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Let's define $\hat{q},\ \hat{p}$ the positon and momentum quantum operators, $\hat{a}$ the annihilation operator and $\hat{a}_1,\ \hat{a}_2$ with its real and imaginary part, such that $$ \hat{a} = \hat{a}_1 + j \hat{a}_2$$ with $$\hat{a}_1 = \sqrt{\frac{\omega}{2 \hbar}}\hat{q},\ \hat{a}_2 = \sqrt{\frac{1}{2 \hbar \omega}}\hat{p}$$ (for a reference, see Shapiro Lectures on Quantum Optical Communication, lect.4)

Define $|a_1 \rangle,\ |a_2\rangle,\ |q\rangle,\ |p\rangle$ the eigenket of the operator $\hat{a}_1,\ \hat{a}_2,\ \hat{q},\ \hat{p}$ respectively.

From the lecture, I know that $$ \langle a_2|a_1\rangle = \frac{1}{\sqrt{\pi}} e^{-2j a_1 a_2}$$ but I do not understand how to obtain $$ \langle p|q\rangle = \frac{1}{\sqrt{2\pi\hbar}} e^{-\frac{j}{\hbar}qp}$$

I thought that with a variable substitution would suffice, but substituting ${a}_1 = \sqrt{\frac{\omega}{2 \hbar}}{q},\ a_2 = \sqrt{\frac{1}{2 \hbar \omega}}{p}$, I obtain $$\frac{1}{\sqrt{\pi}} e^{-\frac{j}{\hbar}qp}$$ which does not have the correct factor $\frac{1}{\sqrt{2\pi\hbar}}$.

What am I missing?

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  • $\begingroup$ It depends on what $\langle p|a_i\rangle$ and $\langle q|a_i\rangle$ are. (It is not a test, I can't remember now). You have to put the identities inside $\langle a_1|a_2\rangle$, you can't just substitute. $\endgroup$ – Antonio Ragagnin Jun 16 '14 at 10:20
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Inner scalar producs

Since $\hat{a}_1 = \sqrt{\frac{\omega}{2 \hbar}}\hat{q}$, then $\langle a_1|q\rangle=N_1\delta\left(a_1- \sqrt{\frac{\omega}{2 \hbar}}q\right).$

Also, since $\hat{a}_2 = \sqrt{\frac{1}{2 \hbar\omega}}\hat{p}$, then $\langle a_2|p\rangle=N_2\delta\left(a_2- \sqrt{\frac{1}{2 \hbar\omega}}p\right).$

Normalization constants

We will use this property: $\int dx \delta\left(\alpha x-y\right) f(x)=\frac{f\left(\frac{y}{\alpha}\right)}{\alpha}.$

If we ask that $|a_1\rangle$ are normalized, we are asking that $$\delta\left(a_1- \bar a_2\right)=\langle a_1|\bar a_1\rangle = \int dq \langle a_1| q\rangle \langle q|\bar a_1\rangle=\int dq \left|N_1\right|^2 \delta\left(a_1- \sqrt{\frac{\omega}{2 \hbar}}q\right) \delta\left(\bar a_1- \sqrt{\frac{\omega}{2 \hbar}}q\right).$$

So, $N_1= \left(\frac{\omega}{2\hbar}\right)^\frac{1}{4}.$

Doing the same thing for $|a_2\rangle$ We then obtained that:

$\langle a_1|q\rangle= \left(\frac{\omega}{2\hbar}\right)^\frac{1}{4}\delta\left(a_1- \sqrt{\frac{\omega}{2 \hbar}}q\right).$

$\langle a_2|p\rangle= \left(\frac{1}{2\hbar\omega}\right)^\frac{1}{4}\delta\left(a_2- \sqrt{\frac{1}{2 \hbar\omega}}p\right).$

Computing $\langle p|q \rangle$

Then, as you know, $\langle p| q\rangle =\int da_1 da_2 \langle p| a_2\rangle \langle a_2| a_1\rangle \langle a_1| q\rangle .$

This whould be enough for you to find the right solution.

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  • $\begingroup$ If I solve the integral you put, I get $$\frac{1}{\sqrt{\pi}} e^{-\frac{j}{\pi} q p}$$ -if I do it correctly - which does not have the correct factor $\frac{1}{\sqrt{2 \pi \hbar}}$ (I have just edited the question for clarification). $\endgroup$ – Nicola Jun 16 '14 at 11:54
  • $\begingroup$ Did you used the property of the Dirac delta computed over x multiplied by a constant? $$\int f(x) \delta (\alpha x) = \frac{f(0)}{\alpha}?$$ $\endgroup$ – Antonio Ragagnin Jun 16 '14 at 21:15
  • $\begingroup$ Oh yes and with this same trick you will find that $\langle a_1|p\rangle$ and $\langle a_2|q\rangle$ have a normalization factor $\endgroup$ – Antonio Ragagnin Jun 16 '14 at 21:20
  • $\begingroup$ @Nicola, I edited my answer taking into account the dirac delta property. $\endgroup$ – Antonio Ragagnin Jun 17 '14 at 9:29

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