Understanding the Two Types of Density Operators in Quantum Mechanics

In summary, the conversation is about the confusion regarding the two density operators, \rho=\sum_{i}\delta(r-r_{i}) and \rho=\sum_{i}|\psi_{i}>\rho_{ii}<\psi_{i}|. The person is asking for an explanation and clarification on the relationship between the two operators. The expert is unable to make sense of the first expression, as it does not seem to follow the standard definition for a continuous spectrum. The person is then asked for more details on the first expression and its source.
  • #1
Liao Chen
2
0
I'm confused about the two density operators:

\rho=\sum_{i}\delta(r-r_{i}) and \rho=\sum_{i}|\psi_{i}>\rho_{ii}<\psi_{i}|

Is there anyone explaining this question to me? Thanks very much.
 
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  • #2
The first one is not normalized if your sum has more than one term.
 
  • #3
arkajad said:
The first one is not normalized if your sum has more than one term.

Thanks a lot. Do you mean the two density operators are the same and connected through some transformations? Could you explain with a little more details?
 
  • #4
After some thinking I really do not know what the first expression could mean. It does not make any sense to me. If I consider it as an operator, it would act as

[tex](\rho\psi)(x')=\int \delta(x-x_i)\psi(x)dx=\sum_i\psi(x_i)[/tex]

which is a number and not a function. For a continuous spectrum the formula should look like

[tex](\rho\psi)(x')=\int \rho(x',x)\psi(x)dx[/tex]

So, where did you get it from?
 

Related to Understanding the Two Types of Density Operators in Quantum Mechanics

1. What are the two kinds of density operators?

The two kinds of density operators are pure states and mixed states. A pure state is described by a single wave function, while a mixed state is described by a statistical combination of multiple wave functions.

2. How are pure states and mixed states different from each other?

Pure states have a well-defined quantum state and can be described by a single wave function, while mixed states have a probabilistic distribution of quantum states and require a statistical approach for description.

3. What is the significance of pure states in quantum mechanics?

Pure states are important in quantum mechanics because they represent idealized systems with well-defined quantum states. They allow for precise predictions of the outcome of measurements and can be used to describe simple systems such as single particles.

4. How are mixed states used in quantum information processing?

Mixed states are used in quantum information processing to represent systems with uncertainty or noise. They are also used in quantum cryptography and quantum computing, where the manipulation of mixed states is necessary for performing certain tasks.

5. Can a pure state ever become a mixed state?

No, a pure state cannot become a mixed state. This is because a pure state represents a well-defined quantum state, while a mixed state represents a probabilistic distribution of quantum states. A system can, however, transition from a pure state to another pure state through quantum evolution.

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