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# 物理代写|量子力学代写Quantum mechanics代考|PX3511 Computation of < p|x >

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## 物理代写|量子力学代写Quantum mechanics代考|Computation of < p|x >

We are now in a position to calculate the inner product $\langle p \mid x\rangle$. Quantities of this type, involving the dot product of different basis vectors, are often needed in quantum mechanics. The following computation may be considered a model for the standard method. One sandwiches an operator between the two states such that its action on either state is known. In the present case we know how to apply either the position or momentum operator, so we may use either one. For example, consider $\langle p|\hat{x}| x\rangle$ and apply the operator $\hat{x}$ to either the ket or the bra.
$$=\left{\begin{array}{l} x \ i \hbar \frac{\partial}{\partial p} \end{array}\right.$$
The two expressions are equal, so that the complex function  must satisfy the first order differential equation
$$i \hbar \frac{\partial}{\partial p}-x=0 .$$
This has the solution
$$=c e^{-\frac{i}{h} p x}$$
where $c$ is a constant. Similar steps give the complex conjugate $=$ $c^* e^{\frac{i}{h} p x}$. To find $c$ consider the consistency with the normalization of the basis and insert identity in the form of the completeness relation
\begin{aligned} \delta\left(p-p^{\prime}\right)=&=\int d x\ &=\int d x|c|^2 e^{-\frac{i}{h}\left(p-p^{\prime}\right) x} \ &=|c|^2 2 \pi \hbar \delta\left(p-p^{\prime}\right) \end{aligned}
Therefore, $|c|=\frac{1}{\sqrt{2 \pi \hbar}}$. The phase of $c$ may be reabsorbed into a redefinition of the phases of the states $|x\rangle,|p\rangle$, so that finally
$$\left|x>=\frac{1}{\sqrt{2 \pi \hbar}} \int_{-\infty}^{\infty} d p\right| p>e^{-\frac{i}{\hbar} p x}, \quad\left|p>=\frac{1}{\sqrt{2 \pi \hbar}} \int_{-\infty}^{\infty} d x\right| x>e^{\frac{i}{\hbar} p x}$$

## 物理代写|量子力学代写Quantum mechanics代考|Translations in space and time

In a very general way one can see that the momentum operator $\hat{p}$ is the infinitesimal generator of translations in coordinate space. Consider two observers, one using the basis $\mid x>$ and the other using the basis $\mid x+a>$ because he measures distances from a different origin that is translated by the amount $a$ relative to the first observer. Both bases are complete, and either one may be used to write an expression for a general state vector. How are the two bases related to each other? Consider the Taylor expansion of $\mid x+a>$ in powers of $a$, and rewrite it as follows by taking advantage of $\partial_x\left|x>=-\frac{i}{\hbar} \hat{p}\right| x>$
\begin{aligned} \mid x+a>&=\sum_{n=0}^{\infty} \frac{a^n}{n !} \frac{\partial^n}{\partial x^n} \mid x>\ &=\sum_{n=0}^{\infty} \frac{1}{n !}\left(\frac{-i a \hat{p}}{\hbar}\right)^n \mid x>\ &=e^{-i a \hat{p} / \hbar} \mid x> \end{aligned}
So, a finite translation by a distance $a$ is performed by the translation operator $\widehat{T}_a=\exp (-i a \hat{p} / \hbar)$, and the infinitesimal generator of translations is the momentum operator. That is, an infinitesimal translation $|x+a>=| x>+\delta_a \mid x>$, is expressed in terms of the momentum operator as
$$\delta_a\left|x>=a \frac{\partial}{\partial x}\right| x>=-\frac{i a}{\hbar} \hat{p} \mid x>.$$

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## 物理代写|量子力学代写Quantum mechanics代考|Computation of $\langle p \mid x\rangle$

$\$ \$$=\backslash left {\backslash begin { array }{1} x 1 i \hbar \frac {\backslash partial }{\backslash partial p} lend { 数组 } \backslash 对。 \ \$$

$\backslash$ left $|x\rangle=\backslash$ frac ${1}{\backslash$ sqrt ${2 \backslash$ pi $\backslash$ hbar $}} \backslash$ int_ ${-\backslash$ infty $} \wedge{\backslash$ infty $}$ dp \right| $p>e \wedge{-\backslash$ frac ${i}{\backslash$ hbar $}$ px $}$,
$\backslash$ quad $\backslash$ left $|p\rangle=\backslash$ frac ${1}{\backslash$ sqrt ${2 \backslash$ pi $\backslash$ hbar $}} \backslash$ int_ ${-\backslash$ infty $} \wedge{\backslash$ infty $} d x \backslash$ 右| $x>e \wedge{\backslash$ frac ${i}{\backslash$ hbar $}$ px $}$
$\$ \$$## 物理代写|量子力学代写Quantum mechanics代考|Translations in space and time 以一种非常一般的方式，我们可以看到动量算符 \hat{p} 是坐标空间中平移的无穷小生成器。考虑两个观察者，一个使用基 \mid x> 另一个 使用基础 \mid x+a> 因为他测量了从不同原点的距离，这个原点被转换成 a 相对于第一个观察者。两个基都是完整的，其中任何一 个都可以用来编写一般状态向量的表达式。这两个基地如何相互关联? 考虑泰勒展开 \mid x+a> 在权力的 a, 并利用 \partial_x\left|x>=-\frac{i}{h} \hat{p}\right| x>$$
\left|x+a>=\sum_{n=0}^{\infty} \frac{a^n}{n !} \frac{\partial^n}{\partial x^n}\right| x>\quad=\sum_{n=0}^{\infty} \frac{1}{n !}\left(\frac{-i a \hat{p}}{\hbar}\right)^n\left|x>=e^{-i a \hat{p} / \hbar}\right| x>
$$|x+a>=| x>+\delta_a \mid x>, 用动量算子表示为$$
\delta_a\left|x>=a \frac{\partial}{\partial x}\right| x>=-\frac{i a}{\hbar} \hat{p} \mid x>.


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