Graduate Qualitative Explanation of Density Operator

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The discussion centers on the confusion surrounding the density operator in quantum mechanics, particularly its necessity over state vectors for mixed states. The density operator allows for a more robust representation of quantum states, especially when dealing with probabilities that do not conform to classical interpretations. It is emphasized that while pure states can be described by state vectors, mixed states require the use of operators to maintain the properties of positive probabilities. The transition from state vectors to density operators is crucial for handling the complexities of quantum systems. Understanding this distinction is essential for grasping advanced quantum mechanics concepts.
astrofunk21
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Hey all!

I am prepping myself for a quantum course next semester at the graduate level. I am currently reading through the Cohen-Tannoudji Quantum Mechanics textbook. I have reached a section on the density operator and am confused about the general concept of the operator.

My confusion stems from the question, why can't we continue to state vectors to describe a system all the time. The textbook says that it leads to clumsy calculations if we apply weighting (pk) to a certain state (|ψk>) and sum over k.

In the pure case we can use both the density operator or state vector to describe the system. Yet with a mixed state we cannot. Where do these two methods diverge and the state vector method fails?

Maybe I am being blind, but this concept has seemed to stump me so far. Would appreciate a qualitative explanation to maybe sort this out.

I appreciate any help you guys give!

Textbook image: https://ibb.co/di6MO6
di6MO6
 
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Because when we project a state vector in terms of its (observable) eigenstates: ##|\Psi\rangle = \sum_i a_i |\psi_i\rangle## the ##a_i##, as "components" in a vector space, do not necessarily have the properties required of classical probabilities (non-negative because they are relative frequencies).
 
Vector pure states, which is what you have been exposed to so far, do not depend on phase ie e^ix |u> is the same state. That means to uniquely define the state you need notation that removes the phase. This is done by switching to the operator |u><u| - change the phase and it doesn't change. Now let's suppose you have a set of states that could be any of |ui> with probability pi then you can represent this in the following way U = ∑pi |ui><ui|. This is called a mixed state and states have now changed to operators rather than vectors. They are positive operators of unit trace. The Born Rule then becomes the average of an observable O, E(O) = trace(OU).

Best IMHO, but beginner and some intermediate books, don't do this, but advanced book do, is to start from the definition of states as positive operators of unit trace - but for reasons best known to them they don't do that.

Thanks
Bill
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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