there two?deepalakshmi said:I wrote the number terms in terms of creation operator.
Who is your "guide"?deepalakshmi said:the initial state and hamiltonian is given to me by my guide.
Guide - teacherPeterDonis said:You are still not answering my question. Your formula for the Hamiltonian has two sets of ladder operations, the a's and the b's. But you only have one set of number eigenstates ##\ket{n}##, so you should only have one set of ladder operators. Why are there two?
Who is your "guide"?
I'm saying I don't understand why there are two sets of ladder operators in your Hamiltonian since you only have one set of number eigenstates. If you don't understand that either, perhaps you should ask your teacher.deepalakshmi said:Are you saying my question is wrong?
What do the two sets of operators operate on?deepalakshmi said:take beam splitter with input state as coherent and vacuum state. There also you have two sets of operator in Hamiltonian
Note: I've said this several times now, but you should think carefully about whether you agree with it.PeterDonis said:you only have one set of number eigenstates
Ok. I will ask my teacher about it.PeterDonis said:Note: I've said this several times now, but you should think carefully about whether you agree with it.
There those two set of operator will act on two states. One is coherent state which I will write in terms of vacuum state and other is vacuum state. So basically the two operator will act on two vacuum statePeterDonis said:I'm saying I don't understand why there are two sets of ladder operators in your Hamiltonian since you only have one set of number eigenstates. If you don't understand that either, perhaps you should ask your teacher.What do the two sets of operators operate on?
What do those two states represent, physically--for example, in the beam splitter scenario?deepalakshmi said:There those two set of operator will act on two states.
They represent vacuum state .i.e zero fock state where there is no photonsPeterDonis said:What do those two states represent, physically--for example, in the beam splitter scenario?
First, only one of the two kets in your initial state is the vacuum state, the other is the coherent state.deepalakshmi said:They represent vacuum state .i.e zero fock state where there is no photons
I can write coherent state in terms of vacuum state. So I have two vacuum state. Secondly for this initial state, the final state is not entangled. Lastly I don't know anything about degrees of freedom.PeterDonis said:First, only one of the two kets in your initial state is the vacuum state, the other is the coherent state.
Second, you are missing the point. You have an initial state ##\ket{0} \ket{\alpha}##. This is a product of two kets. That means you have two physical degrees of freedom in the quantum system that this state describes. What are those two physical degrees of freedom, in, for example, the beam splitter scenario?
You then time evolve that initial state using a Hamiltonian to get a final state. That final state will also describe a quantum system with two physical degrees of freedom (but in general it won't be a product state, it will be entangled). What are those two physical degrees of freedom in the final state, in, for example, the beam splitter scenario? (Hint: they're not the same as the two physical degrees of freedom in the initial state, because of the beam splitter.)
Thanks for mentioning degrees of freedom in Quantum states. I will learn about it.deepalakshmi said:I can write coherent state in terms of vacuum state. So I have two vacuum state. Secondly for this initial state, the final state is not entangled. Lastly I don't know anything about degrees of freedom.
No, you can't. Only one term in the infinite sum of number eigenstates in the formula for a coherent state is the ##n = 0## term, i.e., the vacuum state. All the other terms have ##n > 0## so they contain some nonzero number of photons.deepalakshmi said:I can write coherent state in terms of vacuum state.
Degrees of freedom are what kets refer to. If you have a product of two kets, you have two degrees of freedom. Physically, kets generally describe physical systems that can interact with other physical systems.deepalakshmi said:I don't know anything about degrees of freedom.
PeterDonis said:No, you can't. Only one term in the infinite sum of number eigenstates in the formula for a coherent state is the ##n = 0## term, i.e., the vacuum state. All the other terms have ##n > 0## so they contain some nonzero number of photons.
Ok, so this is what you meant by writing a coherent state "in terms of" the vacuum state. I can similarly write any state "in terms of" any other state by just finding the appropriate operator to apply. So this "in terms of" doesn't seem very useful to me.deepalakshmi said:I can write coherent state as (displacement operator) ( vacuum state).
You still have not replied to the most important question I have asked, several times now: what do the two kets in your initial state ##\ket{0} \ket{\alpha}## refer to in the beam splitter scenario? To put it as simply as possible: this state says you have one "thing" that is in the vacuum state, and another "thing" that is in the coherent state ##\ket{\alpha}##. What are these two "things" in the beam splitter scenario?deepalakshmi said:I will reply later if you ask any question