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?

  • $\begingroup$ 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

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.

| cite | improve this answer | |
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.