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10

Call $u_1, u_2, u_3, u_4$ the eigenvectors described by you, respectively. Your claims are all right, but realize that both $u_1$ and $u_2$ share the same eigenvalue, that is $1$, i.e., $Pu_1=u_1$ and $Pu_2=u_2$. Hence, any linear combination of $u_1$ and $u_2$ will also be eigenvectors with the same eigenvalue $1$. Try to find eigenvectors of $H$ of the ...


8

Hint: When an eigenvalue for an operator $P$ is degenerate, there are more than one way to chose a set of eigenvectors. If the other commuting operator $H$ lifts that degeneracy, there will be a preferred choice of common eigenvectors. More generally, a set of diagonalizable operators commutes if and only if the set is simultaneously diagonalizable.$^1$ ...


3

Really a good question (unfortunately Ī came too late to actually help). All this stuff perfectly makes mathematical sense, but its fundamentals lie in abstract algebra, and rarely are explained to students, and hence remain concealed from those who lack appropriate imagination. To understand firmly bra and ket, not just multiply rows by columns and ...


0

Since there still isn't an answer but the question has attracted a few upvotes, let me elaborate on my comment. This is more maths, than physics, but anyway. Writing $\sum_{jk} \mathrm{tr}(E_j^{\dagger}E_i)|j\rangle\langle k|$ doesn't give you anything. This is indeed a rank-one decomposition, but the theorem does not tell you that ANY rank-1 decomposition ...


1

Consider for example, a plane vector and two orthogonal unit vectors $\hat x$ and $\hat y$. Any vector in the plane can be expressed as $$\vec v = (\vec v \cdot \hat x) \;\hat x + (\vec v \cdot \hat y) \; \hat y = v_x\; \hat x + v_y\; \hat y$$ So, you're correct, $\vec b \cdot \hat a$ is the component of $\vec b$ in the $\hat a$ direction. And further, ...


1

Two states $\rho_0$ and $\rho_1$ are perfectly distinguishable if there exists a POVM measurement $\{P_0,P_1\}$ such that $$Tr(\rho_0P_0)=1$$ $$Tr(\rho_0P_1)=0$$ $$Tr(\rho_1P_0)=1$$ $$Tr(\rho_1P_1)=0$$ and $P_0+P_1=\mathbf{I}$, where $\mathbf{I}$ is the identity matrix. You can interpret the above equations as follows: If we obtain outcome $P_0$, we know ...


1

Current is charge per time unit, i.e $current=charge/time$. If you use your formula ($t=I/Q$) then the result is $(charge/time)/charge$, which is $1/time$, So you got the inverse of the real result and $1/0.5=2$, which is the correct result. As @Amzoth explained in his comment, $t=Q/I$. Hence$$t=25mC/12.5mA=2$$


2

$D_{\mu\nu} = A g_{\mu\nu}+B k_{\mu} k _{\nu}$ with A and B two unknown functions of the scalar k^2. The two tensor after A and B are the only possible Lorentz invariant tensors . Simply plugin and calculate the unknown functions.



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