# Quantum mechanical notations

## Main Question or Discussion Point

These questions are about the motivations behind notations in quantum mechanics.

First on my list is Dirac notation.

Why do we need to use Dirac notation?

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Fredrik
Staff Emeritus
Gold Member
We don't. It's just convenient.

In QM we spend a lot of time dealing with Hilbert space vectors and their duals, applying linear operators to them, and performing inner products on them. Dirac's bra-ket notation provides a very elegant way of describing these operations.

Fredrik
Staff Emeritus
Gold Member
Yes, but most of the time the calculations will be very similar.

$$(f,g)=(f,\sum_{k=1}^\infty (e_k,g)e_k)=\sum_{k=1}^\infty (f,e_k)(e_k,g)$$

$$\langle\alpha|\beta\rangle=\langle\alpha|\left(\sum_{k=1}^\infty|k\rangle\langle k|\right)|\beta\rangle=\sum_{k=1}^\infty\langle\alpha|k\rangle\langle k|\beta\rangle$$

Sure, but bra-ket has some advantages in other areas. First, it provides an easy way to tell whether we're dealing with a vector or its dual--i.e. $$\langle x|$$ is different than $$|x\rangle$$. Second, it describes the application of operators in a slightly more symmetric way, i.e. $$\langle x|O|y\rangle$$ instead of $$(x, Oy)$$ or $$(Ox, y)$$. That's largely an aesthetic thing, but it becomes more of an issue when you're writing out a big expression like $$\langle 0|a_{k_1}a_{k_2}a^\dagger_{k'_1}a^\dagger_{k'_2}|0\rangle$$.

A. Neumaier
2019 Award
Sure, but bra-ket has some advantages in other areas. First, it provides an easy way to tell whether we're dealing with a vector or its dual--i.e. $$\langle x|$$ is different than $$|x\rangle$$.
The usual linear algebra notation that uses \psi for the ket (ak a column vector) and \psi^* for the bra (aka conjugate transposed row vector) has the same advantages -and the additional one that one needs to be familiar with this notation anyway because of standard matrix algebra.

A real advantage of bras and kets appears only when one has a distinguished basis whose elements are labeled by several different labels. Then matrix elements between these basis states are naturally expressible in terms of bras and kets, while the component notation from linear algebra becomes awkward.

Thanks for the replies. Those were very helpful.

This is another of my questions. Why do we invert the r and the psi in psi = u(r) when we write the function in bra-ket notation?

Fredrik
Staff Emeritus
Gold Member
This is another of my questions. Why do we invert the r and the psi in psi = u(r) when we write the function in bra-ket notation?
I don't understand the question, or the expression psi=u(r). A function equal to a number? What do you mean by "invert the r and the psi"?

A. Neumaier
2019 Award
Thanks for the replies. Those were very helpful.

This is another of my questions. Why do we invert the r and the psi in psi = u(r) when we write the function in bra-ket notation?
You are confusing the notation. To gain understanding, ponder the identity

$$\psi= \int dr \psi(r)|r\rangle,$$

which relates the Schroedinger and the Dirac notation!

Would you please mind explaining the identity $$\psi= \int dr \psi(r)|r\rangle,$$
its origin/derivation, and what the Schrodinger and Dirac notations are?

A. Neumaier
2019 Award
Would you please mind explaining the identity $$\psi= \int dr \psi(r)|r\rangle,$$
its origin/derivation, and what the Schrodinger and Dirac notations are?
Schroedinger uses wave function notation to denote state vectors,
Dirac uses a basis notation.

Given the wave function notation, you can define a ket |x_0> to be the wave function whose value at a point x is the delta function delta(x-x_0). With this identification you can verify that the above relation holds.

Given the Dirac notation, you can turn an arbitrary state |psi> into a wave function by defining
$$\psi(x):=\langle x|\psi\rangle.$$.
Then one can easily verify from the completeness relation in Dirac form that
$$|\psi\rangle= \int dr \psi(r)|r\rangle,$$
which is again the above formula if one identifies psi and |psi>.

Thus the two notations are completely equivalent.

Schroedinger uses wave function notation to denote state vectors,
How did the term state vector obtain its name? Is the wave function notation the usual notation involving functions and algebra, e.g. $$u\left(x\right)$$

Dirac uses a basis notation.
What is the basis notation?

Given the wave function notation, you can define a ket |x_0> to be the wave function whose value at a point x is the delta function delta(x-x_0).
Would you please expand on this point?

How did the term state vector obtain its name? Is the wave function notation the usual notation involving functions and algebra, e.g. $$u\left(x\right)$$
I think it's just the fact that it is a state space (a Hilbert space) vector that specifies the quantum state of a system. Do you realize that a Hilbert space is a vector space? I've always found it useful to draw analogies with regular real vector spaces in order to understand QM.

And sure, a wave function is just a function. Actually it a pretty well-behaved function that takes a point in a real space and makes it correspond to a complex number.

Also, I might be wrong, but I think that the bra-ket notation is more powerful than wave function notation because there's no need to specify a variable dependence when you use bra-ket notation.

Would you please expand on this point?
I think he means that $$\langle x|\psi\rangle = \langle \delta_x |\psi\rangle$$

Last edited:
A. Neumaier
2019 Award
How did the term state vector obtain its name? Is the wave function notation the usual notation involving functions and algebra, e.g. $$u\left(x\right)$$
Piure states are represented in general by vectors in a Hilbert space, called state vectors, because they represent the state.

If the Hilbert space is a space of functions of position x, each state vector psi is a function of position, and psi(x) is the value of this function at x, as everywhere in math.

What is the basis notation?
Dirac's notation for a basis vector.