Representing a vector in terms of eigenkets in a continuos basis

[FedEx]

Representing a vector in terms of eigenkets in continuos basis(stuck here,guys help)

I was reading Dirac and there is this formula which bothers me,

$$|P>= \int{|\right{\xi'd}\rangle{d\xi'}} + \sum{|\right{\xi^{r}b}\rangle}$$

Where $$|\right\xi'\rangle$$ denotes the eigenket corresponding to the eigenvalue $$\xi'$$

Now the integral term is bothering me.

Does it mean, that the integral is summing over the entire range of eigenvalues? But than what is the use of the summation term?

And if it is so than there is a contradiction cause a few pages later Dirac says that "If we have an eigenket $$|\right\xi'\rangle$$ dependent on other eigenkets of xi' then these other eigenkets must belong to the same eigenvalue xi' " What does this mean?? He has given a proof, i am able to understand it. But the start point of the proof is bothering me. From what i get is that in the integral term the |xi'> denotes some specific eigenket integrated over the whole range of eigenvalues. But the why that specific eigenket and not some other eigenket let say |xi''>


And eigenkets and eigenvalues go hand in hand, don't they?? As in you can't define let say some eigenket xi' without metioning the eigenvalue xi'.

 

[tommy01]

Hi.

The whole thing is rather sloppy and mathematically not very well justified by Dirac.
If you have an operator A in the Hilbert space the eigenvalues and eigenkets of the operator satisfy
$$A |a\rangle = a|a\rangle$$
where a is an eigenevalue if you have a continuum of eigenvalues in an (real) interval and
$$A |a_i\rangle = a_i|a_i\rangle$$
if you can count them.
You label the eigenkets by the corrsponding eigenvalue.
There may be operators with a mixed spectrum, i. e. a continuous and a discrete part. A sum over all eigenkets of such an operator must incorporate the sum over the discrete part as well as the "sum" over the continuous part which is an integral. So the integral simply means a sum over a continuous variable.
To your second question:
I think Dirac meant degenerate eigenkets i. e. eigenkets with the same eigenvalue. Linear combinations of such eigenkets can be made and the resulting ket is again an eigenket with the same eigenvalue
$$A|a'\rangle=a|a'\rangle ~~~~ A|a''\rangle=a|a''\rangle$$
then
$$A|\tilde{a}\rangle=a|\tilde{a}\rangle ~~~~ |\tilde{a}\rangle = |a' \rangle + |a" \rangle$$

I hope this few lines clarify some of your difficulties.

 

[FedEx]

Thanks for the help... But i don't think that he is talking about degeneracy.. If you have Dirac could you just turn to page 42 (top)?

 

[eaglelake]

I was reading Dirac and there is this formula which bothers me,

$$|P>= \int{|\right{\xi'd}\rangle{d\xi'}} + \sum{|\right{\xi^{r}b}\rangle}$$

Where $$|\right\xi'\rangle$$ denotes the eigenket corresponding to the eigenvalue $$\xi'$$

Now the integral term is bothering me.

Does it mean, that the integral is summing over the entire range of eigenvalues? But than what is the use of the summation term?

And if it is so than there is a contradiction cause a few pages later Dirac says that "If we have an eigenket $$|\right\xi'\rangle$$ dependent on other eigenkets of xi' then these other eigenkets must belong to the same eigenvalue xi' " What does this mean?? He has given a proof, i am able to understand it. But the start point of the proof is bothering me. From what i get is that in the integral term the |xi'> denotes some specific eigenket integrated over the whole range of eigenvalues. But the why that specific eigenket and not some other eigenket let say |xi''>


And eigenkets and eigenvalues go hand in hand, don't they?? As in you can't define let say some eigenket xi' without metioning the eigenvalue xi'.

We want to measure the observable $$\hat E$$ for a system with known state vector $$\left| P \right\rangle$$. One of the quantum postulates tells us to write the state vector in the representation where the eigenvectors of the observable being measured are the basis vectors. If observable $$\hat E$$ has both discrete and continuous eigenvalues, then the state vector becomes $$\left| P \right\rangle = \sum {\left| {\xi ^r } \right\rangle } \left\langle {{\xi ^r }}\mathrel{\left | {\vphantom {{\xi ^r } P}}\right. \kern-\nulldelimiterspace}{P} \right\rangle + \int {d\xi ^' } \left| {\xi ^' } \right\rangle \left\langle {{\xi ^' }}\mathrel{\left | {\vphantom {{\xi ^' } P}} \right. \kern-\nulldelimiterspace}{P} \right\rangle$$. We must include both parts. This is equation 31 in chapter 3 in my copy of Dirac. It is understood that $$E\left| {\xi ^r } \right\rangle = \xi ^r \left| {\xi ^r } \right\rangle$$ in the discrete space where $$\left| {\xi ^r } \right\rangle$$ is the eigenstate corresponding to the eigenvalue $$\xi ^r$$. Likewise, $$\left| {\xi ^' } \right\rangle$$ is the eigenvector corresponding to the eigenvalue $$\xi ^'$$, where the $$\xi ^'$$ constitute a continuous spectrum. In my copy of Dirac, $$\xi ^{''}$$ is just another eigenvalue in the continuous spectrum.
In the integral expansion, we sum over only the continuous range of eigenvalues. $$\xi ^'$$ is now the continuous variable of integration. You are correct: Eigenkets and eigenvalues do go hand in hand. In the integral term, in practice, we integrate any state function over the whole continuous range of eigenvalues. Dirac then writes : $$\left\langle {Q} \mathrel{\left | {\vphantom {Q P}} \right. \kern-\nulldelimiterspace} {P} \right\rangle = \sum {\left\langle {Q} \mathrel{\left | {\vphantom {Q {\xi ^r }}} \right. \kern-\nulldelimiterspace} {{\xi ^r }} \right\rangle } \left\langle {{\xi ^r }} \mathrel{\left | {\vphantom {{\xi ^r } P}} \right. \kern-\nulldelimiterspace} {P} \right\rangle + \int {d\xi ^' } \left\langle {Q} \mathrel{\left | {\vphantom {Q {\xi ^' }}} \right. \kern-\nulldelimiterspace} {{\xi ^' }} \right\rangle \left\langle {{\xi ^' }} \mathrel{\left | {\vphantom {{\xi ^' } P}} \right. \kern-\nulldelimiterspace} {P} \right\rangle$$ where we recognize the functions $$Q^* (\xi ^' ) = \left\langle {Q}\mathrel{\left | {\vphantom {Q {\xi ^' }}}\right. \kern-\nulldelimiterspace} {{\xi ^' }} \right\rangle$$ and $$P(\xi ^' ) = \left\langle {{\xi ^' }} \mathrel{\left | {\vphantom {{\xi ^' } P}}\right. \kern-\nulldelimiterspace} {P} \right\rangle$$. I hope this helps. Best wishes.

 

[FedEx]

Thanks a lot… I was thinking similarly..

But there is one more thing,

Dirac says that <x|xi|x> is the average value of the observable xi for the state x.

Why is it average?? Shouldn't it be the precise value… Cause if x is the eigenstate then we are sure to get the corresponding eigenvalue as our measurement.. So why does he talk about the average??

Furtther more he talks about more than observable and than about summing them and all.. Well position and momentum(though they don't commute) are two observables, well what does the average of this two mean??

 

[Chopin]

Dirac says that <x|xi|x> is the average value of the observable xi for the state x.

Why is it average?? Shouldn't it be the precise value… Cause if x is the eigenstate then we are sure to get the corresponding eigenvalue as our measurement.. So why does he talk about the average??

Because x might not be an eigenstate of xi. If it is, then <x|xi|x> will be the precise value, but imagine that it's a linear combination of many eigenstates. To change your notation slightly, imagine we have an operator $$A$$, which has eigenstates $$A|m\rangle = m|m\rangle$$, and $$A|n\rangle = n|n\rangle$$. Then, if you have a state $$|x\rangle = |m\rangle$$, then $$\langle x|A|x\rangle$$ will be $$\langle m|A|m\rangle = (\langle m|) m (|m\rangle) = m\langle m|m\rangle$$. Since $$\langle m|m\rangle = 1$$, this means $$\langle x|A|x\rangle = m$$ exactly.

But if $$|x\rangle = c|m\rangle + d|n\rangle$$, where c and d are arbitrary complex numbers, then $$\langle x|A|x\rangle = (c^*\langle m| + d^*\langle n|) A (c|m\rangle + d|n\rangle) = (c^*\langle m| + d^*\langle n|) (cm|m\rangle + dn|n\rangle) = c^*cm\langle m|m\rangle + c^*dn\langle m|n\rangle + d^*cm\langle n|m\rangle + d^*dn\langle n|n\rangle$$. Since $$\langle m|m\rangle = 1$$ and $$\langle n|n\rangle = 1$$, and $$\langle m|n\rangle = 0$$ and $$\langle n|m\rangle = 0$$, this means $$\langle x|A|x\rangle = c^*cm + d^*dn$$, which is the average of $$m$$ and $$n$$, weighted by the magnitudes of the respective scaling constants. This can be generalized to a combination of any number of eigenstates, or even an integral over a continuous range of eigenstates. In all cases, it's the weighted average of all the respective eigenvalues.

 

[FedEx]

Because x might not be an eigenstate of xi. If it is, then <x|xi|x> will be the precise value, but imagine that it's a linear combination of many eigenstates. To change your notation slightly, imagine we have an operator $$A$$, which has eigenstates $$A|m\rangle = m|m\rangle$$, and $$A|n\rangle = n|n\rangle$$. Then, if you have a state $$|x\rangle = |m\rangle$$, then $$\langle x|A|x\rangle$$ will be $$\langle m|A|m\rangle = (\langle m|) m (|m\rangle) = m\langle m|m\rangle$$. Since $$\langle m|m\rangle = 1$$, this means $$\langle x|A|x\rangle = m$$ exactly.

But if $$|x\rangle = c|m\rangle + d|n\rangle$$, where c and d are arbitrary complex numbers, then $$\langle x|A|x\rangle = (c^*\langle m| + d^*\langle n|) A (c|m\rangle + d|n\rangle) = (c^*\langle m| + d^*\langle n|) (cm|m\rangle + dn|n\rangle) = c^*cm\langle m|m\rangle + c^*dn\langle m|n\rangle + d^*cm\langle n|m\rangle + d^*dn\langle n|n\rangle$$. Since $$\langle m|m\rangle = 1$$ and $$\langle n|n\rangle = 1$$, and $$\langle m|n\rangle = 0$$ and $$\langle n|m\rangle = 0$$, this means $$\langle x|A|x\rangle = c^*cm + d^*dn$$, which is the average of $$m$$ and $$n$$, weighted by the magnitudes of the respective scaling constants. This can be generalized to a combination of any number of eigenstates, or even an integral over a continuous range of eigenstates. In all cases, it's the weighted average of all the respective eigenvalues.

Cheers :D.

 

[FedEx]

We want to measure the observable $$\hat E$$ for a system with known state vector $$\left| P \right\rangle$$. One of the quantum postulates tells us to write the state vector in the representation where the eigenvectors of the observable being measured are the basis vectors. If observable $$\hat E$$ has both discrete and continuous eigenvalues, then the state vector becomes $$\left| P \right\rangle = \sum {\left| {\xi ^r } \right\rangle } \left\langle {{\xi ^r }}\mathrel{\left | {\vphantom {{\xi ^r } P}}\right. \kern-\nulldelimiterspace}{P} \right\rangle + \int {d\xi ^' } \left| {\xi ^' } \right\rangle \left\langle {{\xi ^' }}\mathrel{\left | {\vphantom {{\xi ^' } P}} \right. \kern-\nulldelimiterspace}{P} \right\rangle$$. We must include both parts. This is equation 31 in chapter 3 in my copy of Dirac. It is understood that $$E\left| {\xi ^r } \right\rangle = \xi ^r \left| {\xi ^r } \right\rangle$$ in the discrete space where $$\left| {\xi ^r } \right\rangle$$ is the eigenstate corresponding to the eigenvalue $$\xi ^r$$. Likewise, $$\left| {\xi ^' } \right\rangle$$ is the eigenvector corresponding to the eigenvalue $$\xi ^'$$, where the $$\xi ^'$$ constitute a continuous spectrum. In my copy of Dirac, $$\xi ^{''}$$ is just another eigenvalue in the continuous spectrum.
In the integral expansion, we sum over only the continuous range of eigenvalues. $$\xi ^'$$ is now the continuous variable of integration. You are correct: Eigenkets and eigenvalues do go hand in hand. In the integral term, in practice, we integrate any state function over the whole continuous range of eigenvalues. Dirac then writes : $$\left\langle {Q} \mathrel{\left | {\vphantom {Q P}} \right. \kern-\nulldelimiterspace} {P} \right\rangle = \sum {\left\langle {Q} \mathrel{\left | {\vphantom {Q {\xi ^r }}} \right. \kern-\nulldelimiterspace} {{\xi ^r }} \right\rangle } \left\langle {{\xi ^r }} \mathrel{\left | {\vphantom {{\xi ^r } P}} \right. \kern-\nulldelimiterspace} {P} \right\rangle + \int {d\xi ^' } \left\langle {Q} \mathrel{\left | {\vphantom {Q {\xi ^' }}} \right. \kern-\nulldelimiterspace} {{\xi ^' }} \right\rangle \left\langle {{\xi ^' }} \mathrel{\left | {\vphantom {{\xi ^' } P}} \right. \kern-\nulldelimiterspace} {P} \right\rangle$$ where we recognize the functions $$Q^* (\xi ^' ) = \left\langle {Q}\mathrel{\left | {\vphantom {Q {\xi ^' }}}\right. \kern-\nulldelimiterspace} {{\xi ^' }} \right\rangle$$ and $$P(\xi ^' ) = \left\langle {{\xi ^' }} \mathrel{\left | {\vphantom {{\xi ^' } P}}\right. \kern-\nulldelimiterspace} {P} \right\rangle$$. I hope this helps. Best wishes.

This makes complete sense.. But have a look chapter 2 eqn 25. He writes,

$$|{\right{P}\rangle} = \int{|\right{\xi'd}\rangle{d\xi'}} + \sum{|\right{\xi^{r}b}\rangle}$$

He doesnot use coefficients... As in he doesnot weigh the eigen kets.. Thats worth noting..

 

[eaglelake]

This makes complete sense.. But have a look chapter 2 eqn 25. He writes,

$$|{\right{P}\rangle} = \int{|\right{\xi'd}\rangle{d\xi'}} + \sum{|\right{\xi^{r}b}\rangle}$$

He doesnot use coefficients... As in he doesnot weigh the eigen kets.. Thats worth noting..

Equation 25 chapter 2 and equation 31 in chapter 3 are equivalent: We must have $$\left| {\xi ^' d} \right\rangle = \left| {\xi ^' } \right\rangle \left\langle {{\xi ^' }} \mathrel{\left | {\vphantom {{\xi ^' } P}} \right. \kern-\nulldelimiterspace}{P} \right\rangle$$ and $$\left| {\xi ^r b} \right\rangle = \left| {\xi ^r } \right\rangle \left\langle {{\xi ^r }} \mathrel{\left | {\vphantom {{\xi ^r } P}} \right. \kern-\nulldelimiterspace} {P} \right\rangle$$
Dirac does not define the unit operator until Chapter 3 Equation 22. I suspect he wanted to do this before writing equation 25. Of course, I am just guessing here. I always thought the unit operator, $$1 = \sum {\left| {\xi ^r } \right\rangle } \left\langle {\xi ^r } \right|$$ for discrete eigenvalues, and $$1 = \int {d\xi ^' } \left| {\xi ^' } \right\rangle \left\langle {\xi ^' } \right|$$ for a continuous spectrum was one of the most powerful calculating tools given to budding physicists. The expansion coefficients become inner products $$\left\langle {\xi } \mathrel{\left | {\vphantom {\xi P}} \right. \kern-\nulldelimiterspace} {P} \right\rangle$$
in both cases. One of the postulates of quantum mechanics tells us to expand the state vector (discrete case), or the state function (continuous case), in terms of the eigenvectors (eigenfunctions) of the observable being measured. Dirac's definition of the unit vector gives us a straightforward way to do this: We simply operate on the state vector (function) with the unit operator!
Best wishes.