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Peskin and Schroeder state something which I'm not fully understanding. More specificially I think it's just phrased in a way I'm not understanding.

In the Schrodinger picture we can expand the real scalar field $\phi(x)$ which satisfies the Klein-Gordon equation as

$$\phi(\textbf{x})=\displaystyle\int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_p}}(a_pe^{i\textbf{p.x}}+a_p^\dagger e^{-i\textbf{p.x}}).$$

Then of course we find $\phi(x)=\phi(\textbf{x},t)$ by switching to the Heisenberg picture.

Now, on page 83 they say

At any fixed time $t_0$ we can of course expand $\phi$ in terms of ladder operators $$\phi(\textbf{x},t_0)=\displaystyle\int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_p}}(a_pe^{i\textbf{p.x}}+a_p^\dagger e^{-i\textbf{p.x}}).$$ Then to obtain $\phi(\textbf{x},t)$ for $t\neq t_0$ we just switch to the Heisenberg picture $$\phi(\textbf{x},t)=e^{iH(t-t_0)}\phi(\textbf{x},t_0)e^{-iH(t-t_0)}.$$

The first problem is that they say we switch to the Heisenberg picture, implying we were in the Schrodinger picture to begin with. But then how can the $\phi$ be time-dependent i.e. why is it depending on $t_0$, even though $t_0$ appears nowhere in the expansion?

Are they just saying somewhat awkwardly that $\phi$ is (obviously) not time-independent in the Schrodinger picture, we pick a certain time slice (where our states are now time fixed) and then time evolve from there? It shouldn't matter since I'd imagine we should have $\phi(t_0)=\phi(t')$ for some other time $t'$

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    $\begingroup$ What Peskin and Schroeder call the operator at $t_0$ is actually the operator at $t_0=0$, and that's why you don't see any time dependence (you evaluate it at $t_0=0$); the book continues this misuse of terminology also later on, and it's a mistake. About the rest: fields are always to be intended in the Heisenberg picture, because they are operators and not element of a Fock space (on which they act instead); as such, there is no Schrödinger picture and this is one of those other things where Peskin and Schroeder is wrong. $\endgroup$
    – gented
    Commented Jun 27, 2015 at 20:47
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    $\begingroup$ @GennaroTedesco $t_0 = 0$ has nothing special. You can evaluate at $t_0= 100$s or $t_0 = 57168.12$s or any $t_0$ that you want, but it has to be a fixed time. And any operator can be in the Schrödinger pucture (constant) or in the Heisenberg picture or the interaction picture (in both cases depends on time), you only have to make an unitary transform to switch between them. $\endgroup$
    – Bosoneando
    Commented Jun 27, 2015 at 22:02
  • $\begingroup$ @Bosoneando $t_0=0$ does have something special, namely the exponential $\textrm{e}^{-ik_0t}$ becomes $1$ and thus you don't see it in the expansion anymore. If you choose any other time you cannot get rid of such term in the formula. This is precisely why you start from the field at $t=0$ and then evaluate the later times by using the $U(t,0)$. $\endgroup$
    – gented
    Commented Jun 27, 2015 at 22:07
  • $\begingroup$ @GennaroTedesco Indeed based on the other stuff in Peskin, in particular equation 2.47 which is the field in the Heisenberg picture. The only way this agrees with the Schrodinger picture is with t=0. $\endgroup$
    – Okazaki
    Commented Jun 28, 2015 at 8:56

1 Answer 1

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The upshot is that we need one condition to specify how the operators in the Schrödinger and Heisenberg picture are connected. This is usually done by declaring that the two pictures agree at some fixed instant $t_0$.

To summarize: The Schrödinger operator $\phi(\textbf{x},t_0)$ does not depend on time $t$, while the Heisenberg operator $\phi(\textbf{x},t)$ does depend on time $t$. For kets and bras it is the other way around.

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    $\begingroup$ Saying that the operator at time $t_0$ does not depend on time is wrong. It does of course, but the dependency is hidden after you integrate out the $k_0$ variable and express everything in terms of the energy. Also, this is true only for the free field, because commutators of the ladders operators and independent of time in this case (which is not true for any other non-free field). $\endgroup$
    – gented
    Commented Jun 27, 2015 at 20:34
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    $\begingroup$ @Qmechanic I think the notation may just be poor. Can I define $\phi^{S}(\textbf{x})=\phi(t_0,\textbf{x})$ and then expand $\phi^{S}(\textbf{x})$ in terms of ladder operators? This would give the expansion in Peskin and by defining $\phi^{H}(\textbf{x},t)=e^{iH(t-t_0)}\phi^{S}(\textbf{x})e^{-iH(t-t_0)}$ everything would agree. $\endgroup$
    – Okazaki
    Commented Jun 28, 2015 at 12:26

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