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So I am a newbie to QM, and coming from math, I believe I am not understanding some key points in bra-ket notation.

So given a quantum state $\psi$, I understand that $|\psi \rangle$ is a just a Hilbert space notation for a function. Now, we say that $$\langle \psi_n | \psi_m \rangle = \int \psi_n^* \psi _m dr$$ Which leads me to believe that the $\langle \cdot | \cdot \rangle$ notation seems to imply that it is a inner product defined on the space. However, then I see that $$\psi (r) = \langle r | \psi \rangle$$ and $$\psi^* (r) = \langle \psi | r \rangle$$ for some position vector $r$, with dimension $3$. Well, if that is the case, of course it isn't just a normal inner product defined on functions. It is something else, not quite sure yet. It does seem like $\langle \psi |$ seems to be the conjugate transpose of the function/vector/state $\psi$. Then I see the following identity, $$\int dr | r \rangle \langle r | = I$$ where $I$ is the identity operator. I don't understand this. How do you get this statement from the above statements? Imagining $r$ to be vectors in $\mathbb{R}^3$, I can see that $rr^T$ leads to a matrix, but I don't see it leading to the identity matrix. What am I missing here? What kind of a product is this final identity?

I also see these identities later in the book: $$|\psi \rangle = \int dr |r\rangle \langle r| \psi \rangle \\ |\psi \rangle = \int dr |p\rangle \langle p| \psi \rangle \\ |\psi \rangle = \int dr | \psi_E \rangle \langle \psi_E| \psi \rangle $$ followed by expectation values of operators, like $$\langle \hat{A} \rangle = \langle \psi | \hat{A} | \psi \rangle$$

Expectation value is $\langle Y \rangle = \int p(x) Y dx$, where $p(x)$ is the probability distribution function. I see that $\langle \psi | \psi \rangle$ gives me the probability distribution, but does that mean the operator $\hat{A}$ is just chilling there? Doing nothing really?

I understand that this looks like forming the vector from the basis, but I don't seem to understand how these operators work, and how I can draw analogies to standard vector spaces like $\mathbb{C} ^n$ or something.

Any advice you have would be appreciated.

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  • $\begingroup$ if $|r\angle$ is normed then$|r\rangle\langle r|=I$ is a projection operator and if you consider the sum of all projections, it will yield the identity operator. $\endgroup$ – Mehmet Bütün Sep 16 '20 at 4:47
  • $\begingroup$ I understand that $|\psi \rangle$ is a just a Hilbert space notation for a function. This isn’t always the case. For example, a “spin-up” state $|\uparrow\rangle$ doesn’t correspond to a function. It’s just a vector in the Hilbert space of spin states. Not all Hilbert spaces for quantum systems are spaces of functions, but all are abstract vector spaces. $\endgroup$ – G. Smith Sep 16 '20 at 4:58
  • $\begingroup$ my answer to a similar confusion physics.stackexchange.com/questions/289108/bra-ket-notation/… $\endgroup$ – physicopath Sep 16 '20 at 6:32
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A complex Hilbert space $\mathcal{H}$ is just a vector space over $\mathbb{C}$ equipped with a (complete) inner product. To help your intuition, it may be useful to recall that every finite dimensional complex Hilbert space is isomorphic to $\mathbb{C}^n$ for some $n$.

Now, assume $\mathcal{H}$ is finite with dimension $n$. Pick a basis and label its elements with $1$, $\cdots$, $n$. Then, in Dirac's notations $|n\rangle$ is an element of your basis (and more generally an element of $\mathcal{H}$) while $n$ is a mere label. A generic element $|\psi\rangle$ of $\mathcal{H}$ can be written $$\sum_{i=1}^n \alpha_i |i\rangle\,.$$

In Dirac's notations the inner product is denoted by $\langle \cdot | \cdot \rangle$, hence $\langle i | \psi\rangle$ just denotes the inner product of $|i\rangle$ with $|\psi\rangle$, namely $\alpha_i$. More generally, if $|\psi\rangle$ and $|\phi\rangle$ denote two elements of $\mathcal{H}$ with $|\phi\rangle = \sum_i \beta_i |i\rangle$, then $\langle \psi|\phi\rangle$ denotes their inner product, namely $\sum_i \alpha_i^*\beta_i$.

Moreover, as a vector space $\mathcal{H}$ has also a dual space $\mathcal{H}^{\dagger}$. Note that $\mathcal{H}^{\dagger}$ is the space of linear maps from $\mathcal{H}$ to $\mathbb{C}$. There is a one-to-one correspondence between $\mathcal{H}$ and $\mathcal{H}^{\dagger}$ that one can define with the help of the inner product. In Dirac's notations, the image of $|\psi\rangle$ under this correspondence is denoted $\langle \psi|$, so that $\langle \psi|$ applied to $|\phi\rangle$ is the inner product $\langle \psi|\phi\rangle$.

Finally, an operator $\hat{A}$ is just a linear map $\mathcal{H}\rightarrow \mathcal{H}$ on states $|\psi\rangle$. Given a state $|\psi\rangle$, $\hat{A}|\psi\rangle$ is just another state, i.e. another element of $\mathcal{H}$. Then, $\langle \psi|\hat{A}|\psi\rangle$ is simply the inner product of $|\psi\rangle$ with $\hat{A}|\psi\rangle$. This is how the expectation value of $\hat{A}$ in the state $|\psi\rangle$ is defined.

Once you understand the discrete case where your basis is finite or at least countable, you can try to understand the case where your basis is not countable anymore. In this case the label $\psi$ becomes continuous and $\sum$ is replaced by an integral $\int$, but conceptually the situation has not changed, you are still dealing with a complex vector space equipped with an inner product.

Indeed, if the elements $|r\rangle$'s, where $r$ is now a continuous label, denote the basis elements of your space, then a generic element $|\psi\rangle$ is of the form $\int dr\, \psi(r)|r\rangle$ where $\psi(r)$ simply denotes the inner product $\langle r|\psi\rangle$.

$\int dr\, |r\rangle \langle r|$ is just a notation for the operator that maps $|\psi\rangle$ to $\int dr\, |r\rangle \langle r|\psi\rangle$. As noted above $\int dr\, |r\rangle \langle r|\psi\rangle$ is just $\int dr\, \psi(r)|r\rangle$, namely $|\psi\rangle$ itself. In other words, $\int dr\, |r\rangle \langle r|$ is just the identity operator $I$.

As a consequence, $I$ applied to $|\psi\rangle$ is $|\psi\rangle$, but it can also be written as $\int dr\, |r\rangle \langle r|\psi\rangle$, hence the first equality of the three equalities found in your book and mentioned in your question. The second and third equalities express the same idea just using different bases (for instance the momentum basis in the second equality instead of the position basis in the first one).

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  • $\begingroup$ Thanks for your answer @A. Bordg! I think I just wanted a quick clarification: from your statement, $|\psi \rangle = \int dr \psi (r) |r\rangle$ right? $\endgroup$ – megamence Sep 16 '20 at 21:14
  • $\begingroup$ Yes, you're right. $\endgroup$ – A. Bordg Sep 17 '20 at 4:10
  • $\begingroup$ @megamence please don't forget to accept one of the answers if you're satisfied. $\endgroup$ – A. Bordg Sep 18 '20 at 15:50
  • $\begingroup$ Is "$\langle\psi|\phi\rangle$ denotes their inner product, namely $\sum_i \alpha_i^*\beta_i$." true in any basis? $\endgroup$ – Yashas Oct 6 '20 at 13:58
  • $\begingroup$ Given any basis of a finite-dimensional vector space $V$, if you definite a map from $V\times V$ to $\mathbb{C}$ this way, you can always check this indeed defines an inner product. $\endgroup$ – A. Bordg Oct 7 '20 at 4:46
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$\langle \cdot | \cdot \rangle$ is an inner product.

First let's rename $r$ as $r_0$ and lets think of it as a fixed constant (i.e. one meter or whatever you want). When you write $\langle r_0 | \psi \rangle$, you are taking the inner product between your state $| \psi \rangle$ and $\langle r_0 |$. But $| r_0 \rangle$ is an element of the basis $\{| r \rangle :$ $r \in \mathbb{R}, r>0\}$, and so we can also say that we are projecting your state (vector) onto the basis element $| r_0 \rangle$. In euclidean space, this is equivalent to projecting your vector onto the x, y or z axis (or really any other direction). The complete basis of our Hilbert space is the set of all position states on the positive (I am aussuming $r$ stands for the radial direction) real line, so the basis is REALLY big, as big as the entire set of real numbers. For every distinct $r_j$, you can think of $|r_j \rangle$ as an axis in euclidean space. But euclidean space only has 3 dimensions, so three axes, where our Hilbert space has an infinite amount of axes.

In euclidean space, you can decompose any vectors into components of a basis. Formerly, given an (orthonormal) basis $x_0,x_1,x_2$, we can write our vector $x$ as $x= x \cdot x_0 \cdot e_0 + x \cdot x_1 \cdot e_1+ x \cdot x_2 \cdot e_2 = \sum_{i=0}^2 x_i\cdot x \cdot e_i$ where $e_i$ is in the directionnal vector of $x_i$. This is equivalent to saying that a vector is the sum of all its components (projections) in all basis directions. Since $\langle r_j| r_i \rangle =0 $ if $r_i\neq r_j$, and $\langle r_j| r_j \rangle =1 $, $|r\rangle$ is an orthonormal basis.

For our Hilbert space we thus have $|\psi \rangle = \sum_{r\in\mathbb{R^+}} \langle r \cdot| \psi \rangle |r\rangle = \sum_{r\in\mathbb{R^+}} |r\rangle\langle r| \psi \rangle $ where we have the direct correspondence $e_i ->|r\rangle , x_i -> \langle r|$ and $x -> |\psi \rangle$.

Since we are summing over the entire positive real line, we can transform that summation into an integral and thus we get $|\psi \rangle = \int |r\rangle\langle r| \psi \rangle dr$

But $|\psi \rangle$ does not depend on r, and so we can take it out of the integral, which yield that $\int |r\rangle\langle r|dr$ must be the identity as the above equation is true for all $|\psi\rangle$.

We can now tackle the expectation value of an operator. $\langle \psi | \hat{A} | \psi \rangle = \langle \psi | \hat{A} \int |x\rangle\langle x|dx| \psi \rangle = \int \langle \psi | \hat{A} |x\rangle\langle x|\psi\rangle dx = \int \langle \psi | \hat{A} |x\rangle\Psi(x) dx$

Now, if you're smart, you use a basis such that $\hat{A}$ is diagonal in that basis, in which case you have $\hat{A}|x\rangle=|x\rangle A(x)$, where $A(x)$ is the eigenvalue of the eigenvector $|x\rangle$ of $\hat{A}$. (For the position operator $\hat{x}$ $\hat{x}|x\rangle=|x\rangle x$. This is actually how the basis $|x\rangle$ is defined, as the basis that diagonalizes the position operator.)

Thus we have $\langle \psi | \hat{A} | \psi \rangle = \int \langle \psi | x\rangle A(x) \Psi(x) dx=\int \Psi(x)^* A(x) \Psi(x) dx$

Note that as in euclidean space, there is more than one orthonormal basis, and so you can project your state $|\psi\rangle$ into another basis other than the position basis. For example you can express your wave function $\Psi$ into position or momentum space, which simply correspond to projecting your state $|\psi \rangle$ into the position or momentum basis respectively. You project your state into the basis that is most useful to you. For example, you project into the momentum basis if you want to know the average momentum of your particle because then the momentum operator is diagonal and so the expected value of the momentum of your particle is easy to compute. $\langle\hat{p}\rangle = \int p|\Psi(p)|^2 dp$

Another example, you can express any spin state by the z, x or y component of the state, which are all equivalently valid basis.

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