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# 物理代写|量子力学代考Quantum mechanics代考|PHYS519 The Schrödinger Versus the Heisenberg Picture

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## 物理代写|量子力学代考QUANTUM MECHANICS代考|Unitary Operators

In the previous section we introduced the concept of time development by considering the time-evolution operator that affects state kets; that approach to quantum dynamics is known as the Schrödinger picture. There is another formulation of quantum dynamics where observables, rather than state kets, vary with time; this second approach is known as the Heisenberg picture. Before discussing the differences between the two approaches in detail, we digress to make some general comments on unitary operators.

Unitary operators are used for many different purposes in quantum mechanics. In this book we introduced (Section 1.5) an operator satisfying the unitarity property. In that section we were concerned with the question of how the base kets in one representation are related to those in some other representations. The state kets themselves are assumed not to change as we switch to a different set of base kets even though the numerical values of the expansion coefficients for $|\alpha\rangle$ are, of course, different in different representations.

Subsequently we introduced two unitary operators that actually change the state kets, the translation operator of Section $1.6$ and the time-evolution operator of Section 2.1. We have
$$|\alpha\rangle \rightarrow U|\alpha\rangle,$$
where $U$ may stand for $\mathscr{T}(d \mathbf{x})$ or $\mathscr{U}\left(t, t_{0}\right)$. Here $U|\alpha\rangle$ is the state ket corresponding to a physical system that actually has undergone translation or time evolution.

It is important to keep in mind that under a unitary transformation that changes the state kets, the inner product of a state bra and a state ket remains unchanged:
$$\langle\beta \mid \alpha\rangle \rightarrow\left\langle\beta\left|U^{\dagger} U\right| \alpha\right\rangle=\langle\beta \mid \alpha\rangle .$$
Using the fact that these transformations affect the state kets but not operators, we can infer how $\langle\beta|X| \alpha\rangle$ must change:
$$\langle\beta|X| \alpha\rangle \rightarrow\left(\langle\beta| U^{\dagger}\right) \cdot X \cdot(U|\alpha\rangle)=\left\langle\beta\left|U^{\dagger} X U\right| \alpha\right\rangle$$
We now make a very simple mathematical observation that follows from the associative axiom of multiplication.
$$\left(\langle\beta| U^{\dagger}\right) \cdot X \cdot(U|\alpha\rangle)=\left\langle\beta\left|\cdot\left(U^{\dagger} X U\right) \cdot\right| \alpha\right\rangle$$

## 物理代写|量子力学代考QUANTUM MECHANICS代考|State Kets and Observables in the Schrödinger and the Heisenberg Pictures

We now return to the time-evolution operator $\mathscr{U}\left(t, t_{0}\right)$. In the previous section we examined how state kets evolve with time. This means that we were following approach 1 , known as the Schrödinger picture when applied to time evolution. Alternatively we may follow approach 2, known as the Heisenberg picture when applied to time evolution.

In the Schrödinger picture the operators corresponding to observables like $x, p_{y}$, and $S_{z}$ are fixed in time, while state kets vary with time, as indicated in the previous section. In contrast, in the Heisenberg picture the operators corresponding to observables vary with time; the state kets are fixed, frozen so to speak, at what they were at $t_{0}$. It is convenient to set $t_{0}$ in $\mathscr{U}\left(t, t_{0}\right)$ to zero for simplicity and work with $\mathscr{U}(t)$, which is defined by
$$\mathscr{U}\left(t, t_{0}=0\right) \equiv \mathscr{U}(t)=\exp \left(\frac{-i H t}{\hbar}\right) .$$
Motivated by (2.79b) of approach 2, we define the Heisenberg picture observable by
$$A^{(H)}(t) \equiv \mathscr{U}^{\dagger}(t) A^{(S)} \mathscr{U}(t)$$
where the superscripts $H$ and $S$ stand for Heisenberg and Schrödinger, respectively. At $t=0$, the Heisenberg picture observable and the corresponding Schrödinger picture observable coincide:
$$A^{(H)}(0)=A^{(S)}$$
The state kets also coincide between the two pictures at $t=0$; at later $t$ the Heisenberg picture state ket is frozen to what it was at $t=0$ :
$$\left|\alpha, t_{0}=0 ; t\right\rangle_{H}=\left|\alpha, t_{0}=0\right\rangle$$
independent of $t$. This is in dramatic contrast with the Schrödinger picture state ket,
$$\left|\alpha, t_{0}=0 ; t\right\rangle_{S}=\mathscr{U}(t)\left|\alpha, t_{0}=0\right\rangle$$
The expectation value $\langle A\rangle$ is obviously the same in both pictures:
$s\left\langle\alpha, t_{0}=0 ; t\left|A^{(S)}\right| \alpha, t_{0}=0 ; t\right\rangle_{S}=\left\langle\alpha, t_{0}=0\left|\mathscr{U}^{\dagger} A^{(S)} \mathscr{U}\right| \alpha, t_{0}=0\right\rangle$
$={ }{H}\left\langle\alpha, t{0}=0 ; t\left|A^{(H)}(t)\right| \alpha, t_{0}=0 ; t\right\rangle_{H} .$

## 物理代写|量子力学代考QUANTUM MECHANICS代考|Unitary Operators

$$|\alpha\rangle \rightarrow U|\alpha\rangle$$

$$\langle\beta \mid \alpha\rangle \rightarrow\left\langle\beta\left|U^{\dagger} U\right| \alpha\right\rangle=\langle\beta \mid \alpha\rangle .$$

$$\langle\beta|X| \alpha\rangle \rightarrow\left(\langle\beta| U^{\dagger}\right) \cdot X \cdot(U|\alpha\rangle)=\left\langle\beta\left|U^{\dagger} X U\right| \alpha\right\rangle$$

$$\left(\langle\beta| U^{\dagger}\right) \cdot X \cdot(U|\alpha\rangle)=\left\langle\beta\left|\cdot\left(U^{\dagger} X U\right) \cdot\right| \alpha\right\rangle$$

## 物理代写|量子力学代考QUANTUM MECHANICS代考|State Kets and Observables in the Schrödinger and the Heisenberg Pictures

$$\mathscr{U}\left(t, t_{0}=0\right) \equiv \mathscr{U}(t)=\exp \left(\frac{-i H t}{\hbar}\right) .$$

$$A^{(H)}(t) \equiv \mathscr{U}^{\dagger}(t) A^{(S)} \mathscr{U}(t)$$

$$A^{(H)}(0)=A^{(S)}$$

$$\left|\alpha, t_{0}=0 ; t\right\rangle_{H}=\left|\alpha, t_{0}=0\right\rangle$$

$$\left|\alpha, t_{0}=0 ; t\right\rangle_{S}=\mathscr{U}(t)\left|\alpha, t_{0}=0\right\rangle$$

\begin{aligned} &s\left\langle\alpha, t_{0}=0 ; t\left|A^{(S)}\right| \alpha, t_{0}=0 ; t\right\rangle_{S}=\left\langle\alpha, t_{0}=0\left|\mathscr{U}^{\dagger} A^{(S)} \mathscr{U}\right| \alpha, t_{0}=0\right\rangle \ &=H\left\langle\alpha, t 0=0 ; t\left|A^{(H)}(t)\right| \alpha, t_{0}=0 ; t\right\rangle_{H} . \end{aligned}

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