Summer Bradford

2022-06-14

Gaussian State Spread

A measurement device which can be represented by a 1D quantum system (with canonical observables $X$ and $P$) 'is prepared in a Gaussian state with spread $s$'

$|\psi \u27e9=\frac{1}{({\pi}^{2}{s}^{2}{)}^{1/4}}\int \mathrm{exp}[-\frac{{x}^{2}}{2{s}^{2}}]\phantom{\rule{negativethinmathspace}{0ex}}\mathrm{d}x|x\u27e9$

Can somebody tell me what it means that its canonical variables are $X$ and $P$?

A measurement device which can be represented by a 1D quantum system (with canonical observables $X$ and $P$) 'is prepared in a Gaussian state with spread $s$'

$|\psi \u27e9=\frac{1}{({\pi}^{2}{s}^{2}{)}^{1/4}}\int \mathrm{exp}[-\frac{{x}^{2}}{2{s}^{2}}]\phantom{\rule{negativethinmathspace}{0ex}}\mathrm{d}x|x\u27e9$

Can somebody tell me what it means that its canonical variables are $X$ and $P$?

pheniankang

Beginner2022-06-15Added 22 answers

To have canonical observables $\hat{x}$ and $\hat{p}$ means that the eigenvalues of these operators are what you measure (denotes $x,p$), and the operators satisfy the "canonical commutation relation"

$[\hat{x},\hat{p}]\equiv \hat{x}\hat{p}-\hat{p}\hat{x}=i\hslash $

To prepare a system in an initial state $|\psi \u27e9$ means smily that this is the state of the system at $t=0$; it is usually denoted as $|\psi (0)\u27e9$

Now, your initial state is given by

$\mid \psi (0)\u27e9=({\pi}^{2}{s}^{2}{)}^{-1/4}\int dx\text{}{e}^{-\frac{{x}^{2}}{2{s}^{2}}}\mid x\u27e9.$

Aside: A probability distribution given by $f(x)=({\pi}^{2}{s}^{2}{)}^{-1/4}{e}^{-\frac{{x}^{2}}{2{\sigma}^{2}}}$ is said to be a Gaussian with spread $\sigma $, where $\sigma $ is the usual standard deviation from the center of the bell curve.

End Aside

Therefore, the meaning of

$\mid \psi (0)\u27e9=({\pi}^{2}{s}^{2}{)}^{-1/4}\int dx\text{}{e}^{-\frac{{x}^{2}}{2{s}^{2}}}\mid x\u27e9.$

is that your system at $t=0$ is in a superposition of "position eigenstates" $|x\u27e9$ (i.e. eigenvectors of $\hat{x}$), weighted by a gaussian distribution. That is, your system isn't just in any old superposition of position eigenstates, but that the ones near $x=0$ are most likely and as you move away from the origin the probability of the system being in that state decreases like ${e}^{-\frac{{x}^{2}}{2{s}^{2}}}$

If you are familiar with wave functions, recall that the definition of $\u27e8x|\mathrm{\Psi}\u27e9\equiv \mathrm{\Psi}(x)$. Then you can get something a bit more useful. Namely, that

$\u27e8x|\psi (0)\u27e9=({\pi}^{2}{s}^{2}{)}^{-1/4}\int d{x}^{\prime}\text{}{e}^{-\frac{{x}^{\prime 2}}{2{s}^{2}}}\u27e8x|{x}^{\prime}\u27e9=({\pi}^{2}{s}^{2}{)}^{-1/4}\int d{x}^{\prime}\text{}{e}^{-\frac{{x}^{\prime 2}}{2{s}^{2}}}\delta (x-{x}^{\prime})=({\pi}^{2}{s}^{2}{)}^{-1/4}{e}^{-\frac{{x}^{2}}{2{s}^{2}}}$

where I changed the variable of integration to ${x}^{\prime}$ to make things clearer.

At any rate, what this means is that your initial wave function is given by

$\mathrm{\Psi}(x,t=0)=({\pi}^{2}{s}^{2}{)}^{-1/4}{e}^{-\frac{{x}^{2}}{2{s}^{2}}}.$

Where to go from here with whatever you're doing should be familiar at this point (i.e. after finding the initial wave function).

$[\hat{x},\hat{p}]\equiv \hat{x}\hat{p}-\hat{p}\hat{x}=i\hslash $

To prepare a system in an initial state $|\psi \u27e9$ means smily that this is the state of the system at $t=0$; it is usually denoted as $|\psi (0)\u27e9$

Now, your initial state is given by

$\mid \psi (0)\u27e9=({\pi}^{2}{s}^{2}{)}^{-1/4}\int dx\text{}{e}^{-\frac{{x}^{2}}{2{s}^{2}}}\mid x\u27e9.$

Aside: A probability distribution given by $f(x)=({\pi}^{2}{s}^{2}{)}^{-1/4}{e}^{-\frac{{x}^{2}}{2{\sigma}^{2}}}$ is said to be a Gaussian with spread $\sigma $, where $\sigma $ is the usual standard deviation from the center of the bell curve.

End Aside

Therefore, the meaning of

$\mid \psi (0)\u27e9=({\pi}^{2}{s}^{2}{)}^{-1/4}\int dx\text{}{e}^{-\frac{{x}^{2}}{2{s}^{2}}}\mid x\u27e9.$

is that your system at $t=0$ is in a superposition of "position eigenstates" $|x\u27e9$ (i.e. eigenvectors of $\hat{x}$), weighted by a gaussian distribution. That is, your system isn't just in any old superposition of position eigenstates, but that the ones near $x=0$ are most likely and as you move away from the origin the probability of the system being in that state decreases like ${e}^{-\frac{{x}^{2}}{2{s}^{2}}}$

If you are familiar with wave functions, recall that the definition of $\u27e8x|\mathrm{\Psi}\u27e9\equiv \mathrm{\Psi}(x)$. Then you can get something a bit more useful. Namely, that

$\u27e8x|\psi (0)\u27e9=({\pi}^{2}{s}^{2}{)}^{-1/4}\int d{x}^{\prime}\text{}{e}^{-\frac{{x}^{\prime 2}}{2{s}^{2}}}\u27e8x|{x}^{\prime}\u27e9=({\pi}^{2}{s}^{2}{)}^{-1/4}\int d{x}^{\prime}\text{}{e}^{-\frac{{x}^{\prime 2}}{2{s}^{2}}}\delta (x-{x}^{\prime})=({\pi}^{2}{s}^{2}{)}^{-1/4}{e}^{-\frac{{x}^{2}}{2{s}^{2}}}$

where I changed the variable of integration to ${x}^{\prime}$ to make things clearer.

At any rate, what this means is that your initial wave function is given by

$\mathrm{\Psi}(x,t=0)=({\pi}^{2}{s}^{2}{)}^{-1/4}{e}^{-\frac{{x}^{2}}{2{s}^{2}}}.$

Where to go from here with whatever you're doing should be familiar at this point (i.e. after finding the initial wave function).

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