KFC
Assume there is a two level system, two eigenstates are written as

$$|\psi_1\rangle = \cos\theta |1, g\rangle + \sin\theta |0, e\rangle$$
and
$$|\psi_2\rangle = -\sin\theta |1, g\rangle + \cos\theta |0, e\rangle$$

For the density operator of the system is written as

$$\rho = \frac{1}{2}|\psi_1\rangle\langle \psi_1| + \frac{1}{2}|\psi_2\rangle\langle \psi_2| = \frac{1}{2}|1, g\rangle\langle 1, g| + \frac{1}{2}|0, e\rangle\langle 0, e|$$

where $$g$$ stands for ground state, $$e$$ stands for excited state, 0 and 1 stands for the number of photon.

If the inital state of the system is in $$|0, e\rangle$$, what's the probability of transition from $$|0, e\rangle \to |1, g\rangle$$ ? I am quite confuse how to use density operator to find the probability, shoud it be

$$\langle1, g|\rho|0, e\rangle$$

or

$$\left|\langle1, g|\rho|0, e\rangle\right|^2$$ ?

Homework Helper
Gold Member
If you already know the initial state of the system, there is no need to use density operators.

Is $$|1, g\rangle$$ an eigenstate of some observable? If so, then the probability of getting the eigenvalue corresponding to $$|1, g\rangle$$ when you measure that observable is $$\left|\langle1, g|0, e\rangle\right|^2$$ (if the original state is $$|0, e\rangle$$).

KFC
If you already know the initial state of the system, there is no need to use density operators.

Is $$|1, g\rangle$$ an eigenstate of some observable? If so, then the probability of getting the eigenvalue corresponding to $$|1, g\rangle$$ when you measure that observable is $$\left|\langle1, g|0, e\rangle\right|^2$$ (if the original state is $$|0, e\rangle$$).

Yes. $$|1, g\rangle$$ is an eigenstate of some observable. But $$\rho$$ is evolve with time, so we need to use density operator.

Homework Helper
Gold Member
Density operators are used when you have an ensemble of systems in a mixed state. You said that the state of your system is $$|0, e\rangle$$ right? That's a pure state. If you have a system in the pure state $$| \psi \rangle$$, then probability of getting some eigenvalue a of an observable is given by $$|\langle a|\psi\rangle|^2$$ where $$|a\rangle$$ is the eigenstate corresponding to the eigenvalue a.

KFC
Density operators are used when you have an ensemble of systems in a mixed state. You said that the state of your system is $$|0, e\rangle$$ right? That's a pure state. If you have a system in the pure state $$| \psi \rangle$$, then probability of getting some eigenvalue a of an observable is given by $$|\langle a|\psi\rangle|^2$$ where $$|a\rangle$$ is the eigenstate corresponding to the eigenvalue a.

OK. Forget about my previous question. I just want to know, if I know the density operator, how to find the transition probability? For example, consider jumping from $$|a\rangle \to |b\rangle$$, which of the following is the transition probability?

$$\langle a | \rho |b\rangle$$

or

$$\left|\langle a | \rho |b\rangle\right|^2$$

deferro
Density matrices apply when states are unknown (in ensembles, as mentioned). These are mixed states = mixed pure states.

Any help?

Gold Member
OK. Forget about my previous question. I just want to know, if I know the density operator, how to find the transition probability? For example, consider jumping from $$|a\rangle \to |b\rangle$$, which of the following is the transition probability?

$$\langle a | \rho |b\rangle$$

or

$$\left|\langle a | \rho |b\rangle\right|^2$$

The first one.

jrlaguna
None. There is a conceptual mistake. Let's say that you start with |a> and, after a while, the state is given by $$\rho$$. Now you want to know the probability of measuring the state |b>, right? So, forget about |a>. The rule is the trace of the product:

$$Tr(\rho |b><b|)$$

since |b><b| is the projector on |b>, same as in all measurements,

$$<A>=Tr(\rho A)$$

KFC
None. There is a conceptual mistake. Let's say that you start with |a> and, after a while, the state is given by $$\rho$$. Now you want to know the probability of measuring the state |b>, right? So, forget about |a>. The rule is the trace of the product:

$$Tr(\rho |b><b|)$$

since |b><b| is the projector on |b>, same as in all measurements,

$$<A>=Tr(\rho A)$$

I am not sure I understand it correctly, tell me if I am wrong.

First of all, if we start from an initial state and later the density operator evolve into $$\rho(t)$$, according to your statement, the transition probability from initial state to final state is only relate to final state as follow ?

$$Tr(\rho(t) |b><b|) = <b|\rho(t)|b>$$

right?

So, any thing like $$<b|\rho(t)|a>$$ doesn't make any sense in physics?

Staff Emeritus
Gold Member
OK. Forget about my previous question. I just want to know, if I know the density operator, how to find the transition probability? For example, consider jumping from $$|a\rangle \to |b\rangle$$, which of the following is the transition probability?

$$\langle a | \rho |b\rangle$$

or

$$\left|\langle a | \rho |b\rangle\right|^2$$
You already got an answer. I'm just trying to make it more clear (because I didn't immediately understand the answer myself). If the system is in state $\rho=|a\rangle\langle a|$, when you measure an observable B, the probability of getting the result b is

$$|\langle b|a\rangle|^2=\langle b|a\rangle\langle a|b\rangle=\langle b|\rho|b\rangle =\sum_n\langle b|n\rangle\langle n|\rho|b\rangle =\sum_n\langle n|\rho|b\rangle\langle b|n\rangle =\mbox{Tr}(\rho |b\rangle\langle b|)$$

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jrlaguna
They key is... what do you want |a> for? It's gone... The current state is given by $$\rho$$.

Gold Member
$$|<a|\rho |b>|^2=|<a|a><a|b>|^2=|<a|b>|^2$$
cause |<a|a>|=1 if it's normalised.
So the second option.

jrlaguna
You're assuming $$\rho=|a><a|$$ for some reason I don't quite grasp. That was so at the beginning, $$\rho(0)=|a><a|$$, but not any more with time evolution...

Gold Member
Well, For the density matrix in the first post, it doesn't change with time.

deferro
Expectation values is tricky, the measurement is the operator.
If you prepare a mixed state, you can measure a pure state.

A density matrix is an involution of two states as $$|\psi \rangle_a|\phi \rangle_b \otimes {}_a \langle \psi| {}_b\langle \phi|$$
You need the inner product $$\varsigma_a\otimes \varsigma_b$$ and measurement = operator. A conjugate pair of trace products that evolve an outer product. You have the ground state so you derive the inner measure on a bipartite state.
Squaring the outer product will give you the same product. It's like the first involution is a and b, the inner measurement is a or b, a sum you have to find, so you can select a, or select b.

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Staff Emeritus
Gold Member
You're assuming $$\rho=|a><a|$$ for some reason I don't quite grasp. That was so at the beginning, $$\rho(0)=|a><a|$$, but not any more with time evolution...
He didn't specify that time has passed beween the time when the system was prepared in state |a> and the time when B is measured. In fact, he didn't even say that the system was prepared in state |a>. He just asked about the transition $|a\rangle\rightarrow|b\rangle$.

jrlaguna
Then, I don't get the meaning of the question. I prepare a system in a state A ($$\rho$$), then I ask for its probability of jumping from a state B ($$|a>$$) to a state C ($$|b>$$)? What is $$\rho$$, then?

deferro
The only way to prepare a state is by mixing pure state stochastics (with Pauli operators).

your $$\rho$$ is a density matrix. You need 4 probable states for the evolution of the measurement when you use the perturbation (magnetic or photonic for electron spin), which are P00, P01, P10, P11.
The first and last are pure states and you can't measure them except as pure states, with mixed state preparation, but over all electrons in the measurement. The only ones you can mix are the middle two inner exclusive, or "XOR states", you can only select one of 4, the others vanish in the measurement.

Perhaps teach has set a question for homework that the students haven't learned enough yet to do? Probability densities are a bit of a plow-through. You need to do a bit of matrix algebra, and you treat each Pij as a Pauli matrix; all quanta can be prepared in 4 possible states, ensembles or 2-qubit states have more separability - they absorb more entropy than one.

Think about having two coins and the possible outcomes, the coin model is a frequently used one, since states prepared or otherwise evolved have a classical correlation.

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