# State transformation - Position to momentum space (Dirac)

• AwesomeTrains
In summary: Consider ##\langle p | x \rangle = e^{-ipx/\hbar} ##, what will ##exp(-i(\textbf b x' + \textbf p \textbf x' / \hbar))## look like when changed into braket notation?\langle p' | x' \rangle = e^{-ip'x'/\hbar}In summary, The conversation discusses a problem involving quantum mechanics. The student is unsure about the correctness of their solution and asks for help. They provide the homework statement and equations used in their attempt
AwesomeTrains
Hey everyone!
It's my first semester with quantum mechanics and I'm uncertain if my solution of this problem is correct, would be nice if someone could check and let me know

1. Homework Statement

I have to calculate the representation of the state:
$|\alpha \rangle \equiv exp[-i \textbf b \cdot \textbf x]|\phi\rangle$, in momentum space.

Where $\textbf b$ is a constant vector.

## Homework Equations

- Dirac notation
- Position operator in momentum space: $\textbf x=i\hbar\nabla$(1)
- $\langle \textbf p | \textbf p' \rangle = \delta(\textbf p-\textbf p')$ (2)

3. The Attempt at a Solution

First I multiply by $\langle \textbf p |$
$\langle \textbf p |\alpha \rangle = \langle \textbf p | exp[-i \textbf b \cdot \textbf x]|\phi\rangle$

Then I add a one to the equation:
$=\int \langle \textbf p |exp[-i \textbf b \cdot \textbf x]|\textbf p'\rangle \langle \textbf p'|\phi\rangle d^3p'$

With (1):
$=\int \langle \textbf p |exp[-i \textbf b \cdot \textbf i\hbar\nabla]|\textbf p'\rangle \langle \textbf p'|\phi\rangle d^3p'$

I'm unsure about this step. Can I move the bra $\langle \textbf p |$ past the operator $exp[-i \textbf b \cdot \textbf i\hbar\nabla]$ like this:
$\int exp[-i \textbf b \cdot \textbf i\hbar\nabla] \delta(\textbf p-\textbf p') \langle \textbf p'|\phi\rangle d^3p'$ ?

If I can then I would get:
$= exp[-i \textbf b \cdot \textbf i\hbar\nabla] \langle \textbf p|\phi\rangle$

$\Rightarrow \alpha(\textbf p) = exp[-i \textbf b \cdot \textbf i\hbar\nabla]\phi(\textbf p)$

I think I'm doing something wrong somewhere since I didn't do a fouriertransform anywhere.
Is this even the right approach?

Any hints/tips or corrections are very appreciated.

Kind regards
Alex

It might be clearer to you if you do this problem by first calculating the given state in position representation ##\langle \mathbf{x}|\alpha \rangle##. Then make use of the fact that the momentum representation of ##|\alpha \rangle## is just the Fourier transform of ##\alpha (\mathbf{x}) = \langle \mathbf{x}|\alpha \rangle##,
$$\langle \mathbf{p} | \alpha \rangle = \int \langle \mathbf{p} |\mathbf{x'}\rangle \langle \mathbf{x'} |\alpha \rangle d^3\mathbf{x'} = \int e^{-i\mathbf{p}\cdot \mathbf{x'}/\hbar} \alpha (\mathbf{x'}) d^3\mathbf{x'}$$

Upon doing this you may see the property of the operator ##e^{-i\mathbf{b}\cdot \mathbf{x}}##, especially when it acts on ##|\mathbf{p} \rangle##.

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When calculating the position representation of the state do I have to use the series representation of the exponential function? What does the solution look like? Is it an exponential function or some wave function?

The position representation means the state ket be projected onto the eigenstate of position operator ##\hat{\mathbf{x}}##, which means
$$\langle \mathbf{x} | \psi \rangle = \langle \mathbf{x} | e^{-i\mathbf{b}\cdot \mathbf{x}} | \phi \rangle$$
Now since the exponential operator in the middle is only a function of the operator ##\hat{\mathbf{x}}##, you can simply make use of the identity ##f(\hat{\mathbf{x}}) | \mathbf{x'} \rangle = f(\mathbf{x'}) | \mathbf{x'} \rangle ## where the ##f(\mathbf{x'}) ## on the right is just a number, that is one replaces any appearing ##\hat{\mathbf{x}}## in ##f(\hat{\mathbf{x}})## with ##\mathbf{x'}##.

EDIT: there was a mistake in the first equation of this comment, now corrected.

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Thanks for the help.
I think I got it now, let me know what you think
$\langle x' | exp[-i\textbf b \hat x] |\phi \rangle = exp[i\textbf b x] \langle x' |\phi \rangle$ and then I would do the Fourier transform by adding a one:
(Our prof always adds a one )
$exp[i\textbf b x] \langle x' |\phi \rangle = \int exp[i\textbf b x] \langle x' | p' \rangle \langle p' |\phi \rangle d^3p' = \int exp[i\textbf b x] \frac{exp[ix'p']}{\sqrt{2\pi}^3} \langle p' |\phi \rangle d^3p'$ If it is correct how do I continue from here? :)
Is this identity you've given valid because $x$ is an eigenvalue to $\hat x$?

AwesomeTrains said:
$\langle x' | exp[-i\textbf b \hat x] |\phi \rangle = exp[i\textbf b x] \langle x' |\phi \rangle$
AwesomeTrains said:
Is this identity you've given valid because $x$ is an eigenvalue to $\hat x$?
yes

AwesomeTrains said:
$exp[i\textbf b x] \langle x' |\phi \rangle = \int exp[i\textbf b x] \langle x' | p' \rangle \langle p' |\phi \rangle d^3p' = \int exp[i\textbf b x] \frac{exp[ix'p']}{\sqrt{2\pi}^3} \langle p' |\phi \rangle d^3p'$ If it is correct how do I continue from here? :)
What you want to calculate is the state in momentum representation and now what you have is the same state represented in position space, right? This is the point where the equation in comment #2 comes into play.

I got the minus because I thought I had to do the complex conjugate of the operator when using it to the left.
Using the Fourier transform I get this:
$\int exp(-i(\textbf b x + \textbf p \textbf x' / \hbar)) \langle x' | \phi \rangle d^3 x' = \langle \textbf p | \phi \rangle$
Is this $| \alpha \rangle$ in momentum space?
By the way, thanks for guiding me

AwesomeTrains said:
I got the minus because I thought I had to do the complex conjugate of the operator when using it to the left.
Yes you did, but the complex conjugation applies on both sides of the equation.
AwesomeTrains said:
$\int exp(-i(\textbf b x + \textbf p \textbf x' / \hbar)) \langle x' | \phi \rangle d^3 x' = \langle \textbf p | \phi \rangle$
Is this $| \alpha \rangle$ in momentum space?
Except for the ##x## which should be primed ##x'## and the ## | \phi \rangle ## on the RHS which should be ## | \alpha \rangle ##, it's correct. But that expression can actually be cast further in a more compact form by noting that ##exp(-i(\textbf b x' + \textbf p \textbf x' / \hbar))## is actually a wavefunction of momentum operator. Consider ##\langle p | x \rangle = e^{-ipx/\hbar} ##, what will ##exp(-i(\textbf b x' + \textbf p \textbf x' / \hbar))## look like when changed into braket notation?
EDIT: Sorry I forget to write the ##\hbar## in the wavefunction of momentum operator above, I'm getting used to atomic unit.

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Something like this? $\langle \textbf {b} + \textbf p/\hbar | x' \rangle$
Is $\textbf p /\hbar$ the wave vector?

Actually ## \langle \textbf {b}/\hbar + \textbf p | x' \rangle##. After you make this replacement in the integrand and making use of the completeness relation in the ##|x'\rangle ## space which I assume you have been familiar with from you prof's lectures, you can get the braket notation of the momentum representation of ##|\alpha \rangle##. Just for advice, it may be interesting to slightly play with the final form of ##\langle p | \alpha \rangle## in order to investigate how the operator ##e^{-i\mathbf{b} \cdot \mathbf{\hat{x}} }## acts on ##| \mathbf{p} \rangle##.

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$\int \langle \textbf b /\hbar + \textbf p| x' \rangle \langle x' | \phi \rangle d^3 x' = \langle \textbf p | \phi \rangle \Rightarrow \langle \textbf b /\hbar + \textbf p| \phi \rangle = \langle \textbf p | \phi \rangle$ Is it some sort of momentum translation?

AwesomeTrains said:
Is it some sort of momentum translation?
Yes exactly. However be careful you are being sloppy in writing the proper maths here and there, the last relation should be ## \langle \mathbf{b} /\hbar + \mathbf{p}| \phi \rangle = \langle \mathbf{p} | \alpha\rangle = \langle \mathbf{p} | e^{-i\mathbf{b} \cdot \mathbf{\hat{x}}} |\phi\rangle ## otherwise it makes no sense to say that the operator in question is a momentum translation operator.
Needless to say, you can also convince yourself that the operator of the form ##e^{i\mathbf{\hat{p}} \cdot \mathbf{a} }## also does the same behavior but in position space, due to which it's called the position translation operator.

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Sorry, I don't quite understand what you mean there. Could you maybe elaborate?

Ok, first I actually have to correct your first equation in comment #11 as well:
## \int \langle \textbf b /\hbar + \textbf p| x' \rangle \langle x' | \phi \rangle d^3 x' = \langle \mathbf{b}/\hbar+ \mathbf{p} | \phi \rangle ##
Then coming to the second equation next to the right of the arrow sign in comment #11, no matter how you look at it, equation like ##\langle \textbf b /\hbar + \textbf p| \phi \rangle = \langle \textbf p | \phi \rangle## cannot be justified. The LHS is just plainly different from the RHS, isn't it? Remember you got ##\langle \textbf b /\hbar + \textbf p| \phi \rangle ## from the first equation which was obtained after modifying ##\langle \mathbf{p} | \alpha \rangle##, this means they are actually equal, that is ##\langle \mathbf{p} | \alpha \rangle = \langle \textbf b /\hbar + \textbf p| \phi \rangle##. If you are still confused, just carefully trace back your derivation to see which comes from which.

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AwesomeTrains
Hey, I think I've understood it now. Thanks for staying with me through this problem
Did the same transformation for another similar operator into position space:
$|\beta \rangle \equiv e^{-i \textbf a \hat p} |\phi \rangle$

$\langle p' | \beta \rangle = \langle p' |e^{-i \textbf a \hat p}|\phi \rangle = e^{-i\textbf a p}\langle p' |\phi \rangle$
Then I do the Fourier transform: (Other sign in front of the i, right?)
$\langle x | \beta \rangle = \int e^{i(p'x/\hbar-\textbf a p)}\phi(p')d^3p' = \langle x-\textbf a / \hbar | \phi \rangle$
Then $\langle x | \beta \rangle = \langle x-\textbf a / \hbar | \phi \rangle$, which is a position translation in position space.

AwesomeTrains said:
Hey, I think I've understood it now. Thanks for staying with me through this problem
Did the same transformation for another similar operator into position space:
$|\beta \rangle \equiv e^{-i \textbf a \hat p} |\phi \rangle$

$\langle p' | \beta \rangle = \langle p' |e^{-i \textbf a \hat p}|\phi \rangle = e^{-i\textbf a p}\langle p' |\phi \rangle$
Then I do the Fourier transform: (Other sign in front of the i, right?)
$\langle x | \beta \rangle = \int e^{i(p'x/\hbar-\textbf a p)}\phi(p')d^3p' = \langle x-\textbf a / \hbar | \phi \rangle$
Then $\langle x | \beta \rangle = \langle x-\textbf a / \hbar | \phi \rangle$, which is a position translation in position space.
That's right.

AwesomeTrains

## 1. What is state transformation from position to momentum space?

State transformation from position to momentum space is a mathematical process used in quantum mechanics to convert a system's wavefunction from the position representation to the momentum representation. This transformation allows for the analysis and prediction of a system's behavior in terms of its momentum instead of its position.

## 2. Why is state transformation important in quantum mechanics?

State transformation is important in quantum mechanics because it allows for the understanding and prediction of a system's behavior in both position and momentum space. This is essential in quantum mechanics, as particles can simultaneously exist in multiple states and have uncertain positions and momenta.

## 3. How is state transformation performed?

State transformation is performed using the Dirac notation, also known as bra-ket notation. This notation involves the use of a bra vector $\langle x|$, representing the position state, and a ket vector $|p\rangle$, representing the momentum state. The transformation is then carried out using the Fourier transform.

## 4. What is the significance of the Dirac notation in state transformation?

The Dirac notation is significant in state transformation because it provides a concise and elegant way of representing the mathematical operations involved. It also allows for the seamless transition between the position and momentum representations, making it a valuable tool in quantum mechanics.

## 5. Are there any physical implications of state transformation from position to momentum space?

Yes, there are physical implications of state transformation from position to momentum space. For example, the uncertainty principle states that there is a fundamental limit to the precision with which the position and momentum of a particle can be known simultaneously. This uncertainty is a direct consequence of the transformation between position and momentum representations.

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