# QM formalism is one big confusion - lack of geometrical explaination with images

I have been trying to learn QM and it went well (all untill harmonic oscilator) until i had to face the formalism:

1. Hilbert space- As a novice to QM i am very sad that in none of the books i have read i found the reasons for using Hilbert space $\mathcal H$ at first place followed by a full geometrical explaination of this space and how we build this space out of $\mathbb R$. It goes same for for its dual space $\mathcal H_d$. Where can i get this? Every single author starts this topic by just bombard novices with bunch of rules for $\mathcal H, \mathcal H_d$ which i can't just trust and this is only pointless learning by heart.

2. Dirac's notation ... because i don't understand Hilbert space i don't understand what i am allowed to do with kets $|~ \rangle$ and bras $\langle~ |$ and for example:

• Why do we have to write an operator on the left side of kets $| \hat{A} \psi\rangle = \hat{A}|\psi\rangle$ but for bras it is vice versa (we write it on the right) $\langle \hat{A} \psi| = \langle \psi|\hat{A}$.
• Why can we factor out a constant from kets $| a\psi \rangle = a |\psi \rangle$ and we can only factor out complex conjugate from bras $\langle a \psi | = a^* \langle \psi |$.
3. Hermitean stuff (which i don't even know what it means as it has too many names which totally confuse me - authors should really start writing in a unified language (1 word 1 meaning) otherwise it is a big mess here. For eample here. Just check how many names there is for an conjugate transpose: $A^*,A^{*T},A^{T*},A^\dagger,A^+, A^H$ $\rightarrow$ this leads to a confusion). It seems to me that $A^\dagger$ is the most spread in QM by far, but $A^{*T}$ makes much more reason.

4. And finally i need all above to just understand this one line: $a a^\dagger = n$. I encountered this in my other topic where i tried to explain a harmonic oscilator to myself. How am i suppose to believe this? Well @Eugene B provided a proff but i don't understand it... The proff was:

By definition,

\begin{equation} \hat{a}^\dagger \left| n \right\rangle = \sqrt{n + 1} \left| n + 1 \right\rangle , \end{equation} \begin{equation} \hat{a} \left| n \right\rangle = \sqrt{n} \left| n - 1 \right\rangle , \end{equation} where $\left| n \right\rangle$ is the eigenstate of creation and annihilation operators, as well as of the Hamiltonian (due to the fact that they commute - homework to prove).

Now \begin{equation} \hat{a}^\dagger\hat{a} \left| n \right\rangle = \hat{a}^\dagger \sqrt{n} \left| n - 1 \right\rangle = \sqrt{n} \sqrt{n} \left| n \right\rangle = n \left| n \right\rangle , \end{equation} so conclude that the eigenvalue of a number operator, $\hat{N}$, is just $n$...

I don't understand the meaning of definitions for $a$ and $a^\dagger$ and even if i did, how can the line above even be true?

\begin{equation} \hat{a}^\dagger\hat{a} \left| n \right\rangle = \hat{a}^\dagger \sqrt{n} \left| n - 1 \right\rangle = \sqrt{n} \sqrt{n} \left| n \right\rangle = n \left| n \right\rangle , \end{equation}

If i changed this equation just a bit and used a $|\psi\rangle$ in place of $|n\rangle$, i would get a different result and couldn't conclude that $a a^\dagger = n$.

Long story short i am totaly confused and i need some guidance to understand this. The best for me would be any book with a geometrical explaination of Hilbert space... Is there any?

• what do you mean by a geometrical explanation of Hilbert spaces? Do you know linear algebra? – Dar May 1 '13 at 12:39
• It looks like you need to spend some more time studying linear algebra: linear operators, eigenvalues, eigenvectors, inner products, and the like. Maybe try Axler's book? – user1504 May 1 '13 at 12:39
• Addressing just the funny way of writing bras and kets: it is equivalent to writing vectors in column and row expressions which means that to multiply with operators (the equivalent of non-square matrices and also subject to transposition on taking the concugate) you have to have them on alternate sides. As other have said: this is linear algebra. – dmckee May 1 '13 at 13:02
• Find a set of lecture notes or a book that starts with a qubit system (a Hilbert space with only two dimensions). That way you can write everything out explicitly and see that bras/kets are two-component column/row vectors respectively, operators are 2x2 matrices, the <bra|ket> is just the inner product of column * row vector etc and the Hilbert space is a 3-sphere (4 real components in a 2 component complex vector minus 1 constraint from normalisation). – Michael Brown May 1 '13 at 13:55
• Griffith's book is not a good book to understand QM with. it's a good book to start doing QM from. I'd suggest Shankar's Principles of QM instead. his first chapter deals purely with the linear algebra you need to understand QM with. it's not very long (because in physics you don't -really- need the full machinery of liner algebra), but you should thoroughly understand his first chapter. that will clear up ALL the questions you had in your post. – nervxxx May 1 '13 at 20:52

I'll try to answer your questions while explaining as much basic quantum mechanics as possible, so that you'll be able to fill in the blanks with only linear algebra.

1. Hilbert space- As a novice to QM i am very sad that in none of the books i have read i found the reasons for using Hilbert space H at first place followed by a full geometrical explaination of this space and how we build this space out of R. It goes same for for its dual space Hd. Where can i get this? Every single author starts this topic by just bombard novices with bunch of rules for H,Hd which i can't just trust and this is only pointless learning by heart.

In QM, the state of a system is given by a set of complex amplitudes. If I have a coin that says 0 on one side and 1 on the other, and I flip it, classically I will have a coin in either the state 0 or the state 1, with a probability of 1/2 for each outcome. In QM, instead of probabilities over different states, you have amplitudes. To get the probability you take the square of the amplitude. So, if I flip a quantum mechanical coin, I may end up in the state $$\frac{1}{\sqrt{2}} |0\rangle + \frac{1}{\sqrt{2}}|1\rangle.$$ All this means is that the probability that I'll observe state 0 is $$\left( \frac{1}{\sqrt{2}} \right)^{2}= \frac{1}{2}.$$

So, to describe the state of a quantum system, you need a set of observation outcomes, and an amplitude for each outcome. Above, our outcomes were 0 and 1, and our amplitudes were $1/\sqrt{2}$.

If we were dealing with regular probabilities, the space of states of a system with $n$ possible outcomes would be $\mathbb{R}^{n}$. This is because each outcome has a probability, so you can write the entire state as a set of $n$ probabilities, which are just real numbers. In quantum mechanics it's the same, except instead of probabilities we have $n$ amplitudes. So the space of states is $\mathbb{C}^{n}$.

The reason physicists talk about Hilbert spaces instead of just complex Euclidean spaces is that in general your set of measurement outcomes could be infinitely big, instead of just $n$. The Hilbert space is just a generalization of the complex Euclidean space, and you can often just think of complex Euclidean spaces instead.

Dirac's notation ... because i don't understand Hilbert space i don't understand what i am allowed to do with kets | ⟩ and bras ⟨ | and for example:

Why do we have to write an operator on the left side of kets |A^ψ⟩=A^|ψ⟩ but for bras it is vice versa (we write it on the right) ⟨A^ψ|=⟨ψ|A^. Why can we factor out a constant from kets |aψ⟩=a|ψ⟩ and we can only factor out complex conjugate from bras ⟨aψ|=a∗⟨ψ|.

Now that we understand that states are given by complex vectors, we can understand bra-ket notation.

In my coin flipping example, imagine that I use the amplitudes $1/\sqrt{3}$ and $\sqrt{2/3}$. I can represent the state of my coin as $$\left|\mbox{coin}\right> = \frac{1}{\sqrt{3}} |0\rangle + \sqrt{\frac{2}{3}}|1\rangle = \frac{1}{\sqrt{3}}\left(\begin{matrix} 1 \\ 0 \end{matrix}\right) + \sqrt{\frac{2}{3}}\left(\begin{matrix} 0 \\ 1 \end{matrix}\right) = \left(\begin{matrix} 1/\sqrt{3} \\ \sqrt{2/3} \end{matrix}\right)$$

There is a lot of notation being used in the expression above.

1. As is often done, I wrote the name of the system inside a ket. This is to be read a "the state of the coin" and means nothing else mathematically, other than to say it is a vector in a Hilbert space. This is actually what I've been doing when I wrote $|0\rangle$ and $|1\rangle$.

2. I've also decided that the outcome 0 should correspond to the vector $\left(\begin{matrix} 1 \\ 0 \end{matrix}\right)$, and 1 should correspond to $\left(\begin{matrix} 0 \\ 1 \end{matrix}\right)$. This was an arbitrary decision. I did it only because I know that my system's state is given by a set of complex numbers, and I want to be able to write the entire state down as a simple vector in a Hilbert space.

So, kets are vectors in a Hilbert space. I've decided to write them as column vectors, but this was also arbitrary.

If kets are column vectors in a Hilbert space, then bras are row vectors with the rule that for any $\left|\psi\right>$, $$\left<\psi\right| = \left|\psi\right>^{\dagger}.$$

Here I used $\dagger$ to mean conjugate transpose. It's unfortunate that there are so many different notations for this operation, but you seem to understand what it means.

Okay, so states are vectors in a Hilbert space. Operators are matrices acting on the vectors. A matrix times a vector gives a vector, so this is a good way to model transformations.

Now the confusion about what is allowed and what isn't allowed with bras and kets just comes down to linear algebra. Hopefully you can now answer questions like why does $\alpha^{*}\left<\psi\right| = \left(\alpha|\psi\rangle \right)^{\dagger}$.

This has turned into a long answer so I'll leave it here for now.

• So if i understood right we choose $\mathbb{R}^{n}$ as a space for probabilities, BUT in QM we have amplitudes and an amplitude is a square root of a probability. So i need a complex space $\mathbb{C}^n$ because of the square root which leads to complex numbers? – 71GA May 8 '13 at 8:33
• And one more questions... It is clear to me that if bra's are column vectors then kets are row vectors. But i don't know the physical meaning of bra's (do they even have one?). I know that ket's are QM states. – 71GA May 8 '13 at 9:18
• bra's are just a different way of writing the same states as kets. You could do all of quantum mechanics backwards if you wanted, and use bras where others use kets. – Alex L May 8 '13 at 15:10

I will try to cover all your questions, but I really recommend a book as the topic is too wide spread for such a "short" answer.

You can think of a Hilbert space as an infinite dimensional vector space, that has similar properties as one knows from $\mathbb{R}^3$:

• The elements are vectors and will be denoted by $\phi$
• You can measure angles and lengths with a scalar product $\langle \cdot | \cdot \rangle: H \times H \rightarrow \mathbb{C}$. So it takes two vectors and gets you a complex numeber $\langle \psi | \phi \rangle$. In the real case you assume it is bilinear (which more or less states that $2 v$ has two times the length of $v$) and here as we have complex numbers it is sesquilinear: It is linear in the second argument and complex linear in the first, that is $\langle \psi | a \phi \rangle = a \langle \psi | \phi \rangle$ and $\langle a \psi | \phi \rangle = a^* \langle \psi | \phi \rangle$.

• We can have operators acting on our vectors, e.g. rotate them, to get a new vector. $\hat A: H \rightarrow H$.

• Finally we have some functions $f: H \rightarrow \mathbb{C}$. The set of all linear functions is the dual $H^*$ of $H$.

So far so good. But now comes the real sugar to the notation. There is an important theorem (Riesz-Theorem) which states the following. For every linear function $f \in H^*$ we can find an element $\psi$ of our Hilbert space such that $f(\phi) = \langle \psi | \phi \rangle$. That is, every dual elements acts by "scalarproducting". Now we denote this function $f$ by the left part of the scalarproduct $\langle \psi |$ and identify $\phi = |\phi \rangle$. So instead of $f(\phi)$ we write $\langle \psi | \phi \rangle$ and Riesz-theorem gets hidden in the notation.

The adjoint of an operator is defined by the relationship $\langle \psi | \hat A \phi \rangle = \langle \hat A^\dagger \psi | \phi \rangle$. Note that this here is really the scalarproduct. If we have $\hat A = \hat A^\dagger$ (Hermitian!) than $\langle \psi | \hat A \phi \rangle = \langle \hat A \psi | \phi \rangle$ and we write this as $\langle \psi | \hat A | \phi \rangle$.

Futhermore the "factorization" of constants from bra and kets follows directly from the above property of the scalarproduct.

Finally,

I don't understand the meaning of definitions for $a$ and $a^\dagger$ and even if i did, how can the line above even be true?

\begin{equation} \hat{a}^\dagger\hat{a} \left| n \right\rangle = > \hat{a}^\dagger \sqrt{n} \left| n - 1 \right\rangle = \sqrt{n} > \sqrt{n} \left| n \right\rangle = n \left| n \right\rangle , > \end{equation}

If i changed this equation just a bit and used a $|\psi\rangle$ in place of $|n\rangle$, i would get a different result and couldn't conclude that $a a^\dagger = n$.

If you change $|n\rangle$ to an arbitary vector $|\phi \rangle$ you loose the identities \begin{equation} \hat{a}^\dagger \left| n \right\rangle = \sqrt{n + 1} \left| n + 1 \right\rangle , \end{equation} \begin{equation} \hat{a} \left| n \right\rangle = \sqrt{n} \left| n - 1 \right\rangle , \end{equation} which really hold only for the vector $|n\rangle$.

• Thank you very much this will help for sure. Although it will take some time for me to process these two anwsers. – 71GA May 7 '13 at 7:39