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Let $\Psi(x)$ be the wave function for a one-dimensional quantum-mechanical system, $i=\sqrt{-1}$ and let $p_x$ be the momentum operator in one dimension. Show that for any real number ℓ one has

$$e^{ilp_{x}}\Psi(x)=\Psi(x+l)$$

In the answer sheet it mentions a method with Fourier transform and another method using power series. I'm struggling to solve it either way. For the power series method I guess that I should expand as following $e^{ilp_{x}}=\sum_{n=0}^{\infty} a_n (?-?_0)^n$ however I am unsure what to write instead of "$?$" and "$?_0$" since the left hand side has no variabel? I mostly interested in the power series method but understanding the Fourie method would also be nice.

Thanks in advance!

P.S. I'm unsure if one could call this "phase-shifting" please feel free to change the title if it is not accurate terminology.

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  • $\begingroup$ What definition of the operator $p_x$ do you know? $\endgroup$ – J.G. May 31 at 10:12
  • $\begingroup$ $p_x=-i\hbar \frac{\partial}{\partial x}$. Momentum operator $\vec p$ is always $\vec p=-i\hbar\nabla$ ? $\endgroup$ – ludz May 31 at 10:15
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    $\begingroup$ You're on the right track now, although the question works with the nondimensionalization $\hbar=1$. Can you expand $e^{d/dx}$ as a power series in $d/dx$, then apply it to $\Psi$ by Taylor's theorem? $\endgroup$ – J.G. May 31 at 10:22
  • $\begingroup$ $e^{\frac{d}{dx}} \Psi(x)=\Sigma \frac{(\frac{d}{dx})^n}{n!} \Psi(x)$ ? $\endgroup$ – ludz May 31 at 10:28
  • $\begingroup$ Sorry, that should have been $e^{l\tfrac{d}{dx}}$. $\endgroup$ – J.G. May 31 at 10:29
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This is clearly a home-work style question, so I'll only give you some hints as to how to do it. First, the quantity $$\hat{T}(l) = e^{il p_x}$$ is an operator. More precisely, it is a function of the momentum operator $\hat{p}_x$. Functions of operators are precisely defined by their power series expansions. In other words, suppose some function $f(u)$ has a power series expansion $$f(u) = \sum_{n=0}^\infty a_n u^n,$$ then the operator $f(\hat{p}_x)$ (for example) is defined as $$f(\hat{p}_x) \equiv \sum_{n=0}^\infty a_n \hat{p}_x^n,\tag{1}\label{1}$$ where the coefficients $a_n$ are the same, and $\hat{p}_x^n$ represents the $\hat{p}_x$ operator acting $n$ times in succession.

In your case, your function $f(u) = e^{ilu}$, whose series expansion is very well known. If you write the series expansion of this function using the form of the momentum operator $\hat{p}_x = -i\hbar \partial_x$, the answer should be obvious. (If it isn't, go back and revise the Taylor Series.)

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The power series "proof" is popular, appealing, but fallacious. It requires the wavefunction to be analytic -- i.e to have a Taylor series that actually converges to the function. There is no reason for the wavefunction to have that property. If the wavefunction vanishes outside some closed interval there is no way that $e^{i\hat p_x a}\psi$ with the $e^{i\hat p_x a}$ defined as $$ e^{i\hat p_x a}\stackrel{?}=1+ia\hat p_x+\frac{a^2}{2}(i\hat p_x)^2+\dots =1+a \partial_x+\frac {a^2}{2} \partial^2_{xx}+\ldots $$ can ever make $\psi$ nonzero where it was originally zero.

On the other hand, if $\psi(x)= \langle x|\psi\rangle$, and $\tilde \psi(p) =\langle p|\psi\rangle$, with $\hat p|p\rangle=p|p\rangle $ then
$$ \psi(x)= \int \frac{dk}{2\pi} e^{ikx}\tilde \psi(k) $$ and $$ \psi(x+a)=\int \frac{dk}{2\pi} e^{ika} e^{ikx} \tilde \psi(k) $$ and this can be turned into valid proof of the claim.

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