This thread has the following 2 goals:
- Knowing how to properly write the below considered cases.
- To help individuals who might want to know the correct expression and calculation for a specific case.
It will be a lengthy one, but I believe it benefits, me personally, but also other members of the community as well. I believe my writing and derivation is not 100% correct, that's why I welcome any suggestion to make it better.
I am trying to find the probabilities of measuring eigenvalues and the expectation values of operators, depending on the spectrum and on state in which the system is found. We have the following cases to consider:
- Discrete non degenerate spectrum + pure state.
- Discrete degenerate spectrum + pure state.
- Discrete non degenerate spectrum + mixed state.
- Discrete degenerate spectrum + mixed state.
- Continuous non degenerate spectrum + pure state.
- Continuous degenerate spectrum + pure state.
- Continuous non degenerate spectrum + mixed state.
- Continuous degenerate spectrum + mixed state.
NOTE: $\rho$ is the probability density function, while $\hat\rho$ is the density operator.
Case 1: Discrete non degenerate spectrum + pure state.
$A|\phi_n\rangle=a_n|\phi_n\rangle$
$|\psi\rangle=\sum_n c_n|\phi_n\rangle$
$P(a_n)=|c_n|^2=|\langle \phi_n|\psi\rangle|^2$
$P(a_n)=|c_n|^2=|\langle \phi_n|\hat\rho|\phi_n\rangle|^2$ where $\hat\rho=|\psi\rangle\langle\psi|$
$\langle A \rangle=Tr(\hat\rho A)$
$\langle A \rangle=\langle\psi|A|\psi\rangle$
*Note: As seen here, I am expressing both, the probability and the expectation value once by using the density operator and once by not using it.
Case 2: Discrete degenerate spectrum + pure state.
$A|\phi_n^i\rangle=a_n|\phi_n^i\rangle$
$|\psi\rangle=\sum_n\sum_i^{g_n} c_n^i|\phi_n^i\rangle$
$P(a_n)=\sum_i^{g_n}|c_n^i|^2=\sum_i\langle \phi_n^i|\hat\rho|\phi_n^i\rangle$
$P(a_n)=\langle \psi|P_n|\psi\rangle$
$\langle A \rangle=\sum_n a_n P(a_n)=...=\langle \psi|A|\psi\rangle$
$\langle A \rangle=Tr(\hat\rho A)$
Case 3: Discrete non degenerate spectrum + mixed state.
$A|u_n\rangle=a_n|u_n\rangle$
$\hat\rho=\sum_k P_k|\psi_k\rangle\langle\psi_k|$ where $|\psi_k\rangle=\sum_n c_{n(k)}|u_n\rangle$
$P(a_n)_{mixed}= \sum_k P_k P(a_n)_k$
$P(a_n)_{mixed}=\langle u_n|\hat\rho|u_n\rangle$
$\langle A \rangle=\sum_na_nP(a_n)_{mixed}=Tr(A\hat\rho)$
$\langle A \rangle=\sum_kP_k Tr(A|\psi_k\rangle\langle\psi_k|)=\sum_kP_k Tr(A\hat \rho_k)$
*Note: The trace is over eigenstate of $A$, meaning $|u_n\rangle$
*Note: $P(a_n)_k$ represents the probability of measuring the eigenvalue $a_n$ when the system is in the pure state $|\psi_k\rangle$ that is included in the mixed state.
Case 4: Discrete degenerate spectrum + mixed state.
$A|u_n^i\rangle=a_n|u_n^i\rangle$
$\hat\rho=\sum_k P_k|\psi_k\rangle\langle\psi_k|$ where $|\psi_k\rangle=\sum_n c_{n(k)}|u_n\rangle$
$P(a_n)_{mixed}=\sum_k P_k P(a_n)_k=\sum_k P_k \sum_i^{g_n}|\langle u_n^i|\psi_k\rangle|^2=\sum_k\sum_i^{g_n} P_k \langle \psi_k|u_n^i\rangle\langle u_n^i|\psi_k\rangle=\sum_i^{g_n} \langle u_n^i|\hat\rho|u_n^i\rangle$
$P(a_n)_{mixed}=\sum_i^{g_n} \langle u_n^i|\hat\rho|u_n^i\rangle$
$P(a_n)_{mixed}=Tr(\hat\rho P_n)$
$\langle A \rangle=\sum_n a_n P(a_n)_{mixed}=Tr(\hat\rho A) $
$\langle A \rangle=\sum_n a_n P(a_n)_{mixed}=\sum_kP_k Tr(A|\psi_k\rangle\langle\psi_k|=\sum_kP_k Tr(A\hat\rho_k) $
Case 5: continuous non degenerate spectrum + pure state.
$A|\nu_\alpha\rangle=\alpha|\nu_\alpha\rangle$
$|\psi\rangle=\int c(\alpha)|\nu_\alpha\rangle d\alpha$
$dP(\alpha)=\rho(\alpha)d\alpha=|\langle \nu_\alpha|\psi\rangle|^2d\alpha$
$dP(\alpha)=\rho(\alpha)d\alpha=\langle \nu_\alpha|\hat\rho|\nu_\alpha\rangle d\alpha$
$\langle A \rangle=\int \alpha \rho(\alpha)d\alpha=...=\langle \psi|A|\psi\rangle$
$\langle A \rangle=\int \langle \nu_\alpha|\hat\rho A|\nu_\alpha\rangle d\alpha=Tr(\hat\rho A)$
Case 6: continuous degenerate spectrum + pure state.
*Note: From here on out I am not 100% about my notations.
$A|\nu_\alpha^i\rangle=\alpha|\nu_\alpha^i\rangle$
$|\psi\rangle=\sum_i \int c(\alpha)^i|\nu_\alpha^i\rangle d\alpha$
$dP(\alpha)=\rho(\alpha)d\alpha=\sum_i |\langle \nu_\alpha^i|\psi\rangle|^2d\alpha$
$dP(\alpha)=\sum_i \langle \nu_\alpha^i|\hat\rho|\nu_\alpha^i\rangle d\alpha $ where $\hat\rho=|\psi\rangle\langle\psi|$
$\langle A \rangle=\int \alpha \rho(\alpha)d\alpha=...=\sum_i\int\langle\psi|A|\nu_\alpha^i\rangle\langle \nu_\alpha^i|\psi\rangle d\alpha=\langle \psi|A|\psi\rangle$
$\langle A \rangle=\sum_i\int \langle \nu_\alpha^i|\hat\rho A|\nu_\alpha^i\rangle d\alpha=Tr(\hat\rho A)$
Case 7: continuous non degenerate spectrum + mixed state.
$A|\nu_\alpha\rangle=\alpha|\nu_\alpha\rangle$
$\hat\rho=\sum_k P_k|\psi_k\rangle\langle\psi_k|$ where $|\psi_k\rangle=\int c_k(\alpha)|\nu_\alpha\rangle d\alpha$
$dP(\alpha)_{mixed}=\sum_kP_kdP(\alpha)_k$
*Note:$dP(\alpha)_k$ is the probability of measuring a value between $\alpha$ and $\alpha+d\alpha$ for when the system is in the pure state $|\psi_k\rangle$.
Then we have:
$dP(\alpha)_{mixed}=\sum_k P_k |\langle \nu_\alpha|\psi_k\rangle|^2 d\alpha$
$dP(\alpha)_{mixed}=\langle\nu_\alpha|\hat\rho|\nu_\alpha\rangle d\alpha$
I am not sure as to why the two above written lines are not shown with the according symbols. When I copy paste them in another,new thread, it looks fine.
$\langle A \rangle=\sum_kP_k \langle A \rangle_k= \sum_kP_k \int \alpha \rho(\alpha)d\alpha=...\int \langle \nu_\alpha|A\hat\rho|\nu_\alpha\rangle d\alpha=Tr(A\hat\rho)$
$\langle A \rangle=\sum P_kTr(A|\psi_k\rangle\langle\psi_k|)=\sum P_kTr(A|\hat\rho_k)$
Case 8: continuous degenerate spectrum + mixed state.
$A|\nu_\alpha^i\rangle=\alpha|\nu_\alpha^i\rangle$
$\hat\rho=\sum_k P_k|\psi_k\rangle\langle\psi_k|$ where $|\psi\rangle=\sum_i \int c_k(\alpha)^i|\nu_\alpha^i\rangle d\alpha$
$dP(\alpha)_{mixed}=\sum_k P_k dP(\alpha)_k=\sum_k P_k \sum_i | \langle \nu_\alpha^i| \psi_k\rangle|^2d\alpha$
$dP(\alpha)_{mixed}=\sum_i \langle \nu_\alpha^i|\hat\rho|\nu_\alpha^i\rangle d\alpha$
I am not sure as to why the two above written lines are not shown with the according symbols. When I copy paste them in another,new thread, it looks fine.
$\langle A \rangle=\sum_kP_k \langle A \rangle_k=\sum_kP_k \int \alpha \rho(\alpha)d\alpha=...=Tr(\hat\rho A)$
$\langle A \rangle=\sum_kP_kTr(A|\psi_k\rangle\langle\psi_k|)=\sum_kP_kTr(A \hat\rho_k)$
