I don't understand about infinitesimal translation

  • #1
Another
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I don't understand about infinitesimal translation
QM1.1.png


and
QM1.png


But

QM2.png


The infinitesimal translation denoted by equ 1.6.12 and 1.6.32

And then i understand about equation 1.6.35 but equation 1.6.36

Why they take limit N go to inf ? , multiply 1/ N ? and power N ?

So is the relationship below still true?

## F(Δ x'\hat{x}) = 1 - \frac{i p_x \dot{} Δ x'\hat{x} }{\bar{h}} = lim_{N → ∞} (1 - \frac{i p_x Δ x'}{N\bar{h}})^N ## ?
 
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  • #2
If you want to displace a system by ##\Delta x##, you can do that in ##N## smaller steps of ##\frac{\Delta x}{N}##. Each of these steps is carried out by the operator that translates a state by ##\frac{\Delta x}{N}##. Carrying out one operation after another is represented by operator multiplication. So, in simple notation:
$$T(\Delta x) = T(\frac{\Delta x}{N})^N$$
You can then imagine taking the limit of this as ##N \rightarrow \infty##. And, in this case, looking at the form that ##T(\frac{\Delta x}{N})## takes, the limit is an exponential.
 
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  • #3
Another, maybe simpler, way is to realize that
$$\mathcal{T}(x+\mathrm{d} x) =\mathcal{T}(\mathrm{d} x) \mathcal{T}(x) $$
and thus (using natural units with ##\hbar=1##)
$$\mathcal{T}(x+\mathrm{d}x)-\mathcal{T}(x) = [\mathcal{T}(\mathrm{d} x)-1]\mathcal{T}(x) =-\mathrm{i}\hat{p} \mathrm{d} x \mathcal{T}(x)$$
or
$$\mathcal{T}'(x)=-\mathrm{i} \hat{p} \mathcal{T}(x) \; \Rightarrow \; \mathcal{T}(x)=\exp(-\mathrm{i} \hat{p} x).$$
It's also realized easily in the position representation of a wave function [EDIT: Corrected in view of #4]:
$$\mathcal{T}(a) \psi(x)=\exp(-\mathrm{i} \hat{p} a) \psi(x) = \exp(-a \mathrm{d}_x) \psi(x) = (1-a \mathrm{d}x + a^2/2 \mathrm{d}_x^j+\cdots) \psi(x) = \sum_{j=0}^{\infty} \frac{1}{j!} (-a)^j \mathrm{d}_x^2 \psi(x)=\psi(x-a).$$
 
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  • #4
vanhees71 said:
It's also realized easily in the position representation of a wave function:
$$\mathcal{T}(a) \psi(x)=\exp(-\mathrm{i} \hat{p} a) \psi(x) = \exp(-a \mathrm{d}_x) \psi(x) = (1-a \mathrm{d}x + a^2/2 \mathrm{d}_x^2 +\cdots) \psi(x) = \sum_{j=0}^{\infty} \frac{1}{j!} (-a)^j \mathrm{d}_x^2 \psi(x)=\psi(x-a).$$
Shouldn't the last part read as follows?

$$\mathcal{T}(a) \psi(x)=\exp(-a \mathrm{d}_x) \psi(x) = (1-a \mathrm{d}_x + (a^2/2!) \, \mathrm{d}_x^2 +\cdots) \psi(x) = \sum_{j=0}^{\infty} \frac{1}{j!} (-a)^j \mathrm{d}_x^j \psi(x)=\psi(x-a).$$
 
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  • #5
Sure, I'll correct it.
 
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