I'm trying to understand a proof my lecturer gave. We define $\hat{T}(a)=e^{-ia\hat{p}/\hbar}$ and write

$\hat{x}\hat{T}(a)|x\rangle =\hat{x}(\mathbb{1}-i\frac{a}{\hbar}\hat{p}+\frac{1}{2}(-i\frac{a}{\hbar}\hat{p})^2+\dots)|x\rangle=(\hat{x}-i\frac{a}{\hbar}\hat{p}\hat{x}+\frac{1}{2}(-i\frac{a}{\hbar}\hat{p})^2\hat{x}+\dots)|x\rangle+a(\mathbb{1}-i\frac{a}{\hbar}\hat{p}+\ldots)|x\rangle=(x+a)e^{-ia\hat{p}/\hbar}|x\rangle=(x+a)\hat{T}(a)|x\rangle.$

Then he says this implies $\hat{T}(a)|x\rangle\propto|x+a\rangle$. Since $\hat{T}$ is unitary this gives $\hat{T}(a)|x\rangle=|x+a\rangle$

I'm confused on how this is the case. Whats inside bras and kets are just labels, so does this mean that kets are labelled their eigenvalues of the position operator?

  • $\begingroup$ Review your Lagrange shift operator--Taylor expansion. $\endgroup$ – Cosmas Zachos Dec 1 '19 at 15:13
  • $\begingroup$ Why is it that the second and third part in the expression for the operator $\hat{x}\hat{T}(a)$ your equation are equal? $\endgroup$ – descheleschilder Dec 1 '19 at 15:50
  • $\begingroup$ @descheleschilder this follows from repeated applications of $\hat{x}\hat{p}=\hat{p}\hat{x}+i\hbar$ $\endgroup$ – jonathan Dec 2 '19 at 1:10
  • $\begingroup$ Of course. I was so stupid not to notice the order of $\hat{x}$ and $\hat{p}$ were reversed. So $\hat{p}\hat{x}=\hat{x}\hat{p}-i\hbar$. Filling this in gives indeed and extra $a$. After the second equal sign, we can pull $\hat{x}$ out of the first term and put is to the front just as we did with $a$ in the second term, and there you go. Note that if we replace the $\hat{p}$ by $\hat{x}$ in the operator $\hat{T}(a)=e^{-ia\hat{p}/\hbar}$ you'll get an operator (working in the p-representation) that replaces the momentum from $p$ to $p+a$ $\endgroup$ – descheleschilder Dec 5 '19 at 14:13

There is a new state $T(a)|x\rangle$ which we name $|?\rangle$ Turns out, from your computations, that the eigenvalue of the position operators on $|?\rangle$ is (a+x) - so we name it with a more meaningful name $|x+a\rangle$. It is common practice to write inside the bra-kets the result of a given operator or at least a hint of what the result would be. The value you put inside the bra-kets depends on what you want to do, is arbitrary. Since here we are talking about positions, the position's eigenvalue is a smart choice (:

This aside, the computations simply proves that if you take a state whose position eigenvalue is $x$ and you use the operator $T(a)$ you get a new state whose position's eigenvalue now is $x+a$ - i. e. $T(a)$ is the translation operator. If you rename the kets as $|A\rangle$ and $|B\rangle$ such that

$\hat{x}|A \rangle=x|A\rangle$ and $\hat{x}|B\rangle=(a+x)|B\rangle$ and such that $T(a)\hat{x}|A \rangle=\hat{x}|B \rangle$ nothing changes, conceptually. Is just harder to read.

  • $\begingroup$ Are we making assumptions then on what $|x\rangle$ is? Are am I supposed to assume that $|x\rangle$ has $\hat{x}|x\rangle=x$ from the get go? $|x\rangle$ is supposed to be a vector in Hilbert space, so if we limit our discussion to vectors with $\hat{x}|x\rangle=x$ then we're missing some, right? $\endgroup$ – jonathan Dec 1 '19 at 14:24
  • $\begingroup$ It is whatever vector such that $\hat{x}|A\rangle = x|A\rangle$. Posited that you have (any) such vector, then $T$ will translate it. You can just repeat it for any vector with eigenvalue $x_1$ or $x_2$ or whatever - or you can just assume the generality. It is a very normal kind of demonstration! For any x such that... then $T$ translates it. $\endgroup$ – JalfredP Dec 1 '19 at 14:35

First, yes, in the notation $|x\rangle$ and $|p\rangle$ the symbols $x$ and $p$ label the eigenvalue of the eigenstate represented by the ket. In that sense $|x+a\rangle$ is the eigenstate of position with eigenvalue $x+a$. It is very common that when working with Hermitian operators one labels the eigenstates of the basis by the eigenvalue (beware, though, that in the degenerate case an additional label might be required to account for the degeneracy).

Nevertheless, I want to address the title question on how to show that $T_a(P)$ is the generator of translations in, what I believe, is a simpler approach that might help you out.

When you have some operator, which is a function of a Hermitian operator, then you know its action in that operator's basis. Here that is the case with the translation operator $T_a(P)$ a function of the momentum operator $P$. In the momentum basis $|p\rangle$ we have $$P|p\rangle=p|p\rangle,$$

and then by definition $T_a(P)$ is $$T_a(P)|p\rangle=T_a(p)|p\rangle.$$

In that case recalling $T_a(P) = e^{-iaP/\hbar}$ this means $$T_a(P)=e^{-iap/\hbar}|p\rangle.$$

This is the main equation for us. Now let $|x\rangle$ be a position eigenstate. We don't know yet how $T_a(P)$ acts upon it. But since we know how $T_a(P)$ acts upon momentum eigenstates and since $T_a(P)$ is linear, we just need expand $|x\rangle$ in the momentum basis:

$$|x\rangle=\int \langle p|x\rangle |p\rangle dp.$$

Recall, though, that $\langle x|p\rangle = \frac{1}{\sqrt{2\pi \hbar}}e^{ipx/\hbar}$ from which taking the complex conjugate it follows $\langle p|x\rangle=\frac{1}{\sqrt{2\pi \hbar}}e^{-ipx/\hbar}.$

In that case we finally get $$T_a(P)|x\rangle=\frac{1}{\sqrt{2\pi \hbar}}\int e^{-ipx/\hbar}T_a(P)|p\rangle dp =\frac{1}{\sqrt{2\pi\hbar}}\int e^{-iap/\hbar} e^{-ipx/\hbar}|p\rangle=\frac{1}{\sqrt{2\pi\hbar}}\int e^{-i(x+a)p/\hbar}|p\rangle dp$$

but now in the last equality look to the integrand. Recalling the form of general $\langle p|x\rangle$ it is just $\langle p|x+a\rangle$. Thus $$T_a(P)|x\rangle = \int \langle p|x+a\rangle |p\rangle dp = |x+a\rangle,$$

the last equality because $\int dp |p\rangle\langle p| = \mathbf{1}$ is the identity operator.


In "Quantum Mechanics" Leonard I.Schiff, 3rd Edition 1968 :

26$\boldsymbol{\square}$SPACE AND TIME DISPLACEMENTS

We shall work entirely in the Schr$\ddot{\rm o}$dinger picture. As our first symmetry operation, we consider the displacement in space of a physical system in a state represented by the ket $\,\boldsymbol{|}\boldsymbol{a}\boldsymbol{\rangle}\,$ or the wave function $\,\boldsymbol{\psi}_{\boldsymbol{a}}\left(\mathbf{r}\right)$.The displacement is through a vector $\,\boldsymbol{\varrho}\,$ and changes the ket $\,\boldsymbol{|}\boldsymbol{a}\boldsymbol{\rangle}\,$ into the ket $\,\boldsymbol{|}\boldsymbol{a}^{\boldsymbol{\prime}}\boldsymbol{\rangle}\,$ or changes the wave function $\,\boldsymbol{\psi}_{\boldsymbol{a}}\left(\mathbf{r}\right)\,$ into the wave function $\,\boldsymbol{\psi}_{\boldsymbol{a}{\!\;\boldsymbol{\prime}}}\left(\mathbf{r}\right)$. This means that \begin{equation} \boldsymbol{\psi}_{\boldsymbol{a}{\!\;\boldsymbol{\prime}}}\left(\mathbf{r}\boldsymbol{+}\boldsymbol{\varrho}\right)\boldsymbol{=}\boldsymbol{\psi}_{\boldsymbol{a}}\left(\mathbf{r}\right) \tag{26.1}\label{26.1} \end{equation} As an example of \eqref{26.1} we note that, if $\,\boldsymbol{\psi}_{\boldsymbol{a}}\left(\mathbf{r}\right)\,$ is a wave packet that has its maximum value when its argument $\,\mathbf{r}\,$ is equal to $\,\mathbf{r_0}$, then $\,\boldsymbol{\psi}_{\boldsymbol{a}{\!\;\boldsymbol{\prime}}}\left(\mathbf{r}\right)\,$ is a wave packet of the same shape that has its maximum value when its argument $\,\mathbf{r}\,$ is equal to $\,\mathbf{r_0}\boldsymbol{+}\boldsymbol{\varrho}$. Thus $\,\boldsymbol{\psi}_{\boldsymbol{a}{\!\;\boldsymbol{\prime}}}\,$ has its maximum displaced by $\,\boldsymbol{\varrho}\,$ with respect to the maximum of $\,\boldsymbol{\psi}_{\boldsymbol{a}}$.

What we have just described is usually called the active point of view, in which the coordinate system is unchanged and the state function is displaced. The completely equivalent passive point of view consists in leaving the state function unchanged but referring it to a coordinate system that is displaced by the vector $\,\boldsymbol{-\varrho}\,$ with respect to the original coordinate system.



.......We now wish to find a transformation.....that changes the ket $\,\boldsymbol{|}\boldsymbol{a}\boldsymbol{\rangle}\,$ into the ket $\,\boldsymbol{|}\boldsymbol{a}^{\boldsymbol{\prime}}\boldsymbol{\rangle}$; since the norm of the ket should not be affected by displacement, the corresponding operator is expected to be unitary, and this will be shown by explicit construction. We call this operator $\,U_{\boldsymbol{r}}\left(\boldsymbol{\varrho}\right)\,$, where the subscript denotes a displacement in space and the argument is the vector displacement interval.

We thus have \begin{equation} U_{\boldsymbol{r}}\left(\boldsymbol{\varrho}\right)\boldsymbol{|}\boldsymbol{a}\boldsymbol{\rangle}\boldsymbol{=}\boldsymbol{|}\boldsymbol{a}^{\boldsymbol{\prime}}\boldsymbol{\rangle} \quad \text{or} \quad U_{\boldsymbol{r}}\left(\boldsymbol{\varrho}\right)\boldsymbol{\psi}_{\boldsymbol{a}}\left(\mathbf{r}\right)\boldsymbol{=} \boldsymbol{\psi}_{\boldsymbol{a}{\!\;\boldsymbol{\prime}}}\left(\mathbf{r}\right) \tag{26.2}\label{26.2} \end{equation} The second of Eqs.\eqref{26.2} makes use of the coordinate representation of the state vector; together with Eq. \eqref{26.1} it becomes \begin{equation} U_{\boldsymbol{r}}\left(\boldsymbol{\varrho}\right)\boldsymbol{\psi}_{\boldsymbol{a}}\left(\mathbf{r}\right)\boldsymbol{=} \boldsymbol{\psi}_{\boldsymbol{a}}\left(\mathbf{r}\boldsymbol{-\varrho}\right) \tag{26.3}\label{26.3} \end{equation} In evaluating $\,\boldsymbol{\psi}_{\boldsymbol{a}}\left(\mathbf{r}\boldsymbol{-\varrho}\right)\,$ it is convenient to choose the coordinate axes at first so that the $\,x\,$ axis is in the direction of the vector $\,\boldsymbol{\varrho}$. Then a Taylor's series expansion gives \begin{align} \boldsymbol{\psi}_{\boldsymbol{a}}\left(\mathbf{r}\boldsymbol{-\varrho}\right)&\boldsymbol{=}\boldsymbol{\psi}_{\boldsymbol{a}}\left(x\boldsymbol{-}\rho,y,z\right) \nonumber\\ &\boldsymbol{=}\boldsymbol{\psi}_{\boldsymbol{a}}\left(x,y,z\right)\boldsymbol{-}\rho\dfrac{\partial}{\partial x}\boldsymbol{\psi}_{\boldsymbol{a}}\left(x,y,z\right)\boldsymbol{+}\dfrac{\rho^2}{2!}\dfrac{\partial^2}{\partial x^2}\boldsymbol{\psi}_{\boldsymbol{a}}\left(x,y,z\right)\boldsymbol{-}\cdots \nonumber \end{align} The right side may be written in the form \begin{equation} \boldsymbol{e}^{\boldsymbol{-\rho}\boldsymbol{\left(\partial/\partial x\right)}}\boldsymbol{\psi}_{\boldsymbol{a}}\left(x,y,z\right) \nonumber \end{equation} For a general choice of coordinate axes, $\rho\left(\partial/\partial x\right)$ may be replaced by $\,\boldsymbol{\varrho\cdot\nabla}\,$, so that we obtain \begin{equation} \boldsymbol{\psi}_{\boldsymbol{a}}\left(\mathbf{r}\boldsymbol{-\varrho}\right)\boldsymbol{=}\exp\left(\boldsymbol{-\varrho\cdot\nabla}\right)\boldsymbol{\psi}_{\boldsymbol{a}}\left(\mathbf{r}\right)\boldsymbol{=}\exp\left(\dfrac{\boldsymbol{-}i\boldsymbol{\varrho\cdot \mathbf{p}}}{\hbar}\right)\boldsymbol{\psi}_{\boldsymbol{a}}\left(\mathbf{r}\right) \nonumber \end{equation} where the momentum operator $\,\mathbf{p}\boldsymbol{=}\boldsymbol{-}i\hbar\boldsymbol{\nabla}\,$ has been introduced. Thus if we choose \begin{equation} U_{\boldsymbol{r}}\left(\boldsymbol{\varrho}\right)\boldsymbol{=}\exp\dfrac{\boldsymbol{-}i\boldsymbol{\varrho\cdot \mathbf{p}}}{\hbar} \tag{26.4}\label{26.4} \end{equation} Eq.\eqref{26.3} is valid for all state vectors. Further, since the operator $\,\boldsymbol{\nabla}$, which is defined only in the coordinate representation, has been replaced by $\,i\mathbf{p}/\hbar\,$, Eq. \eqref{26.4} is valid in all representations. It is easily verified that $\,U_{\boldsymbol{r}}\left(\boldsymbol{\varrho}\right)\,$ is unitary, since $\,\boldsymbol{\varrho}\,$ is real and $\,\mathbf{p}\,$ is hermitian.


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