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I'm studying QM from Sakurai, and I have a doubt regarding the proof given that in the case of time independent Hamiltonian the expectation value of an observable doesn't change with time. The argument goes as follows:

At any time the expectation value of an observable $B$ is given by:

$$<\alpha_0|U^{\dagger}BU|\alpha_0>$$

Where $U$ is the time evolution operator and $|\alpha_0>$ is the ket in our initial state. Suppose that $|\alpha_0>$ is also an eigenket of some other operator $A$ (which commutes with the Hamiltonian operator), that the Hilbert space is finite-dimensional and that the eigenkets of $A$ form a basis of the Hilbert space. We can use the explicit form of the time evolution operator:

$$U=\exp{\frac{-iHt}{\hbar}}$$

And we know that it acts on the eigenkets of $A$ by multiplication by a phase. Sakurai then writes this:

$$<\alpha_0|\exp{\frac{iE_{\alpha_0}t}{\hbar}}B\exp{\frac{-iE_{\alpha_0}t}{\hbar}}|\alpha_0>$$

My doubt is: the observable $B$ could send the ket $|\alpha_0>$ in a superposition of eigenkets, and then the simple multiplication by $\exp{\frac{iE_{\alpha_0}t}{\hbar}}$ wouldn't hold anymore. What am I missing?

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The left exponential evolves the $\langle \alpha_0 \lvert$ on the left.

This is one of the pitfalls of Dirac notation, it would be unambiguous to write $$ (\mathrm{e}^{-\mathrm{i}E_{\alpha_0} t} \lvert \alpha_0 \rangle,B \mathrm{e}^{-\mathrm{i}E_{\alpha_0} t} \lvert \alpha_0 \rangle)$$ where $(\dot{},\dot{})$ denotes the inner product on the Hilbert space, i.e. $(\lvert\psi\rangle,\lvert\phi\rangle) = \langle \psi\vert\phi \rangle$.

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  • $\begingroup$ Yeah, makes sense. Thank you! (I'll accept as soon as possible and I guess the lesson is: when in doubt, rewrite everything in inner product notation) $\endgroup$ – Gennaro Marco Devincenzis Aug 22 '15 at 15:08

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