Quantum Pictures

This note gives a comparison on the three quantum pictures. Also, the Dyson series are introduced with the help of time-ordering operators.

1. Comparison of three pictures

2. Evolution Operator

The evolution operator U_S(t, t_0) in Shrodinger picture is defined as

(1)   \begin{equation*} |\alpha,t_0;t\rangle_S =U_S(t, t_0)|\alpha,t_0;t_0\rangle_S \end{equation*}

In Interaction picture, we can also defined a evolution operator U_I(t, t_0)

(2)   \begin{equation*} |\alpha,t_0;t\rangle_I =U_I(t, t_0)|\alpha,t_0;t_0\rangle_I \end{equation*}

3.Dyson Series

Combine Schrodinger equation and the definition of evolution operator, we can get the equation for evolution operator

(3)   \begin{equation*} i\hbar\frac{\partial}{\partial t}U_S(t, t_0) =HU_S(t, t_0) \end{equation*}

As for the solution of the above equation, there are 3 cases.

Case 1. H is independent of time

(4)   \begin{equation*} U_S(t, t_0)=\exp\left[\frac{-iH(t-t_0)}{\hbar}\right] \end{equation*}

Case 2. H is time-dependent but the H‘s at different times commute

(5)   \begin{equation*} U_S(t, t_0) =\exp\left[-\frac{i}{\hbar}\int_{t_0}^{t}dt'H(t')\right] \end{equation*}

Case 3. H‘s at different times do not commute

(6)   \begin{align*}  U_S(t,t_0) &=1+\sum_{n=1}^{\infty}\left(\frac{-i}{\hbar}\right)^n\int_{t_0}^{t}dt_1\int_{t_0}^{t_1}dt_2\cdots\int_{t_0}^{t_{n-1}}dt_{n}H(t_1)H(t_2)\cdots H(t_n) \end{align*}

which is just the Dyson series.

As for the interaction picture, we have similar results.

(7)   \begin{equation*} i\hbar\frac{\partial}{\partial t}U_I(t, t_0) =V_IU_I(t, t_0) \end{equation*}

Case 1. V_I is independent of time

(8)   \begin{equation*} U_I(t, t_0)=\exp\left[\frac{-iV_I(t-t_0)}{\hbar}\right] \end{equation*}

Case 2. V_I is time-dependent but the V_I‘s at different times commute

(9)   \begin{equation*} U_I(t, t_0) =\exp\left[-\frac{i}{\hbar}\int_{t_0}^{t}dt'V_I(t')\right] \end{equation*}

Case 3. V_I‘s at different times do not commute

(10)   \begin{align*}  U_I(t,t_0) &=1+\sum_{n=1}^{\infty}\left(\frac{-i}{\hbar}\right)^n\int_{t_0}^{t}dt_1\int_{t_0}^{t_1}dt_2\cdots\int_{t_0}^{t_{n-1}}dt_{n}V_I(t_1)V_I(t_2)\cdots V_I(t_n) \end{align*}

4. Time Order

Introducing the time-oder operator, the Dyson series can be written in a more compact form. The time-order operator is defined as

(11)   \begin{equation*} T\{A_1(t_1)A_2(t_2)\cdots A_n(t_n)\} =\left\{ \begin{array}{ll} A_{p_1}(t_{p_1})A_{p_2}(t_{p_2})\cdots A_{p_n}(t_{p_n}) & \text{for bpsonic operators,} \\ (-1)^p A_{p_1}(t_{p_1})A_{p_2}(t_{p_2})\cdots A_{p_n}(t_{p_n}) & \text{for fermionic operators} \end{array} \right. \end{equation*}

where t_{p_1}>t_{p_2}>\cdots>t_{p_n} and p is the number of permutations.

In our problem, the Hs in Eq.6 or V_Is in Eq.10 are all time-ordered. It’s always even number of Fermionic fields appear in the H and V_I. There is no an extra minus sign for the time ordering.

(12)   \begin{equation*} T\{V_1(t_1)V_2(t_2)\cdots V_n(t_n)\} =V_{p_1}(t_{p_1})V_{p_2}(t_{p_2})\cdots V_{p_n}(t_{p_n}) \end{equation*}

Also, if we define

(13)   \begin{equation*} I_n \equiv \int_{t_0}^{t}dt_1\int_{t_0}^{t}dt_2\cdots\int_{t_0}^{t}dt_n K(t_1,t_2,\cdots,t_n) \end{equation*}

and K is symmetric in its arguments, we can divide the integration region into n! sub regions, in which t_1>t_2>\cdots>t_n, t_2>t_1>\cdots>t_n, etc. They all equals to

(14)   \begin{equation*} S_n \equiv \int_{t_0}^{t}dt_1\int_{t_0}^{t_1}dt_2\cdots\int_{t_0}^{t_{n-1}}dt_n K(t_1,t_2,\cdots,t_n) \end{equation*}


(15)   \begin{equation*} S_n=\frac{1}{n!}I_n \end{equation*}

Then, Eq.10 can be written as

(16)   \begin{align*} U_I(t,t_0) &=1+\sum_{n=1}^{\infty}\frac{1}{n!}\left(\frac{-i}{\hbar}\right)^n\int_{t_0}^{t}dt_1\int_{t_0}^{t}dt_2\cdots\int_{t_0}^{t}dt_{n}T\left\{V_I(t_1)V_I(t_2)\cdots V_I(t_n)\right\}\\ &\equiv T\left\{\exp\left[-\frac{i}{\hbar}\int_{t_0}^{t}dt'V_I(t')\right]\right\} \end{align*}

U_I(t,t_0) can be written out in a similar way.

5. Correction on Section 4

Section 4 is wrong. Exactly, we don’t need to use the symmetry properties of the integrand. On the contrary, the time-ordering operator is introduced in case that the integrand is not symmetric on t_is.

Take the second order term \int_{t_0}^{t}dt_1\int_{t_0}^{t_1}dt_2V_I(t_1)V_I(t_2) in Eq.6 for example. The integration area is the blue shadow area in the following figure. We can change the order of integration \int_{t_0}^{t}dt_1\int_{t_0}^{t_1}dt_2 \to \int_{t_0}^{t}dt_2\int_{t_2}^{t}dt_1 with the value of integration unchanged since it’s still the same integration area. Thus,

(17)   \begin{equation*} \int_{t_0}^{t}dt_1\int_{t_0}^{t_1}dt_2V_I(t_1)V_I(t_2) =\frac{1}{2}\int_{t_0}^{t}dt_1\int_{t_0}^{t_1}dt_2V_I(t_1)V_I(t_2)+\frac{1}{2}\int_{t_0}^{t}dt_2\int_{t_2}^{t}dt_1V_I(t_1)V_I(t_2) \end{equation*}

Notice that t_1>t_2 above. We now change dummy variables in the second term, interchanging the labels t_1 and t_2, and the second term becomes

(18)   \begin{equation*} \frac{1}{2}\int_{t_0}^{t}dt_2\int_{t_2}^{t}dt_1V_I(t_1)V_I(t_2) =\frac{1}{2}\int_{t_0}^{t}dt_1\int_{t_1}^{t}dt_2V_I(t_2)V_I(t_1) \end{equation*}

notice that t_2>t_1 on the RHS and the integration area becomes the red shadow area, but we still have the later time on the left. Thus

(19)   \begin{align*} \int_{t_0}^{t}dt_1\int_{t_0}^{t_1}dt_2V_I(t_1)V_I(t_2) &=\frac{1}{2}\int_{t_0}^{t}dt_1\int_{t_0}^{t_1}dt_2V_I(t_1)V_I(t_2)+\frac{1}{2}\int_{t_0}^{t}dt_1\int_{t_1}^{t}dt_2V_I(t_2)V_I(t_1)\\ &=\frac{1}{2}\int_{t_0}^{t}dt_1\int_{t_0}^{t}dt_2T\left\{V_I(t_1)V_I(t_2)\right\} \end{align*}

It’s an integration on the hole area with the integrand is time-ordered.
Similarly, it can be generalized to higher order.

Leave a Reply