The document summarizes the key differences between pure and mixed states for a density operator. A pure state is described by a density operator ρk = |ψ⟩⟨ψ| representing a single quantum state. A mixed state is described by ρ = ∑ pi |ψi⟩⟨ ψi |, representing a statistical mixture of quantum states. The trace of the density operator is 1 for pure states and less than 1 for mixed states. The rank is also 1 for pure states and greater than 1 for mixed states. The time evolution of the density operator is given by d/dt ρk(t) = (1/iħ)[H

1. PRESENTATION ON
Difference between pure and mixed state for a density operator
&
Density operator for time evaluation
Presented by:
Group: 4
MS-231703, MS-231709, MS-
231718, MS-231722, MS-231727,
MS-231737, MS-231739, MS-
231740, MS-231745

2. Density operator: In quantum physics, the density operator, often denoted by ρ is a mathematical
construct used to describe the state of a quantum system.
Mixed state: The state that we use to describe situations where we only have partial information are
called.
The density operator for a mixed state is ρ = ∑ pi |ψi⟩⟨ ψi |.
Pure state: Full information describe with |Ψ > .
The density operator for a pure state is given , 𝜌𝑘 = |ψ⟩⟨ψ|.
An example of expectation value in pure state,
Density operator equivalent to State vector
< 𝐴 >= 𝑇𝑟 𝐴. 𝐴 < 𝐴 > =< 𝜓𝑙 𝐴 𝜓𝑘 >

3. Also for the measurements ,
The probability of measuring eigen value of 𝜆𝑚 with projector operator 𝑃𝑚 = 𝑢𝑚 >< 𝑢𝑚 is,
𝑃 𝜆𝑚 = 𝑇𝑟 𝜌𝑘𝑃𝑚 is equivalent to 𝑃 𝜆𝑚 = < 𝜓𝑚 𝑃
𝑚 𝜓𝑚 >
State vector are more convenient in pure state.
For mixed state it is more convenient because in all these formulas the density operator appeared in linear manner .
For mixed state, |𝜓𝑘 > with probability 𝜌𝑘 is 𝑘 𝜌𝑘 = 1.
For statistical mixture:
The probability of measuring 𝜆𝑚 in pure state is,
𝑃𝑘 𝜆𝑚 = 𝑇𝑟(𝜌𝑘𝑃
𝑚) where 𝑃𝑚 = |𝑢𝑚 >< 𝑢𝑚| is the projector operator.

4. The total probability of measuring 𝜆𝑚 in statistical mixture is equal to the sum of likelihood of each state
there multiplied the probability 𝜆𝑚.
𝑃 𝜆𝑚 = 𝑘 𝑝𝑘𝑇𝑟(𝜌𝑘𝑃𝑚)
𝑊ℎ𝑒𝑟𝑒 𝑇𝑟 is a linear operation so that 𝑇𝑟 𝑎𝑀 = 𝑎𝑇𝑟 𝑀 = 𝑎𝑇𝑟 𝑀
Now applying this condition above equation,
𝑃 𝜆𝑚 = 𝑇𝑟 𝑘 𝑝𝑘
𝜌𝑘
𝑃𝑚 = 𝑇𝑟 𝜌𝑃𝑚 [ where 𝜌 𝑠𝑐𝑎𝑙𝑎𝑟 𝑚𝑎𝑡𝑟𝑖𝑥]
This is the over all probability of 𝜆𝑚 is mixed state.
Then this is same as the probability outcome of measurement in a pure state.

5. Normalization:
Pure state: Tr (𝜌𝑘) = 1
Let us write an analogous formula for mixed state.
ρ = 𝑘 𝑝𝑘𝜌𝑘
Tr(ρ) = Tr ( 𝑘 𝑝𝑘𝜌𝑘) = 𝑘 𝑝𝑘𝑇𝑟(𝜌𝑘) [ Applying linearity operation]
= 𝑘 𝑝𝑘
= 1
For statistical mixture seem over all probabilities is equal to 1.

6. Difference between pure and mixed state for a density operator:
1.Trace of the Density Operator:
•For a pure state : 𝑇𝑟(𝜌𝑘) = 1.
•For a mixed state : Tr(ρ) < 1.
2.Rank of the Density Operator:
•For a pure state, the rank is 1, corresponds to a single ket vector.
•For a mixed state, the rank is greater than 1, represents a mixture of different pure states.
In summary, a pure state is characterized by a density operator that represents a single, definite quantum
state, while a mixed state is characterized by a density operator that represents a statistical mixture of
multiple quantum states with associated probabilities.

7. Density operator for Time evaluation:
 For pure state, applying Hamiltonian operator,
𝑑
𝑑𝑡
𝜌𝑘 𝑡 =
1
𝑖ℏ
[𝐻 𝑡 , 𝜌𝑘(𝑡)]
 For mixed state, we don’t know what precise state we are in, we know the statistical distribution of
states.
If the Hamiltonian 𝐻 𝑡 of the system is known, then the derivative of the density operator for the mixed
state,
𝑑
𝑑𝑡
𝜌 𝑡 =
𝑘
𝑝𝑘
𝑑
𝑑𝑡
𝜌𝑘 𝑡
= 𝑘 𝑝𝑘
1
𝑖ℏ
𝐻 𝑡 , 𝜌𝑘(𝑡)
=
1
𝑖ℏ
𝐻 𝑡 , 𝑘 𝑝𝑘
𝜌𝑘(𝑡)
∴
𝑑
𝑑𝑡
𝜌 𝑡 =
1
𝑖ℏ
[𝐻 𝑡 , 𝜌(𝑡)]

8. It is very convenient change the density operator for pure state to the mixed state by this relations ,
𝜌𝑘 = Ψ𝑘 × Ψ𝑘 → 𝜌 = 𝑘 𝑝𝑘 𝜌𝑘
Then for pure state: ⟩
|Ψ𝑘 ⇒ 𝜌𝑘 = Ψ𝑘 × Ψ𝑘
This 𝜌 is a projection operator and it is independent, i.e., 𝜌𝑘
2
= 𝜌𝑘
𝑇𝑟 𝜌𝑘
2
= 𝑇𝑟(𝜌𝑘)
= 1 [as normalization]

9. For pure state : 𝑇𝑟 𝜌𝑘
2
= 1
For mixed state, 𝜌 = 𝑘 𝑝𝑘 Ψ𝑘 × Ψ𝑘
𝜌2
= 𝑘,𝑙 𝑝𝑘𝑝𝑙 Ψ𝑘 × Ψ𝑘 Ψ𝑙 × Ψ𝑙|
𝜌2 = 𝑘 𝑝𝑘
2
Ψ𝑘 × Ψ𝑘 [different states are orthogonal]
= 𝑘 𝑝𝑘
2
𝜌𝑘
∴ 𝑇𝑟 𝜌2
= 𝑇𝑟 𝑘 𝑝𝑘
2
𝜌𝑘 = 𝑘 𝑝𝑘
2
𝑇𝑟𝜌𝑘 = 𝑘 𝑝𝑘
2
In mixed state at least two different 𝑝𝑘 are nonzero because if we only had one nonzero 𝑝𝑘 , then that
would be a pure state.
For mixed state 𝑘 𝑝𝑘 = 1 [ 0 ≤ 𝑝𝑘 < 1]

10. If we only had one nonzero 𝑝𝑘 , then that would be a pure state.
So in these condition, 𝑝𝑘 is larger than or equal to zero.
As the two 𝑝𝑘 is nonzero for mixed state, then no 𝑝𝑘 can be 1. i.e. 𝑝𝑘 < 1.
As the sum over all 𝑝𝑘 is 1, then sum over all 𝑃𝑘
2
must be smaller than 1. Then, 𝑝𝑘
2
< 𝑝𝑘.

11. THANK YOU