# $\Delta E\Delta t\geq\hbar/2$是如何推导来的？

1. 7周前

### Kushim Jiang

1楼 1月3日
7周前Kushim Jiang 重新编辑

$\Delta E\Delta t\geq\hbar/2$是如何推导来的？因为$[\hat E,\hat t]$也等于$i\hbar$？

2. ### baishuxu

2楼 1月4日 天文版主
6周前baishuxu 重新编辑

#### Ref: Uncertainty principle

Note: Please read section "Robertson–Schrödinger uncertainty relations".

The Schrödinger uncertainty relation
$$\sigma_A^2\sigma_B^2 \geq \left| \frac{1}{2}\langle\{\hat{A}, \hat{B}\}\rangle - \langle \hat{A} \rangle\langle \hat{B}\rangle \right|^2+ \left|\frac{1}{2i} \langle[ \hat{A}, \hat{B}] \rangle\right|^2$$

Proof of the Schrödinger uncertainty relation

The derivation shown here incorporates and builds off of those shown in Robertson, Schrödinger and standard textbooks such as Griffiths. For any Hermitian operator $\hat{A}$, based upon the definition of variance, we have

$$\sigma_A^2 = \langle(\hat{A}-\langle \hat{A} \rangle)\Psi|(\hat{A}-\langle \hat{A} \rangle)\Psi\rangle$$

we let $|f\rangle=|(\hat{A}-\langle \hat{A} \rangle)\Psi\rangle$ and thus

$$\sigma_A^2 = \langle f\mid f\rangle\,$$

Similarly, for any other Hermitian operator $\hat{B}$ in the same state

$$\sigma_B^2 = \langle(\hat{B}-\langle \hat{B} \rangle)\Psi|(\hat{B}-\langle \hat{B} \rangle)\Psi\rangle = \langle g\mid g\rangle$$

for $|g\rangle=|(\hat{B}-\langle \hat{B} \rangle)\Psi \rangle$

The product of the two deviations can thus be expressed as

$$\sigma_A^2\sigma_B^2 = \langle f\mid f\rangle\langle g\mid g\rangle \tag{1}$$

In order to relate the two vectors $|f\rangle$ and $|g\rangle$, we use the Cauchy–Schwarz inequality which is defined as

$$\langle f\mid f\rangle\langle g\mid g\rangle \geq |\langle f\mid g\rangle|^2$$

and thus Eq. (1) can be written as

$$\sigma_A^2\sigma_B^2 \geq |\langle f\mid g\rangle|^2\tag{2}$$

Since $\langle f\mid g\rangle$ is in general a complex number, we use the fact that the modulus squared of any complex number $z$ is defined as $|z|^2=zz^{*}$, where $z^{*}$ is the complex conjugate of $z$. The modulus squared can also be expressed as

$$|z|^2 = \Big(\operatorname{Re}(z)\Big)^2+\Big(\operatorname{Im}(z)\Big)^2 = \Big(\frac{z+z^\ast}{2}\Big)^2 +\Big(\frac{z-z^\ast}{2i}\Big)^2\tag{3}$$

we let $z=\langle f\mid g\rangle$ and $z^{*}=\langle g \mid f \rangle$ and substitute these into the equation above to get

$$|\langle f\mid g\rangle|^2 = \bigg(\frac{\langle f\mid g\rangle+\langle g\mid f\rangle}{2}\bigg)^2 + \bigg(\frac{\langle f\mid g\rangle-\langle g\mid f\rangle}{2i}\bigg)^2\tag{4}$$

The inner product $\langle f\mid g\rangle$ is written out explicitly as

$$\langle f\mid g\rangle = \langle(\hat{A}-\langle \hat{A} \rangle)\Psi|(\hat{B}-\langle \hat{B} \rangle)\Psi\rangle$$

and using the fact that $\hat{A}$ and $\hat{B}$ are Hermitian operators, we find

\begin{aligned} \langle f\mid g\rangle & = \langle\Psi|(\hat{A}-\langle \hat{A}\rangle)(\hat{B}-\langle \hat{B}\rangle)\Psi\rangle \\ & = \langle\Psi\mid(\hat{A}\hat{B}-\hat{A}\langle \hat{B}\rangle - \hat{B}\langle \hat{A}\rangle + \langle \hat{A}\rangle\langle \hat{B}\rangle)\Psi\rangle \\ & = \langle\Psi\mid\hat{A}\hat{B}\Psi\rangle-\langle\Psi\mid\hat{A}\langle \hat{B}\rangle\Psi\rangle -\langle\Psi\mid\hat{B}\langle \hat{A}\rangle\Psi\rangle+\langle\Psi\mid\langle \hat{A}\rangle\langle \hat{B}\rangle\Psi\rangle \\ & =\langle \hat{A}\hat{B}\rangle-\langle \hat{A}\rangle\langle \hat{B}\rangle-\langle \hat{A}\rangle\langle \hat{B}\rangle+\langle \hat{A}\rangle\langle \hat{B}\rangle \\ & =\langle \hat{A}\hat{B}\rangle-\langle \hat{A}\rangle\langle \hat{B}\rangle. \end{aligned}

Similarly it can be shown that $\langle g\mid f\rangle = \langle \hat{B}\hat{A}\rangle-\langle \hat{A}\rangle\langle \hat{B}\rangle$

Thus we have

$$\langle f\mid g\rangle-\langle g\mid f\rangle = \langle \hat{A}\hat{B}\rangle-\langle \hat{A}\rangle\langle \hat{B}\rangle-\langle \hat{B}\hat{A}\rangle+\langle \hat{A}\rangle\langle \hat{B}\rangle = \langle [\hat{A},\hat{B}]\rangle$$

and

$$\langle f\mid g\rangle+\langle g\mid f\rangle = \langle \hat{A}\hat{B}\rangle-\langle \hat{A}\rangle\langle \hat{B}\rangle+\langle \hat{B}\hat{A}\rangle-\langle \hat{A}\rangle\langle \hat{B}\rangle = \langle \{\hat{A},\hat{B}\}\rangle -2\langle \hat{A}\rangle\langle \hat{B}\rangle$$

We now substitute the above two equations above back into Eq. ({{EquationNote|4}}) and get

$$|\langle f\mid g\rangle|^2=\Big(\frac{1}{2}\langle\{\hat{A},\hat{B}\}\rangle - \langle \hat{A} \rangle\langle \hat{B}\rangle\Big)^2 + \Big(\frac{1}{2i} \langle[\hat{A},\hat{B}]\rangle\Big)^{2}\,$$

Substituting the above into Eq. ({{EquationNote|2}}) we get the Schrödinger uncertainty relation

$$\sigma_A\sigma_B \geq \sqrt{\Big(\frac{1}{2}\langle\{\hat{A},\hat{B}\}\rangle - \langle \hat{A} \rangle\langle \hat{B}\rangle\Big)^2 + \Big(\frac{1}{2i} \langle[\hat{A},\hat{B}]\rangle\Big)^2}.$$

This proof has an issue related to the domains of the operators involved. For the proof to make sense, the vector $\hat{B} |\Psi \rangle$ has to be in the domain of the unbounded operator $\hat{A}$, which is not always the case. In fact, the Robertson uncertainty relation is false if $\hat{A}$ is an angle variable and $\hat{B}$ is the derivative with respect to this variable. In this example, the commutator is a nonzero constant—just as in the Heisenberg uncertainty relation—and yet there are states where the product of the uncertainties is zero. (See the counterexample section below.) This issue can be overcome by using a variational method for the proof, or by working with an exponentiated version of the canonical commutation relations.

Note that in the general form of the Robertson–Schrödinger uncertainty relation, there is no need to assume that the operators $\hat{A}$ and $\hat{B}$ are self-adjoint operators. It suffices to assume that they are merely symmetric operators. (The distinction between these two notions is generally glossed over in the physics literature, where the term ''Hermitian'' is used for either or both classes of operators. See Chapter 9 of Hall's book for a detailed discussion of this important but technical distinction.)

Example: In non-relativistic mechanics, time is privileged as an independent variable. Nevertheless, in 1945, L. I. Mandelshtam and I. E. Tamm derived a non-relativistic time–energy uncertainty relation, as follows. For a quantum system in a non-stationary state $\psi$ and an observable $B$ represented by a self-adjoint operator $\hat{B}$, the following formula holds:

$$\sigma_E \frac{\sigma_B}{\left| \frac{\mathrm{d}\langle \hat B \rangle}{\mathrm{d}t}\right |} \ge \frac{\hbar}{2}$$

where $\sigma_E$ is the standard deviation of the energy operator (Hamiltonian) in the state $\psi$, $\sigma_B$ stands for the standard deviation of $B$. Although the second factor in the left-hand side has dimension of time, it is different from the time parameter that enters the Schrödinger equation. It is a lifetime of the state $\psi$ with respect to the observable $B$: In other words, this is the time interval ($\Delta t$) after which the expectation value $\langle {\hat {B}}\rangle$ changes appreciably.

An informal, heuristic meaning of the principle is the following: A state that only exists for a short time cannot have a definite energy. To have a definite energy, the frequency of the state must be defined accurately, and this requires the state to hang around for many cycles, the reciprocal of the required accuracy. For example, in spectroscopy, excited states have a finite lifetime. By the time–energy uncertainty principle, they do not have a definite energy, and, each time they decay, the energy they release is slightly different. The average energy of the outgoing photon has a peak at the theoretical energy of the state, but the distribution has a finite width called the natural linewidth. Fast-decaying states have a broad linewidth, while slow-decaying states have a narrow linewidth.

The same linewidth effect also makes it difficult to specify the rest mass of unstable, fast-decaying particles in particle physics. The faster the particle decays (the shorter its lifetime), the less certain is its mass (the larger the particle's width).

Hope it helps.

3. 6周前

### 蹑履思登

3楼 1月5日

t是背景参数，不是力学量，没有“时间算符”，所以谈及这个需要先明确何谓“时间不确定度”。对于某个在薛定谔绘景中不含时的力学量A，（它的不确定度）除以（它期望值随时间变化率）定义为“时间不确定度”，计算后便能发现其跟能量不确定度的关系，其实也就是楼上所说的东西。

4. ### ShongLee

4楼 1月5日

曾谨言《量》卷1，第五版，12.4 能量-时间不确定度关系。该式的意义和其他不确定度关系的是不一样的。

5. 5周前

### 薛定谔的喵

5楼 1月12日

根本没有时间算符，argue如下：
假设有 $\hat{t}$，且和 $\hat{H}$ 满足正则关系，则 $\mathrm{e}^{\mathrm{i}a\hat{t}}$ 是能量平移算符，则哈密顿量必有连续谱（事实上，正则对易关系没有有限维表示，可直接推出连续谱），这与已知的现实世界不符。