# Dirac Notation and completeness relation

• 0ddbio
In summary, Dirac notation, also known as bra-ket notation, is a mathematical notation used to represent quantum states in quantum mechanics. It was introduced by British physicist Paul Dirac and is widely used in calculations, expressing physical states, and describing quantum operations. The completeness relation, also known as the resolution of identity, is an important concept in Dirac notation and states that any quantum state can be written as a linear combination of a complete set of basis states. It is used in various calculations and applications in quantum mechanics, including determining probabilities and expectation values. However, it has some limitations, such as only applying to systems with a discrete orthonormal basis and not holding for non-Hermitian operators.
0ddbio
I am confused about two minor things right now.
The following illustrates both which I pulled from my QM book:

$$<x|p_{op}|0>=\int_{-\infty}^{\infty}dp<x|p_{op}|p><p|0>=\int_{-\infty}^{\infty}dp~p<x|p><p|0>$$
$$=\int_{-\infty}^{\infty}dp~p\frac{1}{\sqrt{2\pi\hbar}}e^{ipx/\hbar}<p|0>=\frac{\hbar}{i}\frac{d}{dx}\int_{-\infty}^{\infty}dp\frac{1}{\sqrt{2\pi\hbar}}e^{ipx/\hbar}<p|0>$$
$$=\frac{\hbar}{i}\frac{d}{dx}\int_{-\infty}^{\infty}dp<x|p><p|0>=\frac{\hbar}{i}\frac{d}{dx}<x|0>$$

So, the first thing I am confused about is in the middle expression on the top line.
I know they use the completeness relation $\int_{-\infty}^{\infty}dp|p><p|=\hat{1}$ so I guess they just stick that in the middle and I guess it's ok that they pull the integral to the outside because the other terms don't depend on p.
However, what if it was something like: $<p|p_{op}|0>$ when you stick in that completeness relation, you can't pull the integral to the outside... can you?

The second thing that is bothering me is going from the expression on the right (middle line) to the expression on the left (bottom line). They apparently just change $p\rightarrow\frac{\hbar}{i}\frac{d}{dx}$ which I recognize as the momentum operator... BUT, isn't the p the eigenvalue? It first appears on the expression at the right (top line) from $p_{op}|p>=p|p>$ so the p is the eigenvalue... how on Earth do they justify replacing it with the operator expression?

0ddbio said:
[...]
However, what if it was something like: $<p|p_{op}|0>$ when you stick in that completeness relation, you can't pull the integral to the outside... can you?[...]

Well, you can't, because $\hat{p}\langle p| = p \langle p |$ and the integral is wrt p.
0ddbio said:
[...] The second thing that is bothering me is going from the expression on the right (middle line) to the expression on the left (bottom line). They apparently just change $p\rightarrow\frac{\hbar}{i}\frac{d}{dx}$ which I recognize as the momentum operator... BUT, isn't the p the eigenvalue? It first appears on the expression at the right (top line) from $p_{op}|p>=p|p>$ so the p is the eigenvalue... how on Earth do they justify replacing it with the operator expression?

There's no operator expression, but a bracket $\langle x|p\rangle$ which is picked up from the complex exponential and the numerical factor in front of it.

Thanks for the response.

dextercioby said:
There's no operator expression, but a bracket $\langle x|p\rangle$ which is picked up from the complex exponential and the numerical factor in front of it.
I am so sorry, I mentioned the wrong portion of it. I meant to say that about going from the first expression on the middle line to the second. There was no $\langle x|p\rangle$

I'll just rewrite it here for your convenience. The part I was referring to was this:
$$\int_{-\infty}^{\infty}dp~p\frac{1}{\sqrt{2\pi\hbar}}e^{ipx/\hbar}\langle p|0\rangle=\frac{\hbar}{i}\frac{d}{dx}\int_{-\infty}^{\infty}dp\frac{1}{\sqrt{2\pi\hbar}}e^{ipx/\hbar}\langle p|0\rangle$$

and all they did was change the p into it's operator.. but isn't p there just the eigenvalue? So how are they doing that?

thanks.

All they did was use the algebraic fact that p exp(ipx/h) = h/i d/dx exp(ipx/h). No operators involved in that step.

Bill_K said:
All they did was use the algebraic fact that p exp(ipx/h) = h/i d/dx exp(ipx/h). No operators involved in that step.

ooh, yes of course!
thank you both very much.

0ddbio said:
However, what if it was something like: $<p|p_{op}|0>$ when you stick in that completeness relation, you can't pull the integral to the outside... can you?
Then the p in $\int_{-\infty}^{\infty}dp|p><p|=\hat{1}$ is a dummy variable, and exists only within the expression on the left hand side. In particular, it is not the p in $<p|p_{op}|0>$. Therefore, you can still pull the integral outside.

However, you really should use a different dummy variable so that you don't forget which p's are which.

## 1. What is Dirac notation and how is it used in physics?

Dirac notation, also known as bra-ket notation, is a mathematical notation used to represent quantum states in quantum mechanics. It was introduced by British physicist Paul Dirac as a way to simplify complex mathematical equations. In Dirac notation, a quantum state is represented by a "ket" vector (|ψ>), and its corresponding dual vector, or "bra" (⟨ψ|), represents the conjugate transpose of the ket vector. This notation is widely used in quantum mechanics to perform calculations, express physical states, and describe quantum operations.

## 2. What is the meaning of the completeness relation in Dirac notation?

The completeness relation, also known as the resolution of identity, is an important concept in Dirac notation. It states that any quantum state can be written as a linear combination of a complete set of basis states. In other words, the completeness relation shows that the set of basis states spans the entire space of quantum states. Mathematically, it is represented as ∑|i>⟨i| = 1, where the sum is taken over all possible basis states |i>.

## 3. How is the completeness relation used in quantum mechanics calculations?

The completeness relation is used in many calculations in quantum mechanics. For example, it allows us to express any state vector in terms of a complete set of basis states and perform calculations on them. It is also used in calculating the probability of finding a particle in a particular state, as well as in determining the expectation value of an observable in a given state. Furthermore, the completeness relation is used in deriving equations such as the Schrödinger equation and the Heisenberg uncertainty principle.

## 4. What are some applications of the completeness relation in physics?

The completeness relation has numerous applications in physics, particularly in quantum mechanics. It is used in calculating transition probabilities between quantum states, which is crucial in understanding atomic and molecular spectra. It is also used in studying quantum entanglement, a phenomenon where the state of one particle is affected by the state of another particle, even at a great distance. Additionally, the completeness relation is used in many other areas of physics, including quantum field theory, quantum information theory, and quantum computation.

## 5. Are there any limitations to the completeness relation in Dirac notation?

While the completeness relation is a powerful tool in quantum mechanics, it does have some limitations. One limitation is that it only applies to systems with a discrete orthonormal basis. In systems with a continuous basis, the completeness relation must be modified to include an integral instead of a sum. Additionally, the completeness relation does not hold for non-Hermitian operators, which can pose challenges in certain quantum systems. Despite these limitations, the completeness relation remains a fundamental concept in quantum mechanics and is widely used in various calculations and applications.

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