Notation for two qubit composite product state Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?Compute average value of two-qubit systemWhat can I deduce about $f(x)$ if $f$ is balanced or constant?What does the notation $lvert underlinex rangle$ mean?How do I show that a two-qubit state is an entangled state?How is a single qubit fundamentally different from a classical coin spinning in the air?Why is the state of multiple qubits given by their tensor product?Notation for two entangled registersA question about notation for quantum statesA two qubit state in a special formConcurrence for a two qubit state

The following signatures were invalid: EXPKEYSIG 1397BC53640DB551

Do working physicists consider Newtonian mechanics to be "falsified"?

Passing functions in C++

Can a zero nonce be safely used with AES-GCM if the key is random and never used again?

What LEGO pieces have "real-world" functionality?

Slither Like a Snake

Single author papers against my advisor's will?

Are my PIs rude or am I just being too sensitive?

Classification of bundles, Postnikov towers, obstruction theory, local coefficients

Statistical model of ligand substitution

How to retrograde a note sequence in Finale?

What is the largest species of polychaete?

Complexity of many constant time steps with occasional logarithmic steps

Active filter with series inductor and resistor - do these exist?

Cold is to Refrigerator as warm is to?

Why is "Captain Marvel" translated as male in Portugal?

Simulating Exploding Dice

How do I automatically answer y in bash script?

Need a suitable toxic chemical for a murder plot in my novel

Can the prologue be the backstory of your main character?

Am I ethically obligated to go into work on an off day if the reason is sudden?

What are the performance impacts of 'functional' Rust?

ELI5: Why do they say that Israel would have been the fourth country to land a spacecraft on the Moon and why do they call it low cost?

What computer would be fastest for Mathematica Home Edition?



Notation for two qubit composite product state



Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?Compute average value of two-qubit systemWhat can I deduce about $f(x)$ if $f$ is balanced or constant?What does the notation $lvert underlinex rangle$ mean?How do I show that a two-qubit state is an entangled state?How is a single qubit fundamentally different from a classical coin spinning in the air?Why is the state of multiple qubits given by their tensor product?Notation for two entangled registersA question about notation for quantum statesA two qubit state in a special formConcurrence for a two qubit state



.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








2












$begingroup$


In my lecture notes on quantum information processing my lecturer gives an example of composite systems as $|phirangle=|0rangle |0rangle=|00rangle$. I understand that if we have two qubits then its product state will be in 2n dimensional Hilbert space and I understand the 2 qubit state $|00rangle$ to be represented in matrix representation as $beginpmatrix 1 & 1 \ 0 & 0 endpmatrix$ (if that is wrong please do correct my misunderstanding though). My question is about the notation $|0rangle|0rangle=|00rangle$, how can we calculate this with matrices on the left-hand side we have a 2 by 1 matrix multiplied by a 2 by 1 matrix which cannot be calculated. I thought perhaps it was a matter of direct products but my calculation led to an incorrect result there too.



Could anyone clarify this for me, please?



Edit: It occurred to me that I think I'm mistaken about the matrix representation of $|00rangle$, I think it would make more sense to be $beginpmatrix 1 \ 0\0\0 endpmatrix$ in which case the direct product does work and I should take the notation $|0rangle|0rangle$ to be a shorthand for the direct product not the multiplication of two matrices, is that correct?










share|improve this question









New contributor




can'tcauchy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$


















    2












    $begingroup$


    In my lecture notes on quantum information processing my lecturer gives an example of composite systems as $|phirangle=|0rangle |0rangle=|00rangle$. I understand that if we have two qubits then its product state will be in 2n dimensional Hilbert space and I understand the 2 qubit state $|00rangle$ to be represented in matrix representation as $beginpmatrix 1 & 1 \ 0 & 0 endpmatrix$ (if that is wrong please do correct my misunderstanding though). My question is about the notation $|0rangle|0rangle=|00rangle$, how can we calculate this with matrices on the left-hand side we have a 2 by 1 matrix multiplied by a 2 by 1 matrix which cannot be calculated. I thought perhaps it was a matter of direct products but my calculation led to an incorrect result there too.



    Could anyone clarify this for me, please?



    Edit: It occurred to me that I think I'm mistaken about the matrix representation of $|00rangle$, I think it would make more sense to be $beginpmatrix 1 \ 0\0\0 endpmatrix$ in which case the direct product does work and I should take the notation $|0rangle|0rangle$ to be a shorthand for the direct product not the multiplication of two matrices, is that correct?










    share|improve this question









    New contributor




    can'tcauchy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$














      2












      2








      2





      $begingroup$


      In my lecture notes on quantum information processing my lecturer gives an example of composite systems as $|phirangle=|0rangle |0rangle=|00rangle$. I understand that if we have two qubits then its product state will be in 2n dimensional Hilbert space and I understand the 2 qubit state $|00rangle$ to be represented in matrix representation as $beginpmatrix 1 & 1 \ 0 & 0 endpmatrix$ (if that is wrong please do correct my misunderstanding though). My question is about the notation $|0rangle|0rangle=|00rangle$, how can we calculate this with matrices on the left-hand side we have a 2 by 1 matrix multiplied by a 2 by 1 matrix which cannot be calculated. I thought perhaps it was a matter of direct products but my calculation led to an incorrect result there too.



      Could anyone clarify this for me, please?



      Edit: It occurred to me that I think I'm mistaken about the matrix representation of $|00rangle$, I think it would make more sense to be $beginpmatrix 1 \ 0\0\0 endpmatrix$ in which case the direct product does work and I should take the notation $|0rangle|0rangle$ to be a shorthand for the direct product not the multiplication of two matrices, is that correct?










      share|improve this question









      New contributor




      can'tcauchy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      In my lecture notes on quantum information processing my lecturer gives an example of composite systems as $|phirangle=|0rangle |0rangle=|00rangle$. I understand that if we have two qubits then its product state will be in 2n dimensional Hilbert space and I understand the 2 qubit state $|00rangle$ to be represented in matrix representation as $beginpmatrix 1 & 1 \ 0 & 0 endpmatrix$ (if that is wrong please do correct my misunderstanding though). My question is about the notation $|0rangle|0rangle=|00rangle$, how can we calculate this with matrices on the left-hand side we have a 2 by 1 matrix multiplied by a 2 by 1 matrix which cannot be calculated. I thought perhaps it was a matter of direct products but my calculation led to an incorrect result there too.



      Could anyone clarify this for me, please?



      Edit: It occurred to me that I think I'm mistaken about the matrix representation of $|00rangle$, I think it would make more sense to be $beginpmatrix 1 \ 0\0\0 endpmatrix$ in which case the direct product does work and I should take the notation $|0rangle|0rangle$ to be a shorthand for the direct product not the multiplication of two matrices, is that correct?







      quantum-state tensor-product notation






      share|improve this question









      New contributor




      can'tcauchy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|improve this question









      New contributor




      can'tcauchy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|improve this question




      share|improve this question








      edited 1 hour ago









      Sanchayan Dutta

      6,66641556




      6,66641556






      New contributor




      can'tcauchy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      asked 3 hours ago









      can'tcauchycan'tcauchy

      1135




      1135




      New contributor




      can'tcauchy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.





      New contributor





      can'tcauchy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      can'tcauchy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.




















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          $|0rangle|0rangle$ is actually a shorthand for $|0rangle otimes |0rangle$ or $beginbmatrix 1 \ 0 endbmatrix otimes beginbmatrix 1 \ 0endbmatrix $ where $otimes$ stands for the tensor product or essentially the Kronecker product. To quote Wikipedia:




          In mathematics, the Kronecker product, denoted by $otimes$, is an operation
          on two matrices of arbitrary size resulting in a block matrix. It is a
          generalization of the outer product (which is denoted by the same
          symbol) from vectors to matrices, and gives the matrix of the tensor
          product with respect to a standard choice of basis
          . The Kronecker
          product should not be confused with the usual matrix multiplication,
          which is an entirely different operation
          .




          Now the standard choice of basis for a two-qubit system is:



          $11rangle = beginbmatrix 0 \ 0 \ 0 \ 1 endbmatrix$



          If you wish, you can also represent the basis as (if you strictly take $otimes$ as the outer product):



          $11rangle = beginbmatrix 0 & 0 \ 0 & 1 endbmatrix$



          but then while carrying out calculations like determining the action of a quantum gate on a composite state you'd have to write the state using the vector representation (carefully read the linked Mathematics SE answer).



          The key point here is that don't be bent on thinking of these linear algebraic operations in terms of matrices, but rather think in terms of linear maps. You'll get more comfortable with these things once you learn about tensors!



          P.S: Kronecker product and outer product confusion






          share|improve this answer











          $endgroup$













            Your Answer








            StackExchange.ready(function()
            var channelOptions =
            tags: "".split(" "),
            id: "694"
            ;
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function()
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled)
            StackExchange.using("snippets", function()
            createEditor();
            );

            else
            createEditor();

            );

            function createEditor()
            StackExchange.prepareEditor(
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: false,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: null,
            bindNavPrevention: true,
            postfix: "",
            imageUploader:
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            ,
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            );



            );






            can'tcauchy is a new contributor. Be nice, and check out our Code of Conduct.









            draft saved

            draft discarded


















            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fquantumcomputing.stackexchange.com%2fquestions%2f5911%2fnotation-for-two-qubit-composite-product-state%23new-answer', 'question_page');

            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            $|0rangle|0rangle$ is actually a shorthand for $|0rangle otimes |0rangle$ or $beginbmatrix 1 \ 0 endbmatrix otimes beginbmatrix 1 \ 0endbmatrix $ where $otimes$ stands for the tensor product or essentially the Kronecker product. To quote Wikipedia:




            In mathematics, the Kronecker product, denoted by $otimes$, is an operation
            on two matrices of arbitrary size resulting in a block matrix. It is a
            generalization of the outer product (which is denoted by the same
            symbol) from vectors to matrices, and gives the matrix of the tensor
            product with respect to a standard choice of basis
            . The Kronecker
            product should not be confused with the usual matrix multiplication,
            which is an entirely different operation
            .




            Now the standard choice of basis for a two-qubit system is:



            $11rangle = beginbmatrix 0 \ 0 \ 0 \ 1 endbmatrix$



            If you wish, you can also represent the basis as (if you strictly take $otimes$ as the outer product):



            $11rangle = beginbmatrix 0 & 0 \ 0 & 1 endbmatrix$



            but then while carrying out calculations like determining the action of a quantum gate on a composite state you'd have to write the state using the vector representation (carefully read the linked Mathematics SE answer).



            The key point here is that don't be bent on thinking of these linear algebraic operations in terms of matrices, but rather think in terms of linear maps. You'll get more comfortable with these things once you learn about tensors!



            P.S: Kronecker product and outer product confusion






            share|improve this answer











            $endgroup$

















              1












              $begingroup$

              $|0rangle|0rangle$ is actually a shorthand for $|0rangle otimes |0rangle$ or $beginbmatrix 1 \ 0 endbmatrix otimes beginbmatrix 1 \ 0endbmatrix $ where $otimes$ stands for the tensor product or essentially the Kronecker product. To quote Wikipedia:




              In mathematics, the Kronecker product, denoted by $otimes$, is an operation
              on two matrices of arbitrary size resulting in a block matrix. It is a
              generalization of the outer product (which is denoted by the same
              symbol) from vectors to matrices, and gives the matrix of the tensor
              product with respect to a standard choice of basis
              . The Kronecker
              product should not be confused with the usual matrix multiplication,
              which is an entirely different operation
              .




              Now the standard choice of basis for a two-qubit system is:



              $11rangle = beginbmatrix 0 \ 0 \ 0 \ 1 endbmatrix$



              If you wish, you can also represent the basis as (if you strictly take $otimes$ as the outer product):



              $11rangle = beginbmatrix 0 & 0 \ 0 & 1 endbmatrix$



              but then while carrying out calculations like determining the action of a quantum gate on a composite state you'd have to write the state using the vector representation (carefully read the linked Mathematics SE answer).



              The key point here is that don't be bent on thinking of these linear algebraic operations in terms of matrices, but rather think in terms of linear maps. You'll get more comfortable with these things once you learn about tensors!



              P.S: Kronecker product and outer product confusion






              share|improve this answer











              $endgroup$















                1












                1








                1





                $begingroup$

                $|0rangle|0rangle$ is actually a shorthand for $|0rangle otimes |0rangle$ or $beginbmatrix 1 \ 0 endbmatrix otimes beginbmatrix 1 \ 0endbmatrix $ where $otimes$ stands for the tensor product or essentially the Kronecker product. To quote Wikipedia:




                In mathematics, the Kronecker product, denoted by $otimes$, is an operation
                on two matrices of arbitrary size resulting in a block matrix. It is a
                generalization of the outer product (which is denoted by the same
                symbol) from vectors to matrices, and gives the matrix of the tensor
                product with respect to a standard choice of basis
                . The Kronecker
                product should not be confused with the usual matrix multiplication,
                which is an entirely different operation
                .




                Now the standard choice of basis for a two-qubit system is:



                $11rangle = beginbmatrix 0 \ 0 \ 0 \ 1 endbmatrix$



                If you wish, you can also represent the basis as (if you strictly take $otimes$ as the outer product):



                $11rangle = beginbmatrix 0 & 0 \ 0 & 1 endbmatrix$



                but then while carrying out calculations like determining the action of a quantum gate on a composite state you'd have to write the state using the vector representation (carefully read the linked Mathematics SE answer).



                The key point here is that don't be bent on thinking of these linear algebraic operations in terms of matrices, but rather think in terms of linear maps. You'll get more comfortable with these things once you learn about tensors!



                P.S: Kronecker product and outer product confusion






                share|improve this answer











                $endgroup$



                $|0rangle|0rangle$ is actually a shorthand for $|0rangle otimes |0rangle$ or $beginbmatrix 1 \ 0 endbmatrix otimes beginbmatrix 1 \ 0endbmatrix $ where $otimes$ stands for the tensor product or essentially the Kronecker product. To quote Wikipedia:




                In mathematics, the Kronecker product, denoted by $otimes$, is an operation
                on two matrices of arbitrary size resulting in a block matrix. It is a
                generalization of the outer product (which is denoted by the same
                symbol) from vectors to matrices, and gives the matrix of the tensor
                product with respect to a standard choice of basis
                . The Kronecker
                product should not be confused with the usual matrix multiplication,
                which is an entirely different operation
                .




                Now the standard choice of basis for a two-qubit system is:



                $11rangle = beginbmatrix 0 \ 0 \ 0 \ 1 endbmatrix$



                If you wish, you can also represent the basis as (if you strictly take $otimes$ as the outer product):



                $11rangle = beginbmatrix 0 & 0 \ 0 & 1 endbmatrix$



                but then while carrying out calculations like determining the action of a quantum gate on a composite state you'd have to write the state using the vector representation (carefully read the linked Mathematics SE answer).



                The key point here is that don't be bent on thinking of these linear algebraic operations in terms of matrices, but rather think in terms of linear maps. You'll get more comfortable with these things once you learn about tensors!



                P.S: Kronecker product and outer product confusion







                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited 2 hours ago

























                answered 2 hours ago









                Sanchayan DuttaSanchayan Dutta

                6,66641556




                6,66641556




















                    can'tcauchy is a new contributor. Be nice, and check out our Code of Conduct.









                    draft saved

                    draft discarded


















                    can'tcauchy is a new contributor. Be nice, and check out our Code of Conduct.












                    can'tcauchy is a new contributor. Be nice, and check out our Code of Conduct.











                    can'tcauchy is a new contributor. Be nice, and check out our Code of Conduct.














                    Thanks for contributing an answer to Quantum Computing Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid


                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.

                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function ()
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fquantumcomputing.stackexchange.com%2fquestions%2f5911%2fnotation-for-two-qubit-composite-product-state%23new-answer', 'question_page');

                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    Popular posts from this blog

                    Dapidodigma demeter Subspecies | Notae | Tabula navigationisDapidodigmaAfrotropical Butterflies: Lycaenidae - Subtribe IolainaAmplifica

                    Constantinus Vanšenkin Nexus externi | Tabula navigationisБольшая российская энциклопедияAmplifica

                    Vas sanguineum Index Historia | Divisio | Constructio anatomica | Vasorum sanguineorum morbi (angiopathiae) | Notae | Nexus interniTabula navigationisAmplifica