For a spin one system the matrix: $S_y=\hbar \begin{bmatrix} 0 & -i & 0 \\ i & 0& -i \\ 0 & i& 0 \end{bmatrix}$
Suppose an arbitrary vector $|\psi \rangle=[a,b,c]$ where $aa^*+bb^*+cc^*=1$.
Prove that $\langle\psi|S_y |\psi\rangle$ has real value.
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$\begingroup$ You should show what you tried. Homework problems need to show your effort and where you got stuck so that we can help you through it. SE is not a homework problem solving service. $\endgroup$ – enumaris Mar 5 '18 at 19:05
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$\begingroup$ I used computer to tried several arbitrary values, all came out real. I tried a,b,c all real, the result was $0$ which was a real number, but the complex conjugate was not easy to prove. By theory, those were all valid quantum states thus had to came out as real in states calculation. $\endgroup$ – J C Mar 5 '18 at 19:08
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1$\begingroup$ Calculate it out: it's just matrix multiplication. And remember that the difference between $w$ and $w^*$ (where $w$ is any complex number) is pure imaginary. $\endgroup$ – NickD Mar 5 '18 at 19:29
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HINT: The quantity $\langle \psi | S_y | \psi \rangle$ can easily be calculated in terms of $a$, $b$, $c$, and their complex conjugates. If this quantity is real, then it must be equal to its own complex conjugate.
answered Mar 5 '18 at 19:23
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$\begingroup$ Thank you, I forgot about that, but also, I got the result to be $i\cdot a\cdot conjugate(b) + b\cdot (-i\cdot conjugate(a) + i\cdot conjugate(c)) - i\cdot c\cdot conjugate(b)$, which was a really interesting formula, and it had to be real. I was thinking how to expand it. $\endgroup$ – J C Mar 5 '18 at 19:29
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$\begingroup$ Here's another hint, then: for any complex numbers $x$, $y$, and $z$, we have $(x + y + z)^* = x^* + y^* + z^*$, and $(xyz)^* = x^* y^* z^*$. Try taking the complex conjugate of your quantity above (which is correct) and see what happens. $\endgroup$ – Michael Seifert Mar 5 '18 at 19:31
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Suppose $\langle \psi\vert S_y\vert \psi \rangle = a+ib$ with $a$ and $b$ real. Then, since $S_y$ is hermitian: $$ \langle \psi\vert S_y\vert \psi \rangle^* = a-ib = \langle \psi\vert S^\dagger _y\vert \psi \rangle =\langle \psi\vert S_y\vert \psi \rangle =a +ib $$ from which it follows that $-b=b=0$. Thus, $\langle \psi\vert S_y\vert \psi \rangle = a$ is purely real.
answered Mar 5 '18 at 19:10
