Qualitative Explanation of Density Operator

In summary, the confusion regarding the use of density operators and state vectors arises from the difference between pure and mixed states. While pure states can be described using both the density operator and state vector, mixed states can only be described using the density operator. This is because mixed states involve probabilities, which cannot be represented by state vectors. Instead, they are represented by operators that are positive and have a unit trace. The Born Rule, which calculates the average of an observable, is then applied to these operators. Advanced textbooks often begin with the definition of states as positive operators of unit trace, although many beginner and intermediate books do not.
  • #1
astrofunk21
29
0
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
 
Physics news on Phys.org
  • #2
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).
 
  • #3
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
 

1. What is a density operator?

A density operator, also known as a density matrix, is a mathematical representation of the state of a quantum system. It contains information about the probability amplitudes of all possible states that the system can be in.

2. How is a density operator different from a wave function?

A wave function describes the state of a single quantum particle, while a density operator describes the state of an entire quantum system. The density operator takes into account the probabilities of all possible states of the system, while a wave function only describes the probabilities of a single state.

3. How is a density operator used in quantum mechanics?

The density operator is a fundamental tool in quantum mechanics, used to calculate the probabilities of different outcomes in a measurement. It also allows for the calculation of average values and other statistical properties of a quantum system.

4. What is the physical interpretation of a density operator?

The diagonal elements of a density operator represent the probabilities of the system being in a particular state, while the off-diagonal elements represent the coherences between different states. The trace of the density operator is equal to 1, representing the normalization of probabilities.

5. Can a density operator be used to describe classical systems?

No, a density operator is specific to quantum systems and cannot be used to describe classical systems. In classical physics, all probabilities are known and do not need to be represented in a density operator like they do in quantum mechanics.

Similar threads

  • Quantum Physics
Replies
21
Views
2K
  • Quantum Physics
Replies
1
Views
797
  • Quantum Physics
Replies
5
Views
1K
  • Quantum Physics
Replies
2
Views
971
  • Science and Math Textbooks
Replies
2
Views
250
Replies
27
Views
3K
Replies
16
Views
1K
Replies
3
Views
1K
  • Quantum Physics
Replies
6
Views
2K
Replies
5
Views
1K
Back
Top