# Operator question

1. Oct 3, 2007

### BeauGeste

Can someone tell me exactly how
$$\langle x |\hat{p} | \psi\rangle$$
becomes
$$-i \hbar \frac{\partial \psi}{\partial x}$$?
Is
$$\hat{p} \langle x |\psi\rangle$$
an intermediary step?
Because isn't $$\langle x |\psi\rangle = \psi(x)$$?

2. Oct 3, 2007

### malawi_glenn

that is beacause:

$$\langle x \vert \psi \rangle = \psi (x)$$

Per definition. This is how the wavefunction is defined in dirac - notation.

So you are right, exept that the intermediate step is:

$$-i \hbar \frac{\partial}{\partial x} \langle x |\psi\rangle$$

Last edited: Oct 3, 2007
3. Oct 3, 2007

### Hurkyl

Staff Emeritus
The semantics of this expression haven't been spelled out explicitly, but I would imagine that

$$\hat{p} \langle x | \psi \rangle$$

should be nonsense. If we take $\psi$ to be a given and x to be a dummy variable, the it is literally true that $\langle x | \psi \rangle$ is a complex-valued function on the real line, and thus is technically a wavefunction. However, I believe this to be an accidental side-effect of how things are formuated, and you should not interpret it in this way.

By the way, what is the problem with simply computing $\hat{p} | \psi \rangle$, and then applying $\langle x |$ to the result?

By the way, it would be extremely misleading to write $\hat{p} = -i \hbar \partial / \partial x$. The variable x already has a job in this expression: $\langle x |$ is a ket-valued function of the reals, and x is denoting the dummy variable for that function. Therefore, you should not use the letter x for other purposes. You should use a different symbol to denote the dummy variable for your wavefunctions. e.g.

$$\langle x | = \delta(\xi - x)$$

$$\hat{p} = -i \hbar \frac{\partial}{\partial \xi}$$

$$| \psi \rangle = \psi(\xi)$$

4. Oct 3, 2007

### malawi_glenn

All QM books and teachers I've met uses x as symbol for wavefunction when it is in position space.

5. Oct 3, 2007

### Hurkyl

Staff Emeritus
Well, when x is already assigned a different purpose, you either have to:
(1) Use some other symbol ($\xi$ is a commonly used as a variant for x)
(2) Keep careful track of which x's in your expression represent the dummy variable of your wavefunctions, and which x's represent this other purpose.

Using the letter x for the dummy variable is merely a syntactic convention, not a strict requirement.

6. Oct 3, 2007

### nrqed

Strictly speaking, you can't pull out $\hat{p}$

So to be careful, you must first introduce an identity operator in the following way:

$$\langle x |\hat{p} | \psi\rangle = \int dx' ~ \langle x |\hat{p} | x' \rangle \langle x' | \psi\rangle$$

Now use

$$=\langle x |\hat{p} | x' \rangle = - i \hbar \delta(x-x') \partial_x$$

and integrate over x'. This will give you your final result.

7. Oct 3, 2007

### malawi_glenn

so where can I find a reference that defines:

$$\langle x | = \delta(\xi - x)$$

?

8. Oct 3, 2007

### Hurkyl

Staff Emeritus
The buzzword to look for is "position eigenstate". (Or one of the other eigen___ variants)

You can see it for yourself that it has the desired behavior, by computing the braket:

$$\langle x | \psi \rangle = \int_{-\infty}^{+\infty} \delta(\xi - x)^* \psi(\xi) \, d\xi = \psi(x).$$

9. Oct 3, 2007

### A/4

The classic text of Cohen-Tannoudji, Diu, and LaLoe defines projection to the $$|r\rangle$$ representation as the function $$\xi_{r_0}(r) = \delta(r-r_0)$$. The equivalence is established by the orthonormality relation $$\langle r | r_o \rangle = \delta(r-r_0)$$. So, what he is saying is technically and formally true (note that $$\xi$$ is a function in this case, and not an alternate symbol'' to represent $$|x\rangle$$, which I would call a vector).

That being said, however, there is nothing wrong with the standard statement that the two conjugate representations are $$|x\rangle$$ and $$|p\rangle$$. The "equivalence" is basically a statement that either is appropriate. To suggest otherwise simply confuses the answer to an otherwise straightforward question, in my opinion.

To answer the question of the original poster, the process is as follows. We want to find the action of the momentum operator $$\hat{P}$$ in the position representation $$|x\rangle$$. So, we have:

$$\langle x | \hat{P}|\psi\rangle = \int \langle x | p\rangle \langle p | \hat{P} | \psi\rangle~dp$$

by expanding over the complete set of $$|p\rangle$$ basis states. This gives

$$\langle x | \hat{P}|\psi\rangle = \int p \langle x | p\rangle \langle p | \psi\rangle~dp$$

Since $$\langle x | \psi\rangle$$ and $$\langle p | \psi \rangle$$ are related by a Fourier transform, one can show that $$\langle x| p\rangle = e^{ipx/\hbar}$$.

Noting that $$pe^{ipx/\hbar} = \frac{\hbar}{i} \frac{\partial}{\partial x} e^{ipx/\hbar}$$, we can say that

$$\langle x | \hat{P}|\psi\rangle = \frac{\hbar}{i} \frac{\partial}{\partial x} \int \langle x |p\rangle \langle p | \psi\rangle~dp = \frac{\hbar}{i} \frac{\partial}{\partial x} \langle x | \psi\rangle$$

Last edited: Oct 4, 2007
10. Oct 4, 2007

### BeauGeste

In post 6, nrqed inserts a complete set of x' and in post 9, A/4 inserts a complete set of p. Are these two different ways to get the same result? Or is one 'more' correct than the other?
Thanks for the replies so far - they have been helpful in my understanding.

11. Oct 4, 2007

### nrqed

the two methods are equivalent. what A/4 did was to provide a justification to the fact that the momentum operator is represented by the usual -i hbar d/dx. In my post I took this as an accepted fact. On the other hadn, A/4 had to use as an accepted fact that <x|p> = e^{i px} so because of that both our presentations are equivalent (for example, if I use my starting point I can prove that <x|p> = e^{i px} must hold). The one you prefer will depend on which starting point you prefer to take for granted.

hope this helps

patrick

12. Oct 4, 2007

### nrqed

Don't you mean $$\langle x | \xi \rangle= \delta(\xi - x)$$
and

$$\langle \xi'| \hat{p}| \xi \rangle = -i \hbar \delta(\xi-\xi') \frac{\partial}{\partial \xi}$$
and

$$\langle \xi| \psi \rangle = \psi(\xi)$$

?

13. Oct 4, 2007

### Hurkyl

Staff Emeritus
If you're going to think of wavefunctions as complexed-valued functions (or distributions) on the real line, then at some point you're going to have to buckle down and write them as functions! I suppose I should have explicitly that in all three cases, I was using $\xi$ to denote the dummy variable for the corresponding wavefunction. (or 'wave operator', for the case of $\hat{p}$)

Last edited: Oct 4, 2007
14. Oct 4, 2007

### A/4

In that sense, your notation is a bit careless. You're mixing functions with vectors (albeit infinite-dimensional ones).

15. Oct 7, 2007

### dextercioby

The last equation is correct. It basically asserts that the position wavefunction is nothing but a coeficient of the expansion of an arbitrary ket vector into momentum eigenkets.

$$|\psi\rangle =\int_{\sigma_{p}} dp \ \psi (p) |p\rangle$$

The "intermediary step" is not okay, it has nothing to do with the first equation that you wrote.

As for the first step, the proof, even though it looks mathematically formal, i can assure you it has a tremendous amount of mathematics. Since this $\hat{p}$ operator is actually a dual operator (acting in the space of kets and bras), one basically introduces the completeness relation for the momentum eigenkets $\int |p\rangle\langle p| \ dp=\hat{1}$, (where 1 is the dual of the unit operator), inside the matrix element you're trying to compute.

Last edited: Oct 7, 2007
16. Oct 7, 2007

### quetzalcoatl9

if you like special functions, here is an alternate method that makes use of

$$\int dx \,\psi(x) \frac{\partial}{\partial x} \delta (x - x^{\prime}) = -\frac{\partial \psi}{\partial x^{\prime}}$$

(an elementary identity of the Dirac delta functional) and

$$\delta (x - x^{\prime}) = <x^{\prime}|x> = \frac{1}{2\pi \hbar} \int dp \,e^{\frac{i}{\hbar} p (x - x^{\prime})}$$

the fourier representation is in momentum space rather than the usual frequency space and a factor of $$\hbar$$ has been inserted based upon dimensional analysis. the integrations are over the entire real line.

then:

$$\int dx \,<x|\psi> \frac{\partial}{\partial x} \frac{1}{2\pi \hbar} \int dp \,e^{\frac{i}{\hbar} p (x - x^{\prime})} = -\frac{\partial \psi}{\partial x^{\prime}}$$
$$\int dx \,<x|\psi> \frac{-ip}{\hbar} \delta (x - x^{\prime}) = -\frac{\partial \psi}{\partial x^{\prime}}$$
$$\int dx \,<x|\psi> p <x^{\prime}|\psi> = \frac{\hbar}{i} \frac{\partial \psi}{\partial x^{\prime}}$$
$$p\int dx \,<x^{\prime}|x><x|\psi> = \frac{\hbar}{i} \frac{\partial \psi}{\partial x^{\prime}}$$
$$p<x^{\prime}|\psi> = \frac{\hbar}{i} \frac{\partial \psi}{\partial x^{\prime}}$$

we want the operator whose eigenvalue is the momentum, ie.

$$\hat{p} \,\psi(x^{\prime}) = p \,\psi(x^{\prime}) = \frac{\hbar}{i} \frac{\partial }{\partial x^{\prime}} \,\psi(x^{\prime})$$