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My question is about something I read in Zwiebach's "A First Course in String Theory", 2nd edition.

On page 220, just below eqn (11.24) he says $$\alpha(t) = e^{iHt}\alpha e^{-iHt}.$$ Here, $\alpha$ is an operator in the Schroedinger picture and $\alpha(t)$ is that same operator in the Heisenberg picture. On page 219, just below eqn (11.18) he says $$H(p(t), q(t); t)$$ is the Heisenberg Hamiltonian. Here $H(p, q)$ is the Schroedinger Hamiltonian. I assume he means $$H(p(t), q(t); t) = e^{iHt}H(p, q) e^{-iHt}.$$ But $$H(p(t), q(t); t) = H(e^{iHt}p e^{-iHt}, e^{iHt}q e^{-iHt}; t).$$ In other words, $$e^{iHt}H(p, q) e^{-iHt} = H(e^{iHt}p e^{-iHt}, e^{iHt}q e^{-iHt}; t).$$ How do the exponentials pass from surrounding H, to surrounding $p$ and $q$? Is it possible that the Schroedinger Hamiltonian is $H(p(t), q(t))$ and the Heisenberg Hamiltonian is $H(p(t), q(t); t)$? 3 lines below eqn (11.20) on page 219 he specifies that the Schroedinger Hamiltonian is $H(p, q)$. Am I correct in assuming that whenever $H$ appears in the exponent it is the Schroedinger Hamiltonian?

Qmechanic
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Lifetime Beginner
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  • Can you write the Hamiltonian as some power series over p and q? Then you could just insert $1 = U^\dagger U$ in the middle... – Wihtedeka Dec 14 '21 at 13:12
  • Thank you Wihtedeka. I considered that. The Hamiltonians I've seen so far in the book are either polynomials or rational functions of p. However, the author doesn't explicitly state that the Hamiltonian is a power series in p and q. If he did, I wouldn't have posted this question. Is it the custom in QM to implicitly define Hamiltonians as power series? – Lifetime Beginner Dec 14 '21 at 14:50

1 Answers1

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That's a good question.

  1. If the Hamiltonian $H_S(t)$ in the Schrödinger picture has explicit$^1$ time-dependence, the Hamiltonian at different times might not commute. This means that the formula $e^{-\frac{i}{\hbar}Ht}$ no longer apply! The unitary time-evolution operator is more generally given by the (anti)time-ordered exponentiated Hamiltonian $$ U(t)~=~\left\{\begin{array}{rcl} T\exp\left[-\frac{i}{\hbar}\int_0^t\! dt^{\prime}~H_S(t^{\prime})\right] &\text{for}& t ~\geq~0 \cr\cr AT\exp\left[-\frac{i}{\hbar}\int_0^t\! dt^{\prime}~H_S(t^{\prime})\right] &\text{for}& t ~\leq~0, \end{array}\right.\tag{A}$$ cf. e.g. this related Phys.SE post. It satisfies its own TDSE: $$ i\hbar \frac{d U(t)}{d t} ~=~H_S(t)U(t),\qquad U(t\!=\!0)~=~{\bf 1}. \tag{B} $$

  2. Operators in the Heisenberg picture are given as $$ A_H(t)~=~U(t)^{-1} A_S(t) U(t). \tag{C}$$ They satisfies the Heisenberg EOM $$ i\hbar \frac{d A_H(t)}{d t}~=~[A_H(t),H_H(t)]+ i\hbar\frac{\partial A_H(t)}{\partial t}, \tag{D}$$ see the Wikipedia page for details. In particular, eq. (C) holds if we put the operator $A_H$ equal to the Heisenberg Hamiltonian $H_H$, which OP asked about.

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$^1$ In section 11.2 Zwiebach assumes after eqs. (11.19) that there is no explicit time dependence.

Qmechanic
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  • Thank you Qmechanic. Your post has a lot of useful information. It seems to me that I am working too hard on this problem. Zwiebach simply does not explain what he means in a way that I can understand. A few pages later in problem 11.2 on page 233, he brings up the U operator. He doesn't say it equals $e^{-iHt}$. Instead he says "U(t) bears some nontrivial relation to H". That certainly describes your definition in eqn (A). Perhaps that's what he is thinking of. – Lifetime Beginner Dec 15 '21 at 12:41
  • However, on the pages that I quoted in my question he doesn't use U(t) at all, he uses $e^{-iHt}$. He has no less than 4 different notations without clear explanation of their relationship: $H, H(p, q), H(p(t), q(t))$ and $H(p(t), q(t); t)$. – Lifetime Beginner Dec 15 '21 at 12:43
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    Yes, eq. (A) is almost certainly what Zwiebach has in mind. – Qmechanic Dec 15 '21 at 13:04